See discussions, stats, and author profiles for this publication at: https://www.researchgate.

net/publication/235779780

A text book on Business Mathematics

Book · December 2004

CITATIONS READS

0 29,244

1 author:

S. M. Shahidul Islam
Hajee Mohammad Danesh Science and Technology University, Dinajpur, Bangladesh
14 PUBLICATIONS   9 CITATIONS   

SEE PROFILE

All content following this page was uploaded by S. M. Shahidul Islam on 31 August 2018.

The user has requested enhancement of the downloaded file.

 Business Mathematics
Dr. S. M. Shahidul Islam

About the Author:

All along a brilliant student Dr. S. M. Shahidul Islam
son of Late Tajim Uddin Ahmed was born in 1976 at
Kumria, Dinajpur, Bangladesh. At present, he is an
Associate Professor in the Department of Mathematics
at Hajee Mohammad Danesh Science and Technology F  P(1  i) n
University, Dinajpur. His many research papers are
appeared in many prestigious journals such as
International Journal of Production Economics,
International Journal of Operational Research, etc. His
book named Linear Programming is followed and
appreciated by the teachers and the students of different
universities.
Written in accordance with the new syllabus of the National University, Dhaka University, Chittagong
University, Rajshahi University, Asian University of Bangladesh and East West University for the students
of the Department of Business administrations, Management, Accounting, Marketing and Finance.

Business Mathematics

S. M. Shahidul Islam
Lecturer of Mathematics
School of Business
Asian University of
Bangladesh

ABIR PUBLICATIONS
38/ ka Bangla Bazar (2nd floor), Dhaka-1100

i
Published by : Abir Publication
38/ ka Bangla Bazar
Dhaka-1100

First Edition : January, 2004

Computer Compose : Mst. Bakul Nahar Perveen

House – 22, Road – 12
Rupnagar R/A, Mirpur
Dhaka - 1216

Design by : Graphics Scan Systems

47/1 Bangla Bazar
Dhaka – 1100

Printed by : Bangla Muddranalay

Suttrapur, Dhaka- 1100

Copyright : Author

Price : Tk. 160.00 (One hundred sixty only)

ISBN No. : 984-36-0255-3

ii
 PREFACE

Bismillahir Rahmanir Rahim.

The book titled Business Mathematics has been written with the blessings of
D Almighty Allah. It is clear that the use of mathematical tools has been on the
E increase day by day in business arena and now-a-days business mathematics
D is a compulsory course in all business related disciplines. But the fact is that
our students are always afraid of mathematics and they need proper guidance
I and dependable books. There are text books available in the market on
C Business Mathematics but most of them are inadequate in meeting the
A requirements of the students. So in order to gear up the students in business
mathematics we have attempted to write the book using easy language and
T mathematical techniques.
E The text is written with the basic objects of introducing students of business
D administration to the mathematical concepts that help in decision making.
An attempt has been made to present explanations in such a way that the
underlying mathematical theories are fully exposed and the relation between
T theory and application thoroughly understood. The text will also cover a large
part of Production & Operations Management and Management Science.
O We have included in the book a large number of examples, mostly collected
from question papers of different universities. We have verified every
M question and their solution minutely, and if there is any error yet, we beg
apology for that.
Y I am grateful and indebted to my teachers Professor Dr. A.A.K. Majumdar,
Ritsumeikan Asia-Pacific University, Japan; Professor Dr. Nurul Alam Khan
P and Professor Dr. Makbul Hossen of the department of Mathematics,
Jahangirnagar University; Professor Dr. Abdul Awal Khan, Dean, School of
A Business, Asian University of Bangladesh and many of my colleagues who
R encouraged and helped me in writing the book. I am also grateful to the
E authors whose books I have freely consulted in the preparation of the
manuscript. The authority of Abir Publication deserves thanks for
N undertaking the responsibility to publish the book in these days of hardships
T in publication business.
Finally, special recognition should go to the endless co-operation of my wife
S without which the book could not have seen the light of the day.
Constructive criticisms, corrections, and suggestions towards the
improvement of the book will be thankfully received.

S. M. Shahidul Islam

iii
 National University
Subject: Management (Hons.)
Course: NM-201 Business Mathematics
1. Basic Concepts : Concepts of number system, fractions, exponents, equations, factoring, polynomials, ordered
pairs, relations, functions, types of functions.
2. Set Theory: Sets, set notation, operations with sets, laws of set operations, Venn diagrams, application of set
theory.
3. Logarithms: Rules for logarithms, common logarithms, calculation of logarithm of a number, natural logarithm.
4. Equation system: Solution of equations, simultaneous equation system, solution of simultaneous equation systems
with specific applications to business problems, inequalities.
5. Geometry: Cartesian co-ordinate system, distance between two points, straight line-slopes-intercepts, equation of a
line, application of linear equations.
6. Differential Calculus: Explanation of the concepts of limits and continuity, derivative and differentiation, rules of
differentiation, higher order differentiation, chain order differentiation, exponential and logarithmic differentiation,
partial differentiation, optimization, rate of growth and decays, business applications.
7. Integral Calculus: Meaning of integration, rules of integration, indefinite integral, definite integral, resource
depletion, resource accumulation, area between curves, business applications of integration in business decisions.
8. Matrix Algebra: Vectors, matrices, laws and operations, transposes, inverses, adjoints, Cramer’s rule,
determinants, solution of system of equations, application of matrix algebra in business.
9. Mathematics of Finance: Annuities, sinking fund, discount, simple and compound interest, amortization,
calculation of present value and future value and future value of annuities.

Asian University of Bangladesh

BBA Elementary Mathematics and Statistics
The topics include number system, set theory, functions & equations, exponential & logarithmic functions, collection of
data, presentation of data, measures of central tendency (mean, median & mode).

BBA 2309 Business Mathematics

The topics include coordinate geometry, mathematics of finance, matrix algebra, calculus and linear programming. This
course will familiarize students with the more commonly used types of quantitative techniques used in economics and
business.

MBA 6127 Business Mathematics

The course includes elements of algebra, number fields, linear and non-linear inequalities, functions, set, analytical
geometry, logarithm, limit, differential and integral calculus, matrix and linear programming. The purpose of the course
is to help the students learn mathematical tools which are used in business arena.

East West University

MAT 110 : Mathematics For Business and Economics I
Topics include: Set, liner equations and inequalities in one variable, quadratic equations, Cartesian coordinate system
and straight lines, function, linear and quadratic functions, exponential and logarithmic functions, system of liner
equations, matrices, permutation and combination, binomial theorem, arithmetic and geometric progression.

MAT 311 : Mathematics for Business and Economics II

Topics include: Economic and business models, functions, limits and continuity, concept of derivative, rules of
differentiation and integration, and their use. Constrained optimization with Lagrange multiplier, partial derivatives.
Theory is presented informally and techniques are related to polynomials, logarithmic and exponential functions.

iv
 Chapter – 01 : Number system 1 – 11
1.1 Introduction 1
1.2 Natural numbers 1
1.3 Prime numbers 2
1.4 Integer numbers 2
1.5 Even numbers 3
1.6 Odd numbers 3
1.7 Rational numbers 3
1.8 Fraction numbers 5
1.9 Irrational numbers 5
1.10 Real numbers 7
1.11 Interval 8
1.12 Modulus of real number 8
1.13 Imaginary numbers 9
1.14 Complex numbers 9
1.15 Number systems in chart 10
1.16 Exercise 10
Chapter – 02 : Set Theory 12 – 26
2.1 Introduction 12
2.2 Definition of Set 12
2.3 Elements of a set 12
2.4 Methods of describing a set 13
2.5 Types of sets 13
2.6 Operations on sets 15
2.7 Relation 16
2.8 Venn diagram 16
2.9 Number of elements in a finite set 17
2.10 De-Morgan’s laws 18
2.11 Some worked out example 18
2.12 Exercise 24
Chapter – 03 : Progressions (AP & GP) 27 – 37
3.1 Introduction 27
3.2 Sequence 27
3.3 Series 27
3.4 Arithmetic progression (A.P) 27
3.5 Geometric Progression (G.P) 28
3.6 Theorem of arithmetic series 28
3.7 Theorem of geometric series 29
3.8 Some worked out examples 30
3.9 Exercise 36
Chapter – 04 : Permutations and Combinations 38 – 51
4.1 Introduction 38
4.2 Permutation 38
4.3 Factorial notation 39
4.4 Permutations of n different things 39

v
4.5 Permutations of n things not all different 42
4.6 Circular permutation 43
4.7 Combination 44
4.8 Combinations of n different things taken some or all at a time 46
4.9 Combinations of n things not all different taken some or all at a time 46
4.10 Some worked out examples 47
4.11 Exercise 50
Chapter – 05 : Determinant and Matrix 52 – 83
5.1 Introduction 52
5.2 Definition of determinant 52
5.3 Value of the determinant 53
5.4 Minors and co-factors 54
5.5 Fundamental properties of determinant 54
5.6 Multiplication of two determinants 57
5.7 Application of determinants 57
5.8 Definition of matrix 60
5.9 Types of matrices 60
5.10 Matrices operations 65
5.11 Process of finding inverse matrix 67
5.12 Rank of a matrix 68
5.13 Use of matrix to solve the system of linear equations 68
5.14 Some worked out examples 71
5.15 Exercise 80
Chapter – 06 : Functions and Equations 84 – 108
6.1 Introduction 84
6.2 Formula 84
6.3 Relation 85
6.4 Function 85
6.5 Types of functions 85
6.6 Polynomial 88
6.7 Inequality 88
6.8 Equation 89
6.9 Degree of an equation 89
6.10 Quadratic equation 90
6.11 Formation of quadratic equation 91
6.12 Identity 92
6.13 Linear equation 92
6.14 System of linear equations 92
6.15 Solution methods of a system of linear equations 93
6.16 Break-Even point 103
6.17 Break-Even interpretation 103
6.18 Exercise 106
Chapter – 07 : Exponential and Logarithmic Functions 109 – 125
7.1 Introduction 109
7.2 Exponential function 109
7.3 Surds 113
7.4 Logarithmic function 116
7.5 Some worked out Examples 119
7.6 Exercise 123
Chapter – 08 : Mathematics of Finance 126 – 143
8.1 Introduction 126
8.2 Simple interest and the future value 126

vi
8.3 The yield on the common stock of a company 130
8.4 Bank discount 130
8.5 Compound Interest and the future value 131
8.5 Ordinary annuity 138
8.7 Exercise 141
Chapter – 09 : Limit and Continuity 144 – 154
9.1 Introduction 144
9.2 Limit 144
9.3 Difference between Lim f (x) and f (a) 145
xa
9.4 Methods of evaluating limit of a function 145
9.5 Some important limits 146
9.6 Left hand side and right hand side limits 146
9.7 Continuity 149
9.8 Some solved problems 151
9.9 Exercise 153
Chapter – 10 : Differentiation and its applications 155 – 179
10.1 Introduction 155
10.2 Differential coefficient 155
10.3 Fundamental theorem on differentiation 157
10.4 Meaning of derivatives and differentials 158
10.5 Some standard derivatives 158
10.6 Successive differentiation 162
10.7 Maxima, minima and point of inflection 163
10.8 Determination of maxima & minima 163
10.9 Calculus of multivariate functions 165
10.10 Business application of differential calculus 168
10.11 Some worked out examples 169
10.12 Exercise 177
Chapter – 11 : Integration and its applications 180 – 205
11.1 Introduction 180
11.2 Definition of integration 180
11.3 Indefinite integral 181
11.4 Fundamental theorem on integration 181
11.5 Some standard integrals 181
11.6 Integration by substitution 184
11.7 Integration using partial fractions 186
11.8 Definite integral 188
11.9 Properties of definite integral 188
11.10 Application of integration in business problems 190
11.11 Some worked out examples 194
11.12 Exercise 202
Chapter – 12 : Coordinate Geometry 206 – 224
12.1 Introduction 206
12.2 Directed line 206
12.3 Quadrants 207
12.4 Coordinates 207
12.5 Coordinates of mid point 208
12.6 Distance between two points 208
12.7 Section formula 209
12.8 Coordinates of the centroid 210
12.9 Area of a triangle 211

vii
12.10 Area of a quadrilateral 213
12.11 Straight line 213
12.12 Slope or gradient of a straight line 213
12.13 Different forms of equations of the straight line 214
12.14 Circle 215
12.15 Some worked out examples 215
12.16 Exercise 223
Chapter – 13 : Linear Programming 225 – 250
13.1 Introduction 225
13.2 What is optimization 225
13.3 Summation symbol 225
13.4 Linear programming 226
13.5 Formulation 227
13.6 Some important definitions 229
13.7 Standard form of LP problem 230
13.8 Graphical solution 230
13.9 Simplex 234
13.10 Development of a minimum feasible solution 236
13.11 The artificial basis technique 239
13.12 Duality in linear programming problem 242
13.13 Some worked out examples 243
13.14 Exercise 248
Chapter – 14 : Transportation Problem 251 – 279
14.1 Introduction 251
14.2 Transportation problem 251
14.3 Theorem 252
14.4 Northwest corner rule 255
14.5 Loop 259
14.6 Degeneracy case 262
14.7 Multiple solutions 269
14.8 When total supply exceeds total demand 273
14.9 Maximization problem 275
14.10 Exercise 277
Chapter – 15 : Assignment Problem 280 – 295
15.1 Introduction 280
15.2 Assignment problem 280
15.3 Algorithm of the Hungarian method 281
15.4 Justification of Hungarian method 282
15.5 The dual of the assignment problem 283
15.6 Some worked out example 285
15.7 Exercise 293

Annexure – 1 : Trigonometry i – viii

Annexure – 2 : Bibliography ix

viii
 Number system

01
Chapter
Number System

Highlights:

1.9 Introduction 1.1 Irrational numbers

1.10 Natural numbers 1.2 Real numbers
1.11 Prime numbers 1.3 Interval
1.12 Integer numbers 1.4 Modulus of real number
1.13 Even numbers 1.5 Imaginary numbers
1.14 Odd numbers 1.6 Complex numbers
1.15 Rational numbers 1.7 Number systems in chart
1.16 Fraction numbers 1.8 Exercise

1.1 Introduction: To fulfill daily human needs, man invented counting numbers at the
beginning of the civilization. The successive development of numbers develops the
modern mathematics. Mathematics means games of numbers. In business mathematics, we
know the use of numbers in business problems. So, to clear the concepts of mathematics
or business mathematics, firstly, we have to clear the concepts of number system. Number
system means the nature and properties or various types of numbers. In this chapter, we
shall learn about the system of numbers. After this chapter, number will mean real
number.

1.2 Natural numbers: The positive integer numbers 1, 2, 3, 4, . . . are used for counting.
These numbers are known as natural numbers. The set of natural numbers is denoted by
English capital letter, N. That is, N = {1, 2, 3, 4, 5, . . .}. Negative numbers, zero and
5
fractions are not natural numbers, that is, – 5, 0 and = 2.5 are not natural numbers.
2
Properties:
1. Summation or multiplication of any two or more natural numbers must be a natural
number. That is, if m, n  N then (m + n)  N and (m  n)  N. As for example
7,10  N; 7+10 = 17  N and 7  10 = 70  N.
2. Square of any natural number is also a natural number. That is, if m  N then
m2  N. As for example 4,5  N then 42 =16  N and 52 = 25  N.

1
 S. M. Shahidul Islam

3. Subtraction and Division of any two natural numbers may not be natural number.
m
That is, if m, n  N then m – n and may be or may not be natural number. As
n
10 7
for example 2, 7, 10  N; 10 – 7 = 3  N but 7 – 10 = – 3  N, = 5  N but =
2 10
0.7  N.
4. Square root of a natural number may be or may not be a natural number. That is, if
m  N then m  N or m  N. As for example 4, 5  N then 4 = 2  N but
5  N.
5. Between two different natural numbers one must be greater than another number.
That is, if m, n  N then m > n or n > m. As for example 5, 7  N then here 7 > 5.

1.3 Prime numbers: A natural number other than 1 is a prime number if and only if its
only divisors are 1 and the number itself. Such as 2, 3, 5, 7, 11, . . . etc. are prime numbers.

1.4 Integer numbers: The natural numbers, all natural numbers with negative sign before
and zero together known as integer numbers. That is, the integers are whole numbers
positive, negative and zero. The set of integer numbers is denoted by English capital letter,
Z or I. That is, Z = {. . ., –3, –2, –1, 0, 1, 2, 3, . . .}. Fractions are not integer number, that
5
is, = 2.5 is not integer number. It is clear that, all natural numbers are integer number.
2
Properties:
1. Summation or subtraction or multiplication of any two integer numbers must be an
integer number. That is, if m, n  Z then (m + n)  Z, (m – n)  Z and (m  n)  Z. As
for example 5, – 3  Z then 5 +(– 3) = 2  Z, 5 – (– 3) = 8  Z and 5  (– 3) = –15  Z
2. Square of any integer number is also an integer number. That is, if m  Z then
m2  Z. As for example 4, –5  Z; 42 =16  Z and (–5)2 = 25  Z.
3. Division of any two integer numbers may be or may not be integer number. That
m
is, if m, n  Z; then may be or may not be integer number. As for example 2, 5,
n
8 5
8  Z; then = 4  Z but = 2.5  Z.
2 2
4. Square root of an integer number may be or may not be an integer number. That is,
if m  Z then m  Z or m  Z. As for example 4, 5  Z then 4 = 2  Z but
5  Z.
5. Between two different integer numbers one must be greater than another number,
that is, if m, n  Z then m > n or n > m. As for example 5, 7  Z; here 7 > 5.

2
 Number system

1.5 Even numbers: The integer numbers, which are divisible by 2 are called even
numbers. Such as . . ., –6, –4, –2, 0, 2, 4, 6, . . . are even numbers. Generally, an even
number is denoted by 2n where n  Z. If addition or subtraction or multiplication or square
or square root of even number(s) exists as integer, it must be an even number. But division
of two even numbers may or may not be an even number, if it exists as integer.

1.6 Odd numbers: The integer numbers, which are not divisible by 2 are called odd
numbers. Such as . . ., –5, –3, –1, 1, 3, 5, . . . are odd numbers. Generally, an odd number
is denoted by 2n –1 where n  Z. If multiplication or square or square root of odd
number(s) exists as integer, it must be an odd number.
Example: Prove that, if we divide the square of any odd positive integer by 8, remainder
will be 1 always.
Proof: Let x be any positive odd integer number, that is, odd natural number. So, we can
let
x = 2n – 1, where n  N
 x 2  2n  1  4n 2  4n  1  4n(n  1)  1
2

Here, n and (n –1) are two consecutive natural number, that is, one of them is an even
number.
So, n(n –1) is divisible by 2, that is, 4n(n –1) is divisible by 4  2 = 8.
Therefore, if x 2 = 4n(n –1) +1, for all n  N is divided by 8, remainder will be 1 always.

p
1.7 Rational numbers: The numbers which can be expressed in the form where p, q
q
are integers and q is not equal to zero are called rational numbers. The set of rational
p
numbers is denoted by English capital letter, Q. That is, Q ={ : p, q  Z and q ≠ 0}. 7, –2
q
7 2 5
and 2.5 are rational number because 7 = , – 2 = and 2.5 = . We know that, 4 =
1 1 2
4 4 p
, –4= , that is, every integer number can be expressed in form. So, all integer
1 1 q
numbers are rational number.
Properties:
1. Summation or subtraction or multiplication or division or square of rational
number(s) must be a rational number. That is, if m, n  Q then (m + n)  Q, (m – n)
 Q, (m  n)  Q, m/n  Q and m2  Q. As for example 7, 10  Q then 7 + 10 =
17  Q, 7 – 10 = – 3  Q, 7  10 = 70  Q, 7/10 = 0.7  Q and 72 = 49  Q.
2. Square root of a rational number may be or may not be a rational number. That is,
if m  Q then m  Q or m  Q. As for example 4, 5  Q then 4 = 2  Q but
5  Q.

3
 S. M. Shahidul Islam

3. Between two different rational numbers one must be greater than another number.
That is, if m, n  Q then m > n or n > m. As for example 5.4, 7  Q; here 7 > 5.4.
4. There are infinite rational numbers between any two different numbers. Let us
consider any two different numbers 2 and 3, then all of 2.01, 2.001, 2.0001,
. . .
2.00001, 2.1, 2.2, 2.0000002, 2.5, 2.45, 2. 5 , 2.9452, 2.4 4 7 5 , ... etc. are greater
than 2 and less than 3 but all are rational numbers.

1.7.1 Theorem: Division of any two rational numbers (divisor must be non zero) is a
rational number.
Proof: Let us consider, any m, n  Q and let n be divisor, that is, n ≠ 0.
So, by the definition of rational numbers, we get
p p
m = 1 and n = 2 ; where p1 , p 2 , q1 , q 2  Z and p 2 , q1 , q 2 ≠ 0
q1 q2
p1
m q1 p q pq p
Now division of m and n is  = 1  2 = 1 2 =  Q.
n p2 q1 p2 p 2 q1 q ( 0)
q2
[We know that, multiplication of integers is an integer. So, let p1 q 2 = p  Z, p 2 q1 = q  Z
and p 2 q1 = q ≠ 0 (because of p 2 , q1 ≠ 0).]
Therefore, Division of any two rational numbers is a rational number. (Proved)

1.7.2 Theorem: Between two different rational numbers, there lie an infinite number of
rational numbers.
Proof: Let m, n be any two rational numbers such that m < n.
Now, m<n
 m+n<n+n
 m + n < 2n
1
 (m + n) < n
2
Again m<n
 m+m<m+n
 2m < m + n
1
 m < (m + n)
2
1
So, m < (m + n) < n
2
1
By properties of rational numbers, (m + n) is a rational number between m and n.
2

4
 Number system

We have thus shown that between two different rational numbers there lies a third rational
number. From this it follows that there lie an infinite number of rational numbers between
two different rational numbers. [Proved]

p
1.8 Fraction numbers: The numbers which can be expressed in the form where p 
q
Z–{0}, q  N–{1} and p, q are prime to each other are called fraction numbers. That is, the
5 7
rational numbers, which leave a remainder, are called fraction numbers. Such as , ,
3 10
21
2.1 = are fraction numbers. When we write the fraction number in decimal form is
10
 
called decimal fraction number. Such as 5.3, 2.4, 67.23, 2.4 4 7 5 etc. are decimal fraction
numbers.
 
Example: Convert into the rational form: (i) 3.5, (ii) 2.3 5 2
35 7
Solution: (i) 3.5 =  (Answer)
10 2
  2352  23
(ii) 2.3 5 2 =
990
 All the digits (neglecting decimals )  The digits without recurring decimals 
 
 9 for each digits with recurring decimals & 0 for each digits with decimal but not with recurring decimals 
2329
= (Answer)
990

p
1.9 Irrational numbers: The numbers which can not be expressed in the form where
q
p, q are integers and q not equal to zero are called irrational numbers, that is, the real
numbers which are not rational number are called irrational numbers. The set of irrational
numbers is denoted by Q / , that is, Q / = {the real numbers, which are not rational}. Such
as 2, 5 , Π = 3.1415927... etc. are irrational numbers because we cannot express
3,
p
these numbers in the form .
q
Properties:
1. Summation or subtraction or multiplication or division of different irrational
numbers is an irrational number. That is, if m, n  Q / then (m + n)  Q / , (m – n)
 Q / , (m  n)  Q / and m/n  Q / . As for example 2, 3  Q/ ; ( 2 + 3 ) Q / ,
( 3 – 2 ) Q / , 2 3 = 6  Q / and 3 / 2  Q/ .

5
 S. M. Shahidul Islam

2. Square of any irrational number is a rational number. That is, if m  Q / then

 
m2  Q. As for example 2  Q / then 2 = 2  Q.
2

3. Square root of an irrational number must be an irrational number. That is, if

m  Q / then m  Q / . As for example 2  Q / ; 2  Q/ .
4. Algebraic operations between a rational number and an irrational number make the
results irrational numbers. That is, if m  Q and n  Q / then (m + n), (m – n), m  n
and m/n are irrational numbers. As for example 2  Q and 2  Q / then (2 + 2)
 Q , (2 –
/
2 )  Q , 2 2 = 2 2  Q and 2/ 2 = 2  Q .
/ / /

5. Between two different irrational numbers one must be greater than another
number. That is, if m, n  Q / then m >n or n >m. As for example 2 , 3  Q / ;
here 3 > 2 .
6. There are an infinite irrational numbers between any two different numbers. Let us
consider any two different numbers 2 and 3, then all of 2.010564..., 2.00145379...,
2.00014512679..., 2.000017612389..., 2.14367845... etc. are greater than 2 and less
than 3 but all are irrational numbers.

Note: All integer numbers, finite decimal numbers and recurring decimal numbers are
p
rational numbers because they can be expressed in the form , but all infinite decimal
q
p
numbers are irrational numbers because they can not be expressed in the form . Such as
q
   503475  503 125743  
6, -7, 10, 4.5, 6.04, 10.0001, 0. 5 , 5.03 4 7 5 (= = ) and 237. 9 0 are
99900 24975
rational numbers, but 1.4142136..., 3.6055513... and 3.1415927... are irrational numbers.
Example: Show that 2 is an irrational number. [AUB-2002 BBA]
Proof: 12  1 , 2 2  4 and  22
2
So, 2 is greater than 1 and less than 2, that is, 2 is not an integer number.
p
Thus, let 2 = , where p, q  N, q > 1 and p, q is co-prime.
q
2
p
That is, 2 = 2
q
Or, 2 q2 = p2 - - - (i)
Here, left hand number is an even number, so right hand number p 2 or p must also be an
even number.
Let p = 2n, then from (i) we have

6
 Number system

2 q 2 = 2n 2
Or, 2 q 2 = 4n 2
Or, q 2 = 2n 2
Here, right hand number is an even number, so left hand number q 2 or q must also be an
even number.
It means that both p and q are even numbers, that is, p and q have a common factor at least
2, which contradicts the fact that p and q are integers prime to each other (co-prime).
p
Hence, 2 = is absurd. So, 2 cannot be a rational number, that is, 2 is an
q
irrational number. [Proved]

1.10 Real numbers: Rational and irrational numbers together known as real numbers. The
set of real numbers is denoted by R; that is, R = {x: x  Q or x  Q / }. Such as 2, 5, – 6,
5
3.54, – 7.8903, 2 , – 3 , 45.8769403..., etc. are real numbers. From above
7
discussion, we find the following relation:
NZQR
Thus, natural numbers constitute a proper subset of integers and the integers constitute a
proper subset of rational numbers and the latter constitutes a proper subset of real
numbers.
Properties:
1. Summation or subtraction or multiplication or division or square of real number(s)
must be a real number. That is, if m, n  R then (m + n)  R, (m – n)  R, (m  n)  R,
m/n  R and m2  R. As for example 7, 20  R then 7 + 20 = 27  R, 7–20 = –13  R,
7  20 = 140  R, 7/20 = 0.35  R and 72 = 49  R.
2. Even power of any real number is positive real number. So, in particular, square of
any real number is a positive number. That is, if m  R then m2  R ( R is the set
of positive real numbers.). As for example 3.4  R; 3.4  11.56  R .
2

3. Square root of a real number may be or may not be a real number. That is, if m  R
then m  R or m  R. As for example 4, – 2  R then 4 = 2  R but  2  R.
4. Between two different real numbers one must be greater than another number. That
is, if m, n  R then m > n or n > m. As for example 5, 7  R; here 7 > 5.
5. There are infinite real numbers between any two different real numbers. Let us
consider any two different numbers 2 and 3, then all of 2.1, 2.0021, 2.0001,
. . .
2.00001, 2.71, 2.02, 2.0000002, 2.35, 2.405, 2.0 5 , 2.52, 2.0 4 7 5 , ... etc. are
greater than 2 and less than 3 but all are real numbers.
6. We can represent all real numbers by a straight line. This straight line is known as
numbers line or directed line. The line is as follows:

7
 S. M. Shahidul Islam

–7 –6–5 –4 –3 –2 –1 0 1 2 3 4 5 6 7
Negative Positive
Number line
Figure – 1.1
7. If m, n  R and m  n = 0, then at least one of m and n is 0. As for example 0  5 = 0,
5  0 = 0 and 0  0 = 0 but 4  5 ≠ 0.
Example: Solve the polynomial equation: x2 + 3x – 10 = 0.
Solution: Given that, x2 + 3x – 10 = 0
Or, x2 + 5x – 2x – 10 = 0 [By middle term break-up method]
Or, x(x + 5) – 2(x + 5) = 0
Or, (x + 5)(x – 2) = 0 [We know, if m  n = 0 then m = 0 or n = 0]
So, x + 5 = 0  x = – 5
Or, x – 2 = 0  x = 2
Therefore, the solution, x = – 5, 2 [Answer]

1.11 Interval: The interval or domain of x is denoted by a ≤ x ≤ b or [a, b]; where a ≤ b

and a,b  R. This means the value of x contains any real number from a to b. This interval
is known as closed interval because it starting and ending values are known, that are a and
b respectively. No one knows the immediate before real number of a known real number
as well as the immediate after real number of that number. So, if a and b be omitted from
the interval a ≤ x ≤ b, it is indicated as a < x < b or (a, b). It means the value of x contains
any real number from a to b except a and b. This interval is called opened interval because
it starting and ending values are not known.

1.12 Modulus of real number: (Absolute numerical value) The modulus of a real
number, a is defined as the real number a or – a according as a is non-negative or negative.
The modulus of a real number, a is denoted by a and is defined by
 a, if a is non  negative
a 
 a, if a is negative
As for example 5 = 5,  5 = 5 and 0 = 0.
Properties:
1. The modulus of a real number, a is always non-negative, i.e., a  0 .
2. The modulus of a real number, a is always greater or equal to that number, i.e.,
a  a.
3. The modulus of a and – a is equal, i.e., a =  a .
4. The modulus of a is the maximum value of a and – a, i.e., a = max. {a, – a}.
5. The modulus of a is the positive square root of the square of a, i.e., a = a2 .

8
 Number system

Example: Find the solution set of 3x  2  7 and represent it in the number line.
Solution: Given that 3x  2  7
If (3x + 2) be non-negative, then 3x  2 = 3x + 2
That is, 3x + 2 < 7
Or, 3x + 2 – 2 < 7 – 2 [Subtracting 2 from both sides.]
Or, 3x < 5
5
 x< [Dividing both sides by 3]
3
If (3x + 2) be negative, then 3x  2 = – (3x + 2)
That is, – (3x + 2) < 7
[We know, 5 > 3 but –5 < –3. For this inequality sign (>, <) changes when we multiply by
a negative number]
Or, 3x + 2 > – 7 [Multiplying both sides by –1]
Or, 3x + 2 – 2 > – 7 – 2 [Subtracting 2 from both sides.]
Or, 3x > – 9
 x > – 3 [Dividing both sides by 3]
5
Therefore, the solutions is x < and x > – 3.
3
5
And the solution set, S = {x  R: x < and x > – 3}. The number line representing
3
solution set is as follows:–7 –6–5 –4 –3 –2 –1 0 1 2 3 4 5 6 7

Number line 5
Figure – 1.2 3
1.13 Imaginary numbers: Square root of negative numbers is called imaginary numbers,
because square of any real number is positive only. As for example
 2 ,  3,  4 ,  9 etc. are imaginary numbers. Imaginary number  1 is always
denoted by i, that is, i =  1 . So, i 2 = –1 and i 3  i 2 .i  1.i  i . Therefore,
 9  (1).9   1. 9  i3  3i .

1.14 Complex numbers: If ‘a’ and ‘b’ are two real numbers then the number of the form
a+ib is known as a complex number. It has a real part, ‘a’ and the imaginary part, ‘b’. As
for example 4 + i3, 0 + i2, 2 + i0, 3 + i5 and 3 – i5 are complex numbers. The complex
number z = (a – ib) is the conjugate of z = (a + ib).
We can express every real number as complex number. As for example, complex form of
real numbers 3, 4.5 and – 3.2 are 3 + i0, 4.5 + i0 and – 3.2 + i0 respectively.
Example: Multiply 3 + i5 and 3 – i5.

9
 S. M. Shahidul Islam

Solution: Multiplication: (3 + i5)(3 – i5) = 3.3 – 3.i5 + i5.3 – i5.i5

= 9 – 15i + 15i – 25i2
= 9 – 25(–1) [We know, i 2 = –1]
= 9 + 25
= 34, is a real number.

1.15 Number systems in chart: From the above discussion we now present the various
number systems in the form of following chart:

Number system

Real number Imaginary number

Complex number

Rational number Irrational number

Integer number Fraction number

Negative number Non-negative number

Natural number Zero

Prime number

Figure – 1.3

1.16 Exercise:
1. Define with examples: (i) counting number, (ii) decimal fraction number, (iii) co-
prime numbers.
2. Is it possible to convert every real number as complex number? And how?
3. Is every integer number rational number? Discuss your opinion.

10
 Number system

4. Discuss the differences between: (i) natural and integer numbers;

(ii) rational and irrational numbers;
(iii) imaginary and complex numbers.
5. State whether the following statements are true or false:
(i) Every rational number is a real number.
(ii) Every irrational number is a real number.
(iii) Every real number is a rational number.

(iv) 2.15 is a real number.
(v) 6.4343434343 . . . is an irrational number.
(vi) The product of two rational numbers is rational.
(vii) The product of two odd integers is an odd integer.
[Answer: (i) True, (ii) True, (iii) False, (iv) True, (v) False, (vi) True, (vii) True]
6. Write two rational numbers and two irrational numbers between 3.0001 and
3.0002. [Answer: Rational numbers: 3.00012316767, 3.000143435; Irrational
numbers: 3.0001554684672 . . ., 3.00015463. . .]
  81
7. Convert into the rational form: (i) 4.05, (ii) 7, (iii) 0. 3 , (iv) 2.15 [Answer: (i) ,
20
7 1 97
(ii) , (iii) , (iv) ]
1 3 45
8. Show that 3, 5 and 13 are irrational numbers.
9. Solve the following equations:
(i) (x + 1)(x – 3.5) = 0 [Answer: x = – 1, 3.5]
(ii) x2 + 5x + 6 = 0 [Answer: x = – 2, – 3]
(iii) x2 – 17x + 70 = 0 [Answer: x = 7, 10]
10. Find the solution sets:
(i) 2 x  10  6 [Answer: {– 2, – 8}]
(ii) x  2  7 [Answer: {x  R: - 5 ≤ x ≤ 9}]
3 7
(iii) 5 x  2  5 [Answer: {x  R: x ≥ or x ≤  }]
5 5
(iv) 4 < 2 x < 10 [Answer: {x  R: 2 < x < 5 or – 2 > x > – 5}]
11. Prove that between two different real numbers, there lie an infinite number of real
numbers.
12. Prove that between two different irrational numbers, there lie an infinite number of
irrational numbers.
13. Find the square roots with complex number i:
(i) – 4, (ii) – 8, (ii) – 16 [Answer: (i) 2i, (ii) 2 2 i , (iii) 4i]
14. Find the multiplications of (i) (3 + i8)(3 – i8), (ii) (2 + i5)(5 + i9) [Answer: (i) 73,
(ii) – 35 + 43i]

11
 S. M. Shahidul Islam

02
Chapter
Set Theory

Highlights:
2.1 Introduction 2.7 Relation
2.2 Definition of Set 2.8 Venn diagram
2.3 Elements of a set 2.9 Number of elements in a finite set
2.4 Methods of describing a set 2.10 De-Morgan’s laws
2.5 Types of sets
2.6 Operations on sets
2.11 Some worked out example
2.12 Exercise

2.1 Introduction: Some simple concepts about groups or collections or sets are core ideas
in mathematics. We use braces to indicate a set, and specify the members or elements of
the set within the braces. The concepts of sets are used not only in mathematics but also in
statistics and many other subjects. Here we shall discuss the concepts of sets and a few
applications in business problems.

2.2 Definition of Set: A set is a collection of well-defined and well-distinguished objects.

It is almost a convention to indicate sets by capital letters, like A, B, C etc. and to enclose
its elements by second brackets, {}.
Example: 1) A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 0}
2) B = {x: x is the student of BBA 15th batch of AUB}
3) C = {x: x be the possible outcomes in the toss of a die}
N.B. The basic characteristic of a set is that it should be well defined and its members or
elements should be well distinguished for easy recognition by description.

2.3 Elements of a set: The objects that make up a set are called the elements or members
of the set. If a be element of set A then we write aA and is read as ‘a belongs to A’. And
if c is not an element of set A then we write cA and is read as ‘c does not belong to A.’
Example: Let a set, A = {1, 2, 3, 4}. Here 1, 2, 3 & 4 are the elements of the set A, that is
1 A, 2 A, 3 A and 4A.

12
 Set theory

2.4 Methods of describing a set: There are two methods of describing a set. First one is
Tabular/Roster/Enumeration method and second one is Selector/Property builder/Rule
method.
i) Tabular method: In this method we enumerate or list all the elements of the set within
second brackets.
Example: A = {a, b, c, d}, B = {0, 1, 2, 3}, etc.

ii) Selector method: In this method the elements are not listed but are indicated by
description of their characteristics.
Example: A = {x: x is a vowel in English alphabet}

2.5 Types of sets: There are various types of sets. We describe below a few of them:
1) Finite set: When the elements of a set can be counted by a finite number then the set is
called a finite set.
Example: A = {1, 2, 3, 4}
B = {1, 2, 3, . . ., 1000}
C = {x: 2  x  100 and x is even integer}

2) Infinite set: If the elements of a set cannot be counted in a finite number, the set is
called an infinite set.
Example: N = {1, 2, 3, . . .}
B = {x: x be even integer numbers}

3) Singleton: The set, which contains only one element is called a singleton or a unit set.
Example: A = {1}
B = {}
C = {x: x is an integer neither positive nor negative}
D = {x  6 < x < 8 and x is an integer number}

4) Null or Empty set: The set, which has no element, is called null or empty or void set.
Generally it is denoted by a Greek letter  (phi).
Example: = { }
A = {x: 6 < x < 7 and x is an integer number}

5) Equal set: Two sets A and B are said to be equal if every element of A is also an
element of B and every element of B is also an element of A.
Example: A ={1, 2, 3}
B = {1, 2, 3}
Here every element of A is also element of B and every element of B is also element of A.
So set A and set B is equal set.

13
 S. M. Shahidul Islam

6) Equivalent set: If the element of a set can be put into one to one correspondence with
the elements of another set, then the two set are called equivalent set. The symbol ‘ ’ is
used to denote equivalent sets.
Example: A = {1, 2, 3, 4}
B = {a, b, c, d}
Here, set A  B because it is possible to make one to one correspondence among the
elements of both sets.
N.B: If two sets are infinite sets or contain same number of elements then they are
equivalent sets. Example: Let, A = {x: x  N and x be even integer}
and B = {x: x  N and x be odd integer}
Then A  B

7) Disjoint Sets: Two or more sets having no common element are called disjoint sets.
Example: Set A = {a, b, c, d} and B = {p, q, r, s, t} are disjoint sets because they have no
common element.

8) Subsets: If every element of a set B is also an element of a set A then, set B is called
subset of A and is written as, B  A or B  A or A  B or A  B
Example: Let A = {1, 2, 3, 4, 5, 6}
And B ={2, 4, 6}
Then B  A because of every element of B is an element of A.
N.B: Every set has its 2n number subsets where n is the number of elements of that set.
Two of them are improper subsets and the remaining are proper subsets.

9) Proper Subset: Set B is called proper subset of super set A if each and every element
of set B are the elements of the set A and at least one element of super set A is not an
element of B.
Example: Let A = {1, 2, 3, 4, 5, 6, 7}
B = {1, 3, 5, 7}
So, B  A

10) Improper Subset: A set itself is a subset of that set and Ø is a subset of every set;
these two subsets are called improper subset.
Example: Let us consider A ={a, b, c}
The subsets of A are {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}, Ø
Here, we have 2 n = 23 = 8 subsets of A. Subsets {a, b, c} and Ø are improper subset of A
and the remaining are proper subsets of A.

11) Family of sets: The set which all the elements are sets themselves then it is called a
family of sets or set of sets.
Example: A = {{1}, {2}, {1, 2}, Ø} is a family of sets.

14
 Set theory

12) Power set: If a family of sets contains all subset of a set then this family of sets is
called power set of that set.
Example: Let A = {1, 2, 3}
The subsets of A are Ø, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}
Then power set of A is P(A) = {Ø, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}

13) Universal set: Our discussing sets are subsets of a big set, this big set is known as
universal set of that sets. It is generally denoted by the symbol U.
Example: The set of integers may be considered as a universal set for the set of even
integers and the set of odd integers.

2.6 Operations on sets:

1) Intersection of sets: The Intersection of two sets A and B is the set consisting of all
elements, which belong to both sets A and B. The intersection of sets A and B is denoted
by A  B and is read as “ A intersection B,” or “ the intersection of A and B.”
Symbolically, A  B = {x: x  A and x  B}
Example: Let, A = {1, 2, 3, 4, 5} and B = {2, 4, 6}
 A  B = {2, 4}

2) Union of sets: The union of two sets A and B is the set consisting of all elements,
which belong to either A or B or both. The union of sets A and B is denoted by AB and
is read as ‘A union B,’ or ‘the union of A and B.’ Symbolically, AB = {x: xA or xB}
Example: Let, A = {1, 2, 3, 4, 5} and B = {2, 4, 6}
 A  B = {1, 2, 3, 4, 5, 6}

3) Difference of two sets: The Difference of two sets A and B is the set of all those
elements, which belong to A and not to B. The difference of sets A and B is denoted by
A – B or A ~ B and is read as “ A difference B,” or “ the difference of A and B.”
Symbolically, A – B = {x: x  A and x  B}
Example: Let, A = {1, 3, 5}, B = {4, 5, 6}
 A – B = {1, 3} and B – A = {4, 6}.

4) Symmetric difference of two sets: If A and B are two sets, then the set (A– B)(B–A)
is called the symmetric difference of two sets and is written as A ∆ B. That is, the
symmetric difference of two sets A and B is the set of all those elements of A and B,
which are not common to both A and B. We have to note that A ∆ B = B ∆ A.
Example: If A = {a, b, c, d} and B = {b, c, d, e, f} then
A – B = {a} and B – A = {e, f}
So, A ∆ B = (A – B)  (B – A) = {a, e, f}.

15
 S. M. Shahidul Islam

5) Complement of a set: The complement of a set is the set of all elements, which do not
belong to that set. The compliment of the set A is A/ or Ac = U – A = {x: x  U and xA}.
Example: Let, U = {1, 2, 3, 4, 5, 6} and A = {1, 3, 5}
Then, Ac or A/ = U – A = {2, 4, 6}

6) Cartesian product: If A and B be any two sets, then the set of all ordered pairs whose
first element belongs to set A and second element belongs to set B is called the Cartesian
product of A and B and is denoted by A  B.
i.e. A  B = {(x, y): x  A, y  B}
N.B: An ordered pair of objects consists of two elements a and b written in parentheses
(a, b). The ordered pair (a, b) and (b, a) are not same, i.e., (a, b) ≠ (b, a). Two ordered
pairs (a, b) and (c, d) will be equal if a = c and b = d.
Example: Let A = {1, 2}, B ={a, b}
 A  B = {(1, a), (1, b), (2, a), (2, b)}
And B  A = {(a, 1), (a, 2), (b, 1), (b, 2)}
So, A  B ≠ B  A

2.7 Relation: If A and B be two sets then non empty subset of ordered pairs of Cartesian
product, A  B is called relation of A and B and is denoted by R. If we consider x  A and
y  B then (x, y)  R.
Example (i): If A = {1, 2, 3} and B = {3, 5} then A  B = {(1, 3), (1, 5), (2, 3), (2, 5), (3,
3), (3, 5)}. So, the relation x < y where x  A and y  B is R = {(1, 3), (1, 5), (2, 3), (2, 5),
(3, 5)}.
Example (ii): If A = {$4, $7, $8} is a set of cost of per unit product and B = {$5, $8} is the
set of selling price of per unit product of a production firm. Find the profitable relation
between cost and selling price.
Solution: Here, A  B = {($4, $5), ($4, $8), ($7, $5), ($7, $8), ($8, $5), ($8, $8)}
A firm becomes profitable if its selling price of per unit product is greater than the cost of
per unit product. So, the profitable relation, R = {($4, $5), ($4, $8), ($7, $8)}. (Answer)

2.8 Venn diagram: The Venn diagrams are named after English Logician John Venn. In
this diagram, the universal set U is denoted by a region enclosed by a rectangle and one or
more sets are shown by circles or closed curves within this rectangle. These circles or
closed curves intersect each other if there are any common elements among them if there
are no common elements then they are shown separated. This diagram is useful to
illustrate the set relations, the set operations etc.
Example: Let A = {1, 2, 3}, B = {2, 3, 4} and U = {1, 2, 3, 4, 5, 6}. Various relation and
operations are shown bellow by the Venn diagrams:

16
 Set theory

U
A B

2
1 3
4

5 6

Venn diagram

Figure – 2.1

From diagram, we get

A  B = {2, 3}
A  B = {1, 2, 3, 4}
A – B = {1}
B – A = {4}
U – (A  B) = {5, 6}
Ac = {4, 5, 6}
Bc = {1, 5, 6} etc.

2.9 Number of elements in a finite set: It is very important to find out the number of
elements in a finite set in the solution of the practical problems. Generally, if A be a set
then n(A) means the number of elements of the set A. Such as if A = {a, b, c, d, e, f, g}
then n(A) = 7; if B = {1, 3, 5, 7} then n(B) = 4 and if C = {x: x be even integer numbers
between 3 to 15} then n(C) = 6 etc.
Let A and B be two disjoint sets that means they have no common element then total
elements of the both sets will be n(A  B).
So, n(A  B) = n(A) + n(B) - - - (i).
But, if A and B are not disjoint sets that means they have some common elements then
total elements of both sets will be
n(A  B) = n(A) +n(B) – n(A  B) - - - (ii)
Similarly, if A, B and C be three disjoint sets then,
n(ABC) = n(A) + n(B) + n(C) - - - (iii)
But if A, B and C are not disjoint sets then,
n(ABC) = n(A) + n(B) + n(C)– n(AB)– n(AC) – n(BC)+ n(ABC) - - - (iv)
For disjoint cases, each of n(AB), n(AC), n(BC) and n(ABC) is 0. So, formula
(ii) and (iv) are main.

17
 S. M. Shahidul Islam

2.10 De-Morgan’s laws:

1. Complement of a union of two sets is the intersection of complements of that sets.
i.e. if A and B be any two sets them (A  B)/ = A/  B/
2. Complement of an intersection of two sets is the union of the complements of that sets.
i.e. if A and B be any two sets them (A  B)/ = A/  B/
Proof of 1st law: Let x be any element of (A  B)/.
Then by definition of complement,
x  (A  B)/  x  (A  B)
 x A and x B
 x A/ and x B/
 x A/  B/
i.e. (A  B)/  A/  B/ - - - (i)
Again let y be any element of A/  B/
Then by definition of intersection,
y  A/  B/  y  A/ and y B/
 y  A and y B
 yAB
 y  (A  B)/
i.e. A/  B/  (A  B)/ --- (ii)
From (i) and (ii), we have (A  B) = A  B/
/ /
(Proved)
Proof of 2 law: Let x be any element of (A  B) .
nd /

Then by definition of complement,

x  (A  B)/  xAB
 x  A or x  B
 x  A/ or x  B/
 x  A/  B/
i.e. (A  B)/  A/  B/ - - - (i)
Again let y be any element of A/  B/
Then by definition of union,
y  A/  B/  y  A/ or y  B/
 y  A or y  B
 yAB
 y  (A  B)/
i.e. A/  B/  (A  B)/ - - - (ii)
From (i) and (ii), we have (A  B)/ = A/  B/ (Proved)

2.11 Some worked out examples:

Example (1): If A = {5, 10, 12, 15, 19}, B = {3, 10, 15} and C = {7, 10}, write the sets
A B and B  C by tabular & selector methods.

18
 Set theory

Solution: Given that A = {5, 10, 12, 15, 19}, B = {3, 10, 15} and C = {7, 10}
So, A  B = {5, 10, 12, 15, 19}  {3, 10, 15}
= {3, 5, 10, 12, 15, 19} (Answer)
In the selector method, A  B = {x: x = 3 or 5 or 10 or 12 or 15 or 19}
And B  C = {3, 10, 15} {7, 10}
= {10}
In the selector method, B  C = {x: x = 10} (Answer)

Example (2): Let A ={1, 2, 3, 4}, B ={2, 4, 6, 8} and C = {3, 4, 5, 6}. Determine (i) A–B,
(ii) B – C, (iii) B – B [RU-1993 Mgt.]
Solution: Given that A = {1, 2, 3, 4}, B = {2, 4, 6, 8} and C = {3, 4, 5, 6}.
(i) A – B = {1, 2, 3, 4} – {2, 4, 6, 8} = {1, 3}
(ii) B – C = {2, 4, 6, 8} – {3, 4, 5, 6} = {2, 8}
(iii) B – B = {2, 4, 6, 8} – {2, 4, 6, 8} = 

Example (3): Let A = {a, b, c, d}. Find the power set P(A). [NU-1999 Mgt., AUB-1998
B.B.A, RU-1993 A/C]
Solution: The subsets of A are
{a}, {b}, {c}, {d}, {a, b}, {a, c}, {a, d}, {b, c}, {b, d}, {c, d}, {a, b, c}, {a, b,
d},
{a, c, d}, {b, c, d}, {a, b, c, d}, .
So, the Power Set P(A) ={{a}, {b}, {c}, {d}, {a, b}, {a, c}, {a, d}, {b, c}, {b, d}, {c, d},
{a, b, c}, {a, b, d}, {a, c, d}, {b, c, d}, {a, b, c, d}, }
Example (4): Let A = {1, 2, 3}, B = {1, 2}, C = {2, 3}, and D = {1, 3}. Prove that
P(A) = {A, B, C, D, {1}, {2}, {3}, Ø}.
Solution: Given that A = {1, 2, 3}
B = {1, 2}
C = {2, 3}
D = {1, 3}
The subsets of A are {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}, Ø
 The power set of A, P(A) = {{1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}, Ø}
= {{1, 2, 3}, {1, 2}, {2, 3}, {1, 3}, {1}, {2}, {3}, Ø}
= {A, B, C, D, {1}, {2}, {3}, Ø} (Proved)
Example (5): If set A = {a, b, c}, set B = {b, c, d} and the universal set U ={a, b, c, d, e}
then verify De-Morgan’s laws. [AUB-2002 M.B.A]
Solution: Given that, A = {a, b, c}, B = {b, c, d} and U ={a, b, c, d, e}
Then A/ = U – A = {a, b, c, d, e}–{a, b, c} = {d, e},
B/ = U – B = {a, b, c, d, e}–{b, c, d} = {a, e},
A  B = {a, b, c}{b, c, d} = {b, c},
A  B = {a, b, c}{b, c, d} = {a, b, c, d},
(A  B)/ = U – (A  B) = {a, b, c, d, e} – {a, b, c, d} = {e},

19
 S. M. Shahidul Islam

A/  B/ = {d, e}  {a, e} = {e}

So, (A  B)/ = A/  B/
And (A  B)/ = U – (A  B) = {a, b, c, d, e} – {b, c} = {a, d, e},
A/  B/ = {d, e}  {a, e} = {a, d, e}
Therefore, (A  B)/ = A/  B/ (Verified)

Example (6): If A = {1, 4}, B = {4, 5}, C = {5, 7}, verify that,
A  (B  C) = (A  B)  (A  C)
Solution: Given that A = {1, 4}
B = {4, 5}
C = {5, 7}
B  C = {4, 5}  {5, 7}
= {5}
 L.H.S. = A  (B  C)
= {1, 4}  {5} [Putting the value of (B  C)]
= {(1, 5), (4, 5)}
R.H.S. = (A  B)  (A  C)
= ({1, 4}  {4, 5})  ({1, 4}  {5, 7})
= {(1, 4), (1, 5), (4, 4), (4, 5)}  {(1, 5), (1, 7), (4, 5), (4, 7)}
= {(1, 5), (4, 5)}
 L.H.S. = R.H.S. (Verified)

Example (7): Let A, B and C be any there sets. Prove that A(BC) = (AB)  (AC).
[RU-1981 A/C]
Solution: Let x be any element of A  (B  C)
Then by the definition of intersection,
x  A  (B  C)  x  A and x  (B  C)
 x  A and (x  B or x  C)
 (x  A and x  B) or (x  A and x  C)
 x  (A  B) or x  (A  C)
 x  (A  B)  (A  C)
i.e. A  (B  C)  (A  B)  (A  C) - - - (i)
Again let y be any element of (A  B)  (A  C)
Then by the definition of union,
y  (A  B)  (A  C)  y  (A  B) or y  (A  C)
 (y  A and y  B) or (y  A and yC)
 y  A and (y  B or y  C)
 y  A and y (B  C)
 y  A  (B  C)
i.e. (A  B)  (A  C)  A  (B  C) - - - (ii)

20
 Set theory

From (i) and (ii), we have, A  (B  C) = (A  B)  (A  C) (Proved)

Example (8): If A, B and C be any three sets, then prove that A- (BC) = (A-B)  (A-C)
[NU-1998 A/C]
Solution: Let x be any element of A - (B  C)
Then by the definition of difference,
x  A-(B  C)  x  A and x  (B  C)
 x  A and (x  B and x  C)
 (x  A and x  B) and (x  A and x  C)
 x  (A - B) and x  (A - C)
 x  (A - B)  (A - C)
i.e. A - (B  C)  (A - B)  (A - C) - - - (i)
Again let y be any element of (A - B)  (A - C)
Then by the definition of union,
y  (A - B)  (A - C)  y  (A - B) and y  (A - C)
 (y  A and x  B) and (y  A and x  C)
 y  A and (y  B and y  C)
 y  A and y  (B  C)
 y  A- (B  C)
i.e., (A - B)  (A - C)  A - (B  C) - - - (ii)
From (i) and (ii), we have, A - (B  C) = (A-B)  (A-C) (Proved)

Example (9): If A, B and C be any three sets, then prove that

A  (B  C) = (A  B)  (A  C)
Solution: Under the definition of Cartesian product,
A  (B  C) = {(x, y): x  A, y  (B  C)}
= {(x, y): x  A, (y B or y  C)}
= {(x, y): (x  A, y  B) or (x A, y  C)}
= {(x, y): (x, y)  (A  B) or (x, y)  (A  C)}
= {(x, y): (x, y)  (A  B)  (A  C)}
= (A  B)  (A  C) (Proved)

Example (10): If n(U) = 800, n(A) = 200, n(B) = 300, n(AB) = 150, then find n(A/B/)
where A, B are two sets and U is the universal set. [RU-1984, 85, NU-1998 A/C]
Solution: Given that, n(U) = 800, n(A) = 200, n(B) = 300, n(AB) = 150, n(A/B/) = ?
From De-Morgan’s laws, we know that, (A  B)/ = A/  B/.
So, n(A/B/) = n((A  B)/) and n((A  B)/) = n(U) – n(AB).
We also know, n(AB) = n(A) + n(B) – n(AB)
= 200 + 300 – 150
= 350

21
 S. M. Shahidul Islam

Therefore, n(A/B/) = n((A  B)/)

= n(U) – n(AB)
= 800 – 350 = 450 (Answer)

Example (11): Dhaka city has a total population of 8000000. Out of it 1800000 are
service holders and 1000000 are businessmen while 120000 are in both positions. Indicate
how many people are neither service holders nor businessmen. [AUB-2002 M.B.A]
Solution: Let P, S and B denote the sets of total people, service holders and businessmen
respectively. So, n(P) = 8000000, n(S) = 1800000, n(B) = 1000000 and n(SB) = 120000
So, the number of people who are at least one position is n(S  B).
We know that, n(S  B) = n(S) + n(B) – n(S  B).
= 1800000 + 1000000 – 120000
= 2680000
So, the number of people who are neither service holder nor businessman
= n(P) – n(S  B)
= 8000000 – 2680000
= 5320000. [Answer]

Example (12): A roads and highway construction firm has 33 bulldozer drivers, 22 crane
drivers and 35 cement-mixture drivers. Of the drivers 14 persons can drive both mixture
and dozer, 10 persons can drive both mixture and crane, 10 persons can drive both dozer
and crane and 4 persons can drive all the three machines. Determine the total number of
drivers of the firm. [DU-1987 mgt., RU-1984 mgt.]
Solution: Let, the set of bulldozer drivers be A, the set of crane drivers be B and the set of
cement-mixture drives be C.
So, n(A) = 33, n(B) = 22, n(C) = 35, n(AB) = 10, n(BC) = 10, n(AC) = 14 and
n(ABC) = 4.
 the total number of drivers, n(ABC) = ?
We know that,
n(ABC) = n(A) + n(B) + n(C) – n(AB) – n(AC) – n(BC) + n(ABC)
= 33 + 22 + 35 – 10 – 10 – 14 + 4
= 60
Therefore, the firm has total 60 drivers. (Answer)

Example (13): A company studies the product preferences of 25,000 consumers. It was
found that each of the products A, B and C was liked by 8000, 7000 and 6000 respectively
and all the products were liked by 1500. Products A and B were liked by 3000, products A
and C were liked by 2000 and products B and C were liked by 2200. Prove that the study
results are not correct. [AUB-2003 B.B.A]

22
 Set theory

Solution: Let A, B and C denote the set of consumers who like products A, B and C
respectively. The given data means n(A) = 8000, n(B) = 7000, n(C) = 6000, n(AB) =
3000, n(BC) = 2200, n(AC) = 2000, n(ABC) = 1500 and n(ABC) = 25,000.
We know that
n(ABC) = n(A) + n(B) + n(C) – n(AB) – n(AC) – n(BC) + n(ABC)
= 8000 + 7000 + 6000 – 3000 – 2200 – 2000 + 1500
= 15,300 ≠ 25,000
This shows that the study results are not correct.

Example (14): Out of 1200 students of a college, 400 played cricket, 350 played football
and 512 played table tennis: of the total 100 played both cricket and football; 142 played
football and table tennis; 95 played cricket and table tennis; 50 played all the three games.
(i) How many students did not play any game? (ii) How many students played only one
game? [AUB-2002 B.B.A]
Solution: Let the C, F and T denote the sets of players playing cricket, football and table
tennis respectively. And S represents the set of total students. Now we are given that
n(C) = 400, n(F) = 350, n(T) = 512, n(CF) = 100, n(FT) = 142, n(CT) = 95,
n(CFT) = 50 and n(S) = 1200. U

T
F

T
Figure – 2.2
(i) Number of students who played at least one game is given by
n(CFT) = n(C) + n(F) + n(T) – n(CF) – n(FT) – n(CT) + n(CFT)
= 400 + 350 + 512 – 100 – 142 – 95 + 50
= 975
So, number of students who did not play any game = n(S) – n(CFT)
= 1200 – 975
= 225 (Answer)
(ii) We know, the number of students who played cricket, n(C) = 400. This contains
a) The students who played cricket only.
b) The students who played cricket and football, n(CF)
c) The students who played cricket and table tennis, n(CT)
d) The students who played cricket, football and table tennis, n(CFT)
Now from the Venn diagram we get:
The number of students who played cricket only
= n(C) – n(CF) – n(CT) + n(CFT)

23
 S. M. Shahidul Islam

= 400 – 100 – 95 + 50
= 255
Similarly,
The number of students who played football only
= n(F) – n(CF) – n(FT) + n(CFT)
= 350 – 100 – 142 + 50
= 158
And the number of students who played tennis only
= n(T)–n(FT)–n(CT) + n(CFT)
= 512– 142 – 95 + 50
= 325
So, the number of students who played only one game = 255 + 158 + 325
= 738 (Answer)

10 p
Example (15): Demand function of a product D = and supply function S = p2;
p3
where p means the price in dollar of the product per unit. Using set theory determine
equilibrium price and quantity. [AUB-2003 M.B.A]
10 p
Solution: Substituting p by 1, 2, 3, . . . etc. in D = and S = p2 we get the following
p3
sets of ordered pairs:
{(p, D)} = {(1, – 5), (2, – 20), (3, ∞), (4, 40), (5, 25), (6, 20)}
(Notice that, using p = 7, 8, 9, . . . etc. we can make a larger set.)
{(p, S)} = {(1, 1), (2, 4), (3, 9), (4, 16), (5, 25), (6, 36)}
Let, A = {(p, D)} = {(1, – 5), (2, – 20), (3, ∞), (4, 40), (5, 25), (6, 20)}
And B = {(p, S)} = {(1, 1), (2, 4), (3, 9), (4, 16), (5, 25), (6, 36)}
So, AB = {(5, 25)}
Therefore, equilibrium price, p = $5 and quantity = 25 units.

2.12 Exercise:
1. Define with examples: (i) set, (ii) subset, (iii) power set, (iv) intersection of sets
2. What do you mean by  and {}?
3. What is difference between union and intersection of sets?
4. Discuss the difference between difference of sets and symmetric difference of sets.
5. State and prove the De-Morgan’s laws. [AUB-02, 03, RU-95]
6. What do you mean by ordered pair?
7. Let U = {1, 2, 3, 4, 5, 6, 7}, A = {1, 3, 5, 7}, B = {2, 4, 6} and C = {2, 3, 4, 5} then
find the following sets: a) AB, b) AB, c) BC, d) A/, e) B/, f) A/  B/, g)

24
 Set theory

(AB)/. [Answer: a){1, 2, 3, 4, 5, 6, 7}, b) , c){2, 3, 4, 5, 6}, d){2, 4, 6}, e){1, 3,

5, 7}, f) , g) ]
8. Let A = {5, 10, 15, 20, 25}, B = {5, 15, 21} and C = {10, 20}, write the following
sets by the roster and property builder methods: a) AB, b) AB, c) AC, e)
BC. [Answer: a){5, 10, 15, 20, 21, 25}, {x: x = 5 or 10 or 15 or 20 or 21 or 25}
b){5, 15}, {x: x = 5 or 15} c){5, 10, 15, 20, 25}, {x: x = 5 or 10 or 15 or 20 or 25}
d){}, {x: x has no value}]
9. Write down three sets A, B and C such that A  B ≠ , A  C ≠  and B  C ≠
 but A  B  C = . [Answer: {0, 1, 2}, {3, 4, 5} and {6, 7, 8}]
10. Let A = {-1, 0, 1, 2}. Find its three subsets, each of which contains 3 elements.
[Answer: {-1, 0, 1}, {-1, 0, 2} and {0, 1, 2}]
11. Let A = {1, 2, 3, 4}, find the power set of A, i.e., P(A). [Answer: {{1}, {2}, {3},
{4}, {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}, {1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2,
3, 4}, {1, 2, 3, 4}, }]
12. If A = {1, 2} and B = {1, 5, 7} then find A  B and A  A. [Answer: {(1, 1), (1, 5),
(1, 7), (2, 1), (2, 5), (2, 7)} and {(1, 1), (1, 2), (2, 1), (2, 2)}]
13. If A = {a, b, c} and B = {d, e} then prove that A  B ≠ B  A
14. If A = {2, 3, 5} and B = {3, 7} then find the relation, R as x > y where x  A and
y  B, [Answer: {(5, 3)}]
15. If A = {$3, $5, $8} is a set of cost of per unit product and B = {$5, $10} is the set
of selling price of per unit product of a production firm. Find the profitable relation
between cost and selling price. [Answer:{($3, $5), ($3, $10), ($5, $10), ($8, $10)}]
16. If (x + y, 5) = (15, x – y) then find the value of (x, y). [Answer: (10, 5)]
17. If A, B and C be any three sets, prove that: A  (B  C) = (A  B)  (A  C)
18. If A, B and C be any three sets, then prove that A  (B  C) = (A  B)  (A  C)
19. For any three sets A, B and C prove that A  (B – C) = (A  B) – (A  C).
20. Let A = {1, 2, 3}, B = {2, 4, 6} and Universal set U = {1, 2, 3, 4, 5, 6}. Verify De-
Morgan’s laws for these sets.
21. If A = {1, 2, 3}, B = {2, 3, 5}, C = {1, 3, 5} and D = {2, 5, 6} then prove that
(A  B)  (C  D) = (A  C)  (B  D)
22. There are 100 students in a class. Out of them 50 bring books and 30 bring books
without khatas in the class. How many students bring both books and khatas in the
class? How many students bring khatas without books? [Answer: 20 students, 50
students]
23. In a survey of the Dhaka city, it was found that 65% of the people watched the
news on BTV, 40% read a newspaper and 25% read a newspaper and watched the
news on BTV. What percent of the people survey neither watched the news on
BTV nor read a newspaper? [Answer: 20%]
24. Out of 400 students of a university, 102 studied Business Mathematics, 110 studied
Management and 152 studied Business Law: of the total 27 studied Management
and Business Law; 36 studied Business Mathematics and Business Law; 18

25
 S. M. Shahidul Islam

studied Business Mathematics and Management; 11 studied all the three subjects.
(i) How many students did not study any subject? (ii) How many students studied
only one subject? [Answer: 106, 235]
25. A town has a total population of 50,000. Out of it 28,000 read Ittefaq and 23,000
read Inclub while 4,000 read both the papers. Determine how people read neither
Ittefaq nor Inclub? [Answer: 3,000]
26. A company studies the product preferences of 20,000 consumers. It was found that
each of the products A, B and C was liked by 7500, 6500 and 5500 respectively
and all the products were liked by 1230. Products A and B were liked by 2500,
products A and C were liked by 2300 and products B and C were liked by 2530.
Prove that the study results are not correct.
27. A class of 60 students appeared for an examination of Mathematics, Statistics and
Economics. 25 students failed in Mathematics, 24 failed in Statistics, 32 failed in
Economics, 9 failed in Mathematics alone, 6 failed in Statistics alone; 5 failed in
Statistics and Economics only and 3 failed in Mathematics and Statistics only.
(i)How many students failed in all three subjects? (ii) How many students passed
in all three subjects? [Answer: (i) 10, (ii) 10]
5p  2
28. Demand function of a product D = and supply function S = p2 + 2; where p
p 3
means the price in taka of the product per unit. Using set theory determine
equilibrium price and quantity. [Answer: p = 4 and quantity = 18]

26
 Progressions

03
Chapter
Progressions (AP & GP)

Highlights:
3.1 Introduction 3.6 Theorem of arithmetic series
3.2 Sequence 3.7 Theorem of geometric series
3.3 Series
3.4 Arithmetic progression (A.P)
3.8 Some worked out examples
3.5 Geometric Progression (G.P), 3.9 Exercise

3.1 Introduction: Arithmetic and geometric series are two special types of series
increasing or decreasing by an absolute quantity or a certain ratio. In this chapter we shall
discuss the methods of finding different terms and summations of the series with
sequences and progressions. We shall also try to show some applications.

3.2 Sequence: A set of numbers that are arranged according to some definite law is called
a sequence. Each number of a sequence is called the term so that we have the first term,
second term and so on. If the nth term of a sequence be un then the sequence is denoted by
{un} or, < un > or, < u1, u2, u3, ..., un, . . . >.
Example: i) The sequence of natural number < 1, 2, 3, ..., n, ... > is denoted by < n >
or, {n}
ii) < n2 > means of sequence < 12, 22, 32, ... > or, < 1, 4, 9, ... >
iii) < (-1) n-1.4n > means the sequence < 4, -16, 64, ... >

3.3 Series: A Series is an expression consisting of the sum of the terms in a sequence. The
sequence < un > forms the series u1 + u2 + u3 + ... + un + .... As for example 1 + 2 + 3 + 4
+ ... is a series.

3.4 Arithmetic progression (A.P): An arithmetic progression is a sequence whose terms

increase or decrease by a constant number called the common difference the sequence <1,
5, 9, 13, 17, . . . > is an infinite arithmetic progression with common difference 4. So, the
series (1 + 5 + 9 + 13 + . . .) is called arithmetic series.

27
 S. M. Shahidul Islam

a + (a +d) + (a +2d) + (a + 3d) + . . . +{a +(n – 1)d}+ . . . is the standard form of arithmetic
series where a, d and n are first term, common difference and number of terms
respectively.

3.5 Geometric Progression (G.P): A geometric progression is a sequence whose terms

increases or decreases by a constant ratio called the common ratio. The sequence < 2, 6,
18, 54, . . . > is an infinite geometric progression whose first term is 2 and the common
ratio is 3. This sequence forms the geometries since 2 + 6 + 18 + 54 + . . .
a  ar  ar 2  ar 3  ar 4  . . .  ar n1  . . . is the standard form of geometric series where
a, r and n are first term, common ratio and number of terms respectively.

3.6 Theorem of arithmetic series: If “a” be the first term and “d” be the common
difference of an arithmetic series, then
1) nth term = a + (n – 1)d and
n
2) sum of first n terms, Sn = 2 {2a + (n – 1) d}
Proof: 1) Given that, first term = a
Common difference = d
So, 2nd term = a + d = a + (2 – 1) d
3rd term = a + d + d = a + 2d = a + (3 – 1)d
4th term = a + 2d + d = a + 3d = a + (4 – 1 )d
Thus, let mth term = a + (m –1)d
So, (m + 1)th term = a + (m –1)d + d
= a + (m – 1 +1) d
= A + {(m + 1) – 1} d
That is, for all n Є N, the nth term = a + (n – 1)d. (Proved)

2) Since S n is the sum of first nth terms of the given series,

S n = a + (a + d) + (a + 2d) + . . . +{a + (n – 2)d}+{a + (n –1)d} . . . (i)
Writing the series in the reverse order, we get
S n = {a + (n –1)d} + {a + (n – 2)d}+ . . . + (a + 2d) + (a + d) + a . . . (ii)
Adding (i) and (ii), we get
2 S n = {2a + (n –1)d} + {2a + (n –1)d} + . . . + {2a +(n –1)d}
Or, 2 S n = n{2a + (n –1)d}
n
So, S n  {2a + (n –1)d} (Proved)
2
n( a  l )
Note: S n  ; where l = nth term = a + (n – 1)d
2

28
 Progressions

3.7 Theorem of geometric series: If “a” be the first term and “r” be the common ratio of
a geometric series, then
1) nth term = a r n 1
a(r n  1)
2) Sum of first n terms, S n  , when r >1
r 1
a(1  r n )
= , when r <1
1 r
Proof: 1) Given that, first term = a and common ratio = d
So, 2nd term = a ×d = ad = a d 21
3rd term = ad ×d = a d 2 = ad 31
4th term = a d 2 × d = ad 3  ad 41
Thus, let mth term = ad m1
So, (m + 1)th term = ad m1 × d
= ad m
= ad ( m1)1
That is, for all n Є N, the nth term = a r n 1 (Proved)
2) Since S n is the sum of first nth terms of the given series,
S n = a  ar  ar 2  . . .  ar n2  ar n1 . . . (i)
Multiplying both sides by r, we get
r S n = ar  ar 2  ar 3  . . .  ar n1  ar n . . . (ii)
Subtracting (ii) from (i), we get
S n – r S n = a  ar n
Or, (1 – r) S n = a(1  r n )
a(1  r n )
So, S n  , r≠1 . . . (iii)
1 r
Changing the signs of the numerator and denominator, we can also write
a(r n  1)
Sn  , r≠1 . . . (iv)
r 1
It is convenient to use form (iii) when r < 1 and form (iv) when r >1. So,
a(r n  1)
Sum of first n terms, S n  , when r >1
r 1
a(1  r n )
= , when r <1
1 r
a
Note: Sum of infinite terms, S   , when | r | < 1
1 r
= not found, when | r | > 1

29
 S. M. Shahidul Islam

3.8 Some worked out examples:

Example (1): Find the sum of the series
62 + 60 + 58 + . . . + 40
Solution: The given series is an arithmetic series with first term, a = 62 and common
difference, d = 60 – 62 = – 2.
Let n be the number of terms. Then nth term = 40
That is, a + (n – 1) d = 40
Or, 62 + (n – 1) (– 2) = 40 [Putting value of a and d]
Or, 62 – 2n + 2 = 40
Or, – 2n + 64 = 40
Or, – 2n = 40 – 64
 24
Or, n =
2
So, n = 12
n
We know, sum of n terms, Sn = 2 {2a + (n – 1)d}
12
So, sum of 12 terms, S12 = 2 {2 . 62 + (12-1) (-2)}
= 6 (124 – 22)
= 6.102
= 612 (Answer)

Example (2): The mth term of an A.P is n and the nth term is m. Show that the rth term is
(m + n – r) and the (m + n)th term is 0. [AUB-2003 B.B.A]
Solution: Let a be the first term and d be the common difference of the given series.
So, n = a + (m –1)d . . . (i)
m = a + (n –1)d . . . (ii)
Doing (i) – (ii), we have
n – m = md – nd
Or, (n – m) = (m – n)d
 ( m  n)
Or, d =
( m  n)
So, d =–1
Substituting the value of d in (i), we get
n = a + (m – 1)( – 1)
Or, n=a–m+1
So, a = m + n –1
Therefore, the rth term = a + (r –1)d
= m + n –1 + (r –1)( –1) [Putting value of a & d]
= m + n – 1– r + 1
=m+n–r

30
 Progressions

And the (m + n)th terms = a + {(m + n ) – 1}d

= m + n –1 + (m + n –1) (–1)
= m + n –1– m – n + 1
=0 (Proved)

Example (3): If a, b, c are the sums of p, q, r, terms respectively of an A.P, show that
a(q  r ) b( r  p c( p  q)
+ + =0
p q r
Solution: Let, 1st term = m and the common difference = n. Then
p
Sum of pth terms, a = 2 {2m + (p –1) n}
q
Sum of qth terms, b = 2 {2m + (q – 1) n}
r
And sum of rth terms, c = 2 {2m + (r – 1) n}
a(q  r ) b( r  p c( p  q)
L.H.S = + +
p q r
p (q  r ) q (r  p) r ( p  q)
= 2 {2m + (p –1)n}× + 2 {2m + (q –1)n}× + 2 {2m + (r-1)n}×
p r r
1
= 2 [{2m + (p –1)n}×(q – r) +{2m + (q –1)n}×(r – p) + {2m + (r –1)n}×(p –q)]
1
= 2 (2qm + pqn – qn – 2rm – prn + rn + 2rm + qrn – rn – 2pm – pqn + pn +
2mp + prn – pn – 2qm – qrn + qn)
1
=2 ×0
= 0 = R.H.S
So, L.H.S = R.H.S (Proved)

Example (4): The first term and the last term of an A.P are respectively –4 and 146, and
the sum of the A.P = 7171. Find the number of terms of the A.P and also its common
difference.
Solution: We have, the first term, a = – 4, the last term, l =146.
Let n be the number of terms of the A.P and d be the common difference.
So, the summation of n terms, Sn = 7171.
n( a  l )
We know, Sn 
2
n(4  146)
Or, 7171 =
2

31
 S. M. Shahidul Islam

n  142
Or, 7171 =  71n
2
Or, 7171 = 71n
7171
Or, n =
71
So, n = 101
Therefore, the number of terms of this A.P is 101. (Answer)
We also know that,
nth term = a + (n –1)d
Here, last term means the 101st term.
So, 146 = – 4 + (101 –1)d
Or, 146 = – 4 +100 d
Or, 100d = 146 + 4
Or, 100d = 150
So, d = 1.5
Therefore, the common difference is 1.5. (Answer)

Example (5): A farmer agrees to repay debt of Tk. 6682 in a number of installments, each
installment increasing the previous one by Tk. 5. If the first installment be of Tk. 1, find
how many installments will be necessary to wipe out the loan completely? [AUB-2000]
Solution: Since every installments (terms) increase by 5 taka, the problem is an A.P of
First installment (term), a = 1 taka
Common difference, d = 5 taka
Let n be the number of installments (terms) to wipe out the loan.
So, summation of n installments means total loan, Sn = 6682 taka
n
We know that, Sn = {2a + (n –1) d}
2
n
Or, 6682 = {2×1 + (n-1) 5}
2
Or, 13364 = n (2 + 5n – 5)
Or, 13364 = n (5n – 3)
Or, 5n² – 3n –13364 = 0
Or, 5n² – 260n + 257n –13364 = 0
Or, 5n(n – 52) + 257(n – 52) = 0
Or, (n – 52) (5n + 257) = 0
So, n – 52 = 0  n = 52
257
Or, 5n + 257 = 0  n=– (Not acceptable)
5
Hence, the number of installments is 52 (Answer)

32
 Progressions

Example (6): A man saved $16500 in ten years. In each year after the first he saved $100
more than he did in the preceding year. How much did he save in the first year? [AUB-
2002 BBA]
Solution: Since each savings increases by $100, it is an A.P of
Common difference, d = $100
Number of terms (years), n = 10
Summation of n terms (total savings), Sn = $16500
First term (savings), a =?
n
We know that, Sn = 2 {2a + ( n –1) d}
10
Or, 16500 = 2 {2a + (10 – 1) 100}
Or, 16500 = 5 {2a + 900}
Or, 10a + 4500 = 16500
Or, 10a = 16500 – 4500
Or, 10a = 12000
12000
Or, a = 10
So, a = 1200
Therefore, he saved $1200 in the first year. (Answer)

Example (7): Sum to n terms of the series: 5 + 55 + 555 + . . . [AUB-1999 BBA]

Solution: Let Sn be the sum of n terms.
So, Sn = 5 + 55 + 555 + . . . + nth term
= 5(1 + 11 + 111 + . . . + nth term)
5
= (9 + 99 + 999 + . . . + nth term)
9
5
= [(10 –1) + (100 –1) + (1000 –1) + . . . + nth term]
9
5
= [(10 +100 +1000 + . . . + nth term) – (1 +1 +1 + . . . + nth term)]
9
5
= [(10 +10² +10³ + . . . + nth term) – n]
9
5 10(10 n  1)
= [  n]
9 10  1
[(10 +10² +10³ + . . . + nth term) is a G.P of 1st term = 10 and common ratio = 10]
5
= [10(10n – 1) – 9n]
81
5
= (10n +1 – 10 – 9n) (Answer)
81

33
 S. M. Shahidul Islam

Example (8): Find the sum of the following series

1 2 4 8
 2  3  4  . . . to 
5 5 5 5
Solution: The series is in G.P form, because
2 4 8
 2  4
5  5  5  ... 2
3

1 2 4 5
2 3
5 5 5
1
Here, 1st term, a =
5
2
 2
2 5 2
Common ratio, r = 5   2   
1 5 1 5
5
2 2
And | r | = |  | = <1, so the series has an infinite summation.
5 5
We know that the infinite summation,
a
S 
1 r
1
= 5
 2
1  
 5
1
= 5
2
1
5
1 5
= 
5 7
1
= (Answer)
7

Example (9): If the value of a flat depreciated by 25% annually, what will be its estimated
value at the end of 5 years if its present value is Tk. 15,00,000? [AUB-2001 BBA]
Solution: It is clear that the value of at the end of first, second, third, fourth and fifth years
from a G.P with common ratio, r = (100 – 25)%
= 75%

34
 Progressions

75
=
100
3
=
4
= 0·75
The value at the end of 5th year = the value at the beginning of 6th year
Here, the value at the beginning of 1st year, a =15,00,000 taka.
So, the value at the beginning of 6th year = a r 61
= a r5
= 15,00,000 (0·75)5 taka
=15,00,000 × 0.2373046 taka
=3,55,957·03 taka.
Therefore, the estimated value at the of 5th year is 3,55,957·03 taka. (Answer)

Example (10): If the population of a town increases 2.5% per year and the present
population is 26,24,000, what will be the population in three years time? What was it a
year ago?
Solution: Since, the population increases by the percentage (ratio) 2.5%, it is a G.P of
The first term (present population), a = 26,24,000
Common ratio, r = (100 + 2.5)%
= 102.5%
102.5

100
 1.025
The population in three years = the population at the beginning of 4th year.
We know that nth term = a r n 1
The population will be in three years time = a r 41 [4th term = a r 41 ]
= a r3
 26,24,000  (1.025) 3
= 2,82,55,761 (Answer)
01
The population of a year ago = a r
= a r 1
 26,24,000  (1.025) 1
= 25,60,000 (Answer)

35
 S. M. Shahidul Islam

3.9 Exercises:
1. Define with examples: (i) sequence, (ii) series
2. Discuss the differences of series and sequence.
3. What is difference between arithmetic and geometric series?
4. Find the sum of the following series:
(i) 7 + 10 + 13 + . . . up to 40th term. [Answer: 2620]
1 1
(ii) 5  7  10  12  . . .  25th term. [Answer: 875]
2 2
5. How many natural numbers within 1 and 1000, which are divisible by 5? And find
their sum. [Answer: 200 and 100500]
6. Find the sum of natural numbers from 1 to 200 excluding those divisible by 5.
[Hints: Sum = (1+2+3+ . . . +200) – (5+10+15+ . . . +200)] [Answer: 16000]
7. The 7th and the 9th terms of an A.P are respectively 15 and 27. Find the series of
the A.P and the sum of first 50 terms. [Answer: 3 + 6 + 9 +12 + . . . and 3825]
8. The first term of an A.P series is 2, the nth term is 32 and the sum of first n terms
is 119. Find the series. [Answer: 2 + 7 + 12 + 17 + . . .]
9. The sum of the first n terms of the A.P series 13 + 16.5 + 20 + . . . is the same as
the sum of the first n terms of the A.P series 3 + 7 + 11 + . . .. Calculate the value
of n. [Answer: 41]
10. A man secures an interest free loan of $14500 from a friend and agrees to repay it
in 10 installments. He pays $1000 as first installment and then increases each
installment by equal amount over the preceding installment. What will be his last
installment? [Answer: $1900]
11. A firm produced 500 sets of close circuit TV during its first year and increased its
production each year uniformly. The total sum of the productions of the firm at the
end of 5 years operation was 4500 sets. (i) Estimate by how many units, production
increased each year. (ii) How many TV sets were produced for the 9th year?
[Answer: (i) 200, (ii) 2100]
12. The price of each shares of 100.00 taka of a company increases 10 taka every year.
A man buys some primary shares. After 10 years, the price of his shares will
become 5700.00 taka. What is the price of his primary shares? What will the price
of his shares be after 5 years? [Hints: Let, no. of shares = x, so, 1st term = 100x
taka and c.d = 10x taka. So, 100x + (10 – 1)10x = 5700 => x = 30] [Answer:
3000taka and 4200 taka]
13. Find the 10th term and sum of first 10 terms of the following series:
1+ 3 + 9 + 27 + . . . [Answer: 19683 and 29524]
14. If the sum of n terms of a G.P series is 225, the common ratio is 2 and the last term
(nth term) is 128. Find the value of n. [Answer: 8]

36
 Progressions

15. Find the sum of the following series:

4
(i) 4 + 44 + 444 + . . . + nth term. [Answer: (10n +1 – 10 – 9n)]
81
7
(ii) 7 + 77 + 777 + . . . + nth term. [Answer: (10n +1 – 10 – 9n)]
81
16. Sum the following series to infinity:
1 1 1
(i) 1     . . . [Answer: 2]
2 4 8
4 5 4 5 23
(ii)  2  3  4  . . . [Answer: ]
7 7 7 7 48
4 4 1 4 4 1
[Hints =  ( 2  2 )  4  ( 4  4 )  . . .
7 7 7 7 7 7
4 4 4 1 1 1
= (  2  3  . . . )  ( 2  4  6  . . . )]
7 7 7 7 7 7
17. The population of Dinajpur increases 4% per year. How long time is required to be
the double people in that town? [Answer: 18.67 years]
18. The cost price of a machine is $50000. If the price of this machine depreciated by
10% annually, what will be its estimated price at the end of 7 years? [Answer:
$23914.85]

37
 S. M. Shahidul Islam

04
Chapter
Permutation and Combination
Highlights:
4.1 Introduction 4.7 Combination
4.2 Permutation 4.7.1 Theorem
4.3 Factorial notation 4.8 Combinations of n different things
4.4 Permutations of n different things taken some or all at a time
4.5 Permutations of n things not all 4.9 Combinations of n things not all
different different taken some or all at a time
4.6 Circular permutation 4.10 Some worked out examples
4.11 Exercise

4.1 Introduction: Permutations mean the different arrangements of things from a given
lot taken one or more at a time whereas combinations refer to different sets or groups
made out of a given lot, without repeating any element at a time. The difference between
permutation and combination will be clear by the following illustration of permutations
and combinations made out of a lot of three elements, such as x, y, z.

Permutations Combinations
(i) one at a time: {x}, {y}, {z} {x}, {y}, {z}
(ii) two at a time: {x, y}, {y, x}, {x, z}, {x, y}, {x, z}, {y, z}
{z, x}, {y, z}, {z, y}
(iii) three at a time: {x, y, z}, {x, z, y}, {x, y, z}
{y, x, z}, {y, z, x},
{z, x, y}, {z, y, x}
In the above example we see that every set represents different combination but a set with
different arrangements of its elements represents different permutation. And no element
appears twice in the sets of permutations or combinations such as {x, x}, {y, y} and {z, z}.

4.2 Permutation: Permutations mean the different arrangements of things from a given
lot taken one or more at a time without repetition of any object. Let us consider three
letters a, b, c. The different arrangements of these three letters taking three at a time are
abc, acb, bac, bca, cab and cba. Thus there are 6 different ways of arranging three distinct
objects when each arrangement is of all the three objects without any repetition of objects.

38
 Permutation and Combination

Each of these arrangements is a permutation. We can illustrate the example as follows:

there are three places to be filled, the first can be filled in 3 ways, the second in 2 ways
while for the third in 1 way. Hence, there are 3.2.1 = 6 ways to arrange the 3 letters at a
time.

4.3 Factorial notation: The product of the first n natural numbers, viz., 1, 2, 3, ..., n, is
called factorial n or n factorial and is written as n or n!.
Thus n! = 1  2  3  . . .  (n – 1)  n
It follows that n! = n  {(n – 1)!}
= n  (n – 1)  {(n – 2)!}
... ... ...
= n(n –1)(n – 2) ... (n – r +1){(n – r)!}
6! 6.5.4!
Illustration: = = 6.5 = 30
4! 4!

4.4 Permutations of n different things: Let the number of permutations of n different

things taken r at a time, where r  n is n p r or P(n, r).
The number of permutations of n different things taken r at a time is the same as the
number of different ways in which r places can be filled up by n things.
The first place can be filled up in n ways, for any one of the n things can be put in it.
 n
p1 = n
When the first place has been filled up in any one of the n ways, the second place can be
filled up in (n – 1) different ways, for any one of the remaining (n – 1) things can be put in
it. Since each way of filling up the first place is associated with each way of filling up the
second place, the first two places can be filled up in n(n-1) ways.
 n
p 2 = n(n – 1)
When the first two places have been filled up in any one of the n(n –1) ways, the third
place can be filled up in (n – 2) different ways, for any one of the remaining (n – 2) things
can be put in it. Since the first two places are associated with the third place, the first three
places can be filled up in n(n – 1)(n – 2) ways.
 n
p3 = n(n – 1)(n – 2)
Proceeding in the same way we notice that the number of factors is same as the number of
places to be filled up and each factor is less then the former by 1. So, the r places can be
filled up in n(n – 1)(n – 2) ... (n – r + 1) ways.
That is, n
p r = n(n – 1)(n – 2) ... (n – r + 1)

Remarks:
1. n p r = n(n – 1)(n – 2) ... (n – r + 1)

39
 S. M. Shahidul Islam

n(n  1)(n  2)...(n  r  1){(n  r )!}

=
(n  r )!
n!
=
(n - r)!

n!
So, n p r 
(n - r)!

2. The number of permutations of n different things taken all at a time is

n
p n = n(n – 1)(n – 2) ... 3.2.1 = n!.
n! n!
3. n p n =  , but n p n = n!
(n - n)! 0!
So, 0! = 1.
According to the definition, 0! is meaningless. But when we use it as a symbol its
value is 1.
4. The number of permutations of n different things taken r things at a time excluding
always a particular object is n 1 p r .
5. The number of permutations of n different things taken r objects at a time
including always a particular object is r. n1 p r 1 . (Restricted permutation)
Illustration: Since a particular object belongs to every permutation, one place of
the r places can be filled up in r ways by the particular object. So, the remaining
(r – 1) places can be filled up in n1 p r 1 ways by the remaining (n – 1) objects.
Therefore, the r places can be filled up in r. n1 p r 1 ways. (Restricted permutation)
6. The number of permutations of n different things taken r things at a time in which
each object is repeated r times in any permutation is nr.
Illustration: The first place can be filled up in n ways, for any one of the n things
can be put in it. Since the object that filled up the first place is repeatable, the
second place can be filled up in n ways. Thus, the first two places can be filled up
in n  n = n2 ways. Similarly, the third place can be filled up in n ways and the first
three places can be filled up in n  n2 = n3 ways. Proceeding in the same way we
notice that the exponent of n is same as the number of places to be filled up. So,
the r places can be filled up in nr ways.

Example: Find how many three-letter words can be formed out with the letters of the
word EQUATIONS (the words may not have any meaning).
Solution: There are 9 different letters, therefore, n is equal to 9 and since we have to find
three-letter words, r is 3. Hence the required number of words is

40
 Permutation and Combination

9 9!
p3 =
(9 - 3)!
9.8.7.(6!)
=
6!
= 9.8.7 = 504

Example: Indicate how many 4-digit numbers greater than 8000 can be formed from the
digits 5, 6, 7, 8, 9.
Solution: If the digits are to be greater than 8000, then the first digit must be any one of
the 8 and 9. Now the first digit can be chosen in 2 p1 = 2 ways and the remaining three
digits can be any of the four digits left, which can be chosen in 4 p3 ways. Therefore, the
total number of ways
= 2  4 p3 = 2  4  3  2 = 48

Example: Six papers are set in an examination, of which two are Mathematics. In how
many different orders can the papers he arranged so that the two statistic papers are not
together? [RU-89]
Solution: The total number of arrangements that can be made of 6 papers is 6!.
Now let the mathematics papers be taken together. These taken as one and the remaining 4
can be arranged amongst themselves in 5! ways. The mathematics papers can be arranged
between them in 2! ways.
 The total number of arrangements in which the mathematics papers can come together
is 5!  2!.
 The number of arrangements in which the two particular papers are not together is
6! – 5!  2! = 720 – 240 = 480

Example: Find the numbers less than 1000 and divisible by 5 which can be formed with
digits 0, 1, 2, 3, 4, 5, 6, 7 such that each digit does not occur more than once in each
number.
Solution: The required numbers may be of one digit, two digits or three digits and each of
them must end in 5 or 0, except the number of one digit which must end with 5.
The number of one digit ending in 5 is 1.
The number of two digits ending in 5 is 7 p1 – 1
(Since, the number having 0 as the first position is to be rejected)
The number of two digits ending in 0 is 7 p1
The number of three digits ending in 5 is 7 p 2 – 6 p1
(Since, the numbers having 0 as the first position are 6 p1 )
The number of three digits ending in 0 is 7 p 2

41
 S. M. Shahidul Islam

Hence, the total number of required numbers is 1 + ( 7 p1 – 1) + 7 p1 + ( 7 p 2 – 6 p1 ) + 7 p 2

= 1 + 7 – 1 + 7 + 42 – 6 + 42
= 92

4.5 Permutations of n things not all different: Let us consider three letters a, b, c. The
different permutations (arrangements) of these three letters taking three at a time are abc,
acb, bac, bca, cab and cba. Thus there are 3! = 6 different ways of arranging three distinct
objects taking all at a time. If we change c by b we get three letters a, b, b but two of them
are each similar to b and the arrangements will be abb, abb, bab, bba, bab and bba in
which three arrangements are same to other three arrangements. So, the actual
3!
permutations will be abb, bab and bba and the number of permutations is 3 = .
2!
So, by the above discussion we can say that, the number of permutations taking all at a
time of n things of which p things are of one kind, q things are of a second kind, r things
are of a third kind and all the rest are different is given by
n!
p!  q!  r!

Example: How many numbers greater than 2000000 can be formed with the digits 4, 6, 6,
0, 4, 6, 3?
Solution: Each number must consist of 7 or more digits. There are 7 digits in all, of which
there are 2 fours, 3 sixes and the remaining numbers are different.
7!
 The total numbers are 2 ! 3! = 420
Of these numbers, some begin with zero and are less than one million that must be
rejected.
6!
The numbers beginning with zero are 2 ! 3! = 60
 The required numbers are 420 – 60 = 360

Example: (a) Find the number of permutations of the word PERMUTATION.

(b) Find the number of permutations of letters in the word COMBINATION.
Solution: (a) The word PERMUTATION has 11 letters, of which 2 are Ts and the rest are
different. Therefore, the number of permutations is
11! 11.10.9.8.7.6.5.4.3.2.1
  19958400
2! 2
(b) Since the word COMBINATION consists of 11 letters, in which there are 2 Os, 2 Is, 2
Ns and the remaining letters are different, the total number of permutations is
11! 11.10.9.8.7.6.5.4.3.2.1
  4989600
2!.2!.2! 2.2.2

42
 Permutation and Combination

Example: (i) How many different words can be made out with the letters in the word
ALLAHABAD? (ii) In how many of these will the vowels occupy the even places?
[AUB-01]
Solutions: (i) The word ALLAHABAD consists of 9 letters of which A is repeated four
times, L is repeated twice and the rest all are different.
9!
4 !2! = 7560
(ii) Since the word ALLAHABAD consists of 9 letters, there are 4 even places that can be
filled up by the 4 vowels in 1 way only, since all the vowels are similar. Further, the
remaining 5 places can be filled up by the 5 consonants of which two are similar which
5! 5!
can be filled in 2! ways. Hence the required number of arrangements is 1 2! = 60.

4.6 Circular permutations: The circular permutations are related with arrangement of
objects as in the case of a sitting arrangement of members in a round table conference. In
this case, the arrangement does not change unless the order changes. Let us consider the
following two arrangements of 4 members:
1 3

4 2
2 4

3 1

Figure 4.1

In the above figure we see that they have changed their positions but both are same
permutation 1234 because their order is not changed.
So, in the circular permutation, the relative position of the other objects depend on the
position of the objects placed first. A new permutation is made by the arrangement of the
remaining objects. Thus, the number of circular arrangements of n objects will be (n–1)!
but not n!. Therefore, the circular arrangement of 4 boys will be in 3! = 3.2.1 = 6 ways.
In the circular permutation, the clockwise and anticlockwise arrangements do not make
any difference. If the neighborhood of one or more is restricted, the arrangement will get
restricted to that extent. If the restriction is that no two similar objects are close to each
other then the number of permutations will be ½{(n – 1)!}. For example if 5 boys are
seated around a table so that all of the permutations have not the same neighbors, then the
required number of permutations will be ½{(n – 1)!} or ½(4.3.2.1) = 12.

43
 S. M. Shahidul Islam

Example: In how many ways can 6 science students and 6 arts students be seated around a
table so those no 2 science students are adjacent?
Solution: Let the arts students be seated first. They can sit in 5! ways according to the rule
indicated above. Now since the seats for the science students in between arts students are
fixed. The option is there for the science students to occupy the remaining 6 seats, there
are 6! ways for the science students to fill up the 6 seats in between 6 arts students seated
around a table already. Thus, the total number of ways in which both arts and science
students can be seated such that no 2 science students are adjacent are 5!  6! = 86400
ways.

4.7 Combination: Combinations refer to different sets or groups made out of a given lot
without repeating any object, taking one or more of them at a time neglecting the order. In
other words each of the groups which can be made out of n things taking r at a time
without repeating and regarding the order of things in each group is termed as
n
combination. It is denoted by nCr or C(n, r) or   .
r 

4.7.1 Theorem: The number of combinations of n different things taken r at a time are
given by
n n!
Cr = where (0 ≤ r ≤ n)
r!(n  r )!,
Proof: Let nCr denote the number of combinations of n different things taken r at a time.
So, each of these combinations has r different things.
 If the r different things be arranged among themselves in all possible ways, each
combination would produce r! permutations.
 nCr combinations would produce nCr  r! permutations. But this number is clearly
equal to the number of permutations of n different things taken r things at a time.
Hence n
Cr  r! = nPr
n!
Or, n
Cr  r! =
(n - r)!
n n!
Or, Cr =
r! (n - r)!
n!
So, n
Cr 
r! (n  r )!

Example: Find the value of 10C 7 .

10! 10.9.8.7!
Solution: 10
C7  = = 120
7! (10  7)! 7!.3!

44
 Permutation and Combination

Example: 11 questions are set in the questions paper of Business Mathematics in a year
final examination. In how many different ways can an examinee choose 7 questions?
Solution: The number of different choices is evidently equal to the number of
combinations of 11 different things taken 7 at a time.
11! 11.10.9.8.7!
 the required number of ways = 11C7 =   330
7!.4! 7!.4.3.2.1
n +1
Example: Show that Cr = nCr + nCr –1.
n!
Solution: We know that n C r 
r! (n  r )!
n n
R.H.S = Cr + Cr –1.
n! n!
= 
r! (n  r )! (r - 1)! (n - r  1)!
n! n!
= 
r (r  1)! (n  r )! (r - 1)! (n - r  1) (n - r)!
n! 1 1 

(r  1)! (n  r )!  r (n - r  1) 

=

n!  n 1 
(r  1)! (n  r )!  r(n - r  1) 
=

(n  1)!
=
r! (n  r  1)!
(n  1)!
=
r! (n  1  r )!
n +1
= Cr = L.H.S

Example: A cricket team of 11 players is to be formed from 15 players including 3

bowlers and 2 wicket keepers. In how many different ways can a team be formed so that
the team contains (i) exactly 2 bowlers and 1 wicket keeper, (ii) at least 2 bowlers and at
least 1 wicket keeper.
Solution: (i) A cricket team of 11 players is exactly to contain 2 bowlers and 1 wicket
keeper.
2 bowlers can be selected out of 3 in 3C2 = 3 ways.
1 wicket keeper can be selected out of 2 in 2C1 = 2 ways.
The remaining 8 player can be selected out of remaining 10 players in 10C8 = 45 ways.
So, the total number of ways in which the team can be formed = 3  2  45 = 270.
(ii) A cricket team of 11 players of which at least 2 bowlers and 1 wicket keeper can be
formed in the following ways:

45
 S. M. Shahidul Islam

(a) 2 bowlers, 1 wicket keeper and 8 other players.

(b) 2 bowlers, 2 wicket keeper and 7 other players.
(c) 3 bowlers, 1 wicket keeper and 7 other players.
(d) 3 bowlers, 2 wicket keeper and 6 other players.
We now consider the above 4 cases.
(a) 2 bowlers, 1 wicket keeper and 8 other players can be selected in
3
C2  2C1  10C8 = 3  2  45 = 270 ways.
(b) 2 bowlers, 2 wicket keeper and 7 other players can be selected in
3
C2  2C2  10C7 = 3  1  120 = 360 ways.
(c) 3 bowlers, 1 wicket keeper and 7 other players can be selected in
3
C3  2C1  10C7 = 1  2  120 = 240 ways.
(d) 3 bowlers, 2 wicket keeper and 6 other players can be selected in
3
C3  2C2  10C6 = 1  1  210 = 210 ways.
Therefore, the total number of different ways = 270 + 360 + 240 + 210 = 1080.

4.8 Combinations of n different things taken some or all at a time:

The first thing can be dealt within 2 ways, for it may either be left or taken.
The second thing can also be dealt within 2 ways, for it may either be left or taken.
Since each way of dealing with the first must be associated with the ways of dealing with
the second thing, the total number of ways of dealing with these two things is 2  2 = 22 .
The third thing can also be dealt within 2 ways, for it may either be left or taken.
So, the total number of ways of dealing with the first three things is 2  2  2 = 23.
Proceeding in this way, the total number of ways for n different things = 2 n. But this
number includes one case in which none of the things are taken.
Therefore, the number of combinations of n different things taken some or all at a time is
2n – 1.

Example: 33% marks have to be secured in each of the 10 subjects in order to pass the
S.S.C examination. In how many ways can a student fail?
Solution: Each subject can be dealt within 2 ways, one the student pass in it other he fail
in it. So, the 10 subjects can be dealt within 210 = 1024 ways. But this number includes
the case in which the student passes in all subjects. Excluding this case, the number of
ways to fail the student is 1024 – 1 = 1023.

4.9 Combinations of n things not all different taken some or all at a time : Let us
consider n things of which p things are of one kind, q things are of a second kind, r things
are of a third kind and all the rest are different. Here, p + q + r ≤ n.
Consider the first kind things. The p things can be dealt in (p + 1) ways, for we may take 1
thing or 2 things or 3 things or . . . or p things or none in any selection. Similarly, the q
like things can be dealt in (q + 1) ways and r like things in (r + 1) ways. Since each way of
any kind things must be associated with the ways of other kind things, the number of ways
is (p + 1)(q + 1)(r + 1).

46
 Permutation and Combination

The remaining {n – (p + q + r)} things are different. So, these things can be dealt in
2n–p–q–r ways.
Since every thing is associated to each other, the total number of ways
= (p + 1)(q + 1)(r + 1)( 2n–p–q–r) .
But this number includes the case in which all things are left.
Therefore, the total number of combinations of n things not all different taken some or all
at a time = (p + 1)(q + 1)(r + 1)( 2n–p–q–r) – 1.

Example: Let there are 3 Econo pens, 4 2B pencils, 1 Business Math. book and 1 CD.
Find the number of combinations in which at least one thing is present.
Solution: The 3 Econo pens can be dealt in (3 + 1) = 4 ways, for we may take 1 pen
or 2 pens or 3 pens or none in any selection. Similarly, the 4 2B pencils can be dealt in
(4 + 1) = 5 ways. So, the pens and the pencils can be dealt in 4  5 = 20 ways.
Business Math. book can be dealt within 2 ways, for taken or left. Similarly, the CD can
be dealt within 2 ways, for taken or left. So, the book and the CD can be dealt within
2  2 = 22 = 4 ways.
Since every things is associated with others, the total number of ways = 20  4 = 80.
But this number includes the case in which all things are left. Therefore, the required
number of combinations = 80 – 1 = 79. (Answer)
Another way: Using the above formula, we can find the number of combinations very
easily as (3 + 1)(4 + 1)(22) – 1 = 4  5  4 – 1 = 79.

4.9 Some worked out examples:

Example (1): Find how many three-letter words can be formed out with the letters of the
word LOGARITHM (the words may not have any meaning).
Solution: There are 9 different letters, therefore, n is equal to 9 and since we have to find
three-letter words, r is 3. Hence the required number of words is
9 9!
p3 =
(9 - 3)!
9.8.7.(6!)
=
6!
= 9.8.7 = 504

Example (2): In how many ways can 5 Bengali 3 English and 3 Arabic books he arranged
if the books of each different language are kept together.
Solution: The each language book amongst themselves can be arranged in the following
ways:
Bengali : 5 books in 5p5 = 5! ways
English : 3 books in 3p3 = 3! ways
Arabic : 3 books in 3p3 = 3! ways

47
 S. M. Shahidul Islam

Also arrangement of these groups can be made in 3p3 = 3! ways, hence by the fundamental
theorem, the required arrangements are
5!  3!  3!  3! = 25920

Example (3): Show that n

Pr = n  n –1Pr –1
Solution: R.H.S = n  n –1Pr –1
(n  1)!
=n
{(n  1)  (r  1)}!
n!
= = nPr = L.H.S
(n  r )!

Example (4): How many arrangements can be made with the letters of the word
MATHEMATICS and in how many of them vowels occur together? [RU-88]
Solution: The word MATHEMATICS consists of 11 letters of which 2 are As, 2 Ms, 2 Ts
and the rest all different.
11!
 The total number of arrangements are = 4989600
2!  2!  2!
The word MATHEMATICS consists of 4 vowels A, A, E and I (two are similar). To find
the number of arrangements in which the four vowels occur together, consider the four
vowels as tied together and forming one letter. Thus we are left with 8 letters of which 2
are Ms, 2 are Ts, 1 is H, 1 is C, 1 is S and the vowels as 1 letter. These letters can be
8!
permuted in  10080 ways. The 4 vowels that are tied together can again be
2!.2!
4!
permuted among themselves in  12 ways (since two of the vowels are similar). Hence
2!
the total number of arrangements are 10080  12 = 120960.

Example (5): In how many ways can the letters of word. 'ARRANGE" be arranged? How
many of these arrangements are there in which
(i) the two Rs come together,
(ii) the two Rs do not come together,
(iii) the two Rs and the two As come together ? [AUB–02]
Solution: The word ARRANGE consists of 7 letters of which two are As, two are Rs and
7!
the rest all different. Hence they can be arranged amongst themselves in  1260 ways
2!.2!
(i) The number of arrangements in which the two Rs come together can be obtained by
treating the two Rs as one letter. Thus there are 6 letters of which two (the two As) are
6!
similar and so the total number of arrangements =  360 .
2!.

48
 Permutation and Combination

(ii) The number of arrangements in which the two Rs do not come together can be
obtained by subtracting from the total number of arrangements, the arrangements in which
the two Rs come together. Thus the required number is 1260 – 360 = 900.

(iii) The number of arrangements in which the two Rs and the two As come together can
be obtained by treating the two Rs and the two As a single letter. Thus there are 5 letters
that all are different and so the number of arrangements is 5! = 120.

Example (6): In how many ways can 5 Bangladeshis and 5 Pakistanis be seated at a round
table so that no two Pakistanis may be together?
Solution: First we let one of the Pakistani in a fixed seat and then the remaining 4
Pakistanis arrange their seats as in the 4! ways. After they have taken their seats in any
way, there are five seats for the Bangladeshis each between two Pakistanis. Therefore, the
Bangladeshis can be seated in 5! ways.
 Total number of circular permutations is 4!  5! = 2880.

Example (7): Find the value of n when 4  n p3 = 5  n1

p3
Solution: We are given that 4  p3 = 5 
n n1
p3
 4  n (n – 1)(n – 2) = 5  (n – 1)(n – 2)(n – 3)
[Dividing throughout by (n-l)(n-2), we get]
 4n = 5(n – 3)
 4n = 5n – 15
 – n = – 15
 n = 15

Example (8): In how many different ways can 7 examination papers be arranged in a line
so that the best and worst papers are never together? [DU-85]

Solution: The total number of arrangements that can be made of 7 papers is 7!.
Now let the best and the worst papers be taken together. These taken as one and the
remaining 5 can be arranged amongst themselves in 6! ways. The best and the worst
papers can be arranged between them in 2! ways.
 The total number of arrangements in which the best and the worst papers can come
together is 6!  2!.
 The number of arrangements in which the two particular papers are not together is
7! – 6!  2! = 5040 – 1440 = 3600

Example (9): In how many ways can 5 white and 4 black balls be selected from a box
containing 20 white and 16 black balls.
Solution: This is a problem of combinations.

49
 S. M. Shahidul Islam

5 out of 20 white balls can be selected in

20! 20  19  18  17  16  15!
20
C5 =   15504 ways.
5!  (20 - 5)! 5  4  3  2  1  15!
16  15  14  13
4 out of 16 black balls can be selected in 16C4 =  1820 ways.
4  3  2 1
So, the required number of combinations = 15504  1820 = 28217280.

Example (10): A question paper contains 5 questions, each having an alternative. In how
many ways can an examinee answer one or more questions ? [AUB-02]
Solution: The first question can be dealt with in 3 ways, for the question itself may be
answered, or its alternative may be answered or none of them may be answered.
Similarly, the second question also can be dealt with in 3 ways. Hence, the first two
questions can be dealt with in 3  3 or 32 ways. Proceeding in this way, all the 5 questions
may be dealt with in 35 ways.
But this number includes one case in which none of the questions is answered.
 The required number of ways = 35-1 = 242.

Example (11): A committee of 4 boys and 3 girls is to be formed from 7 boys and 8 girls.
In how many different ways can the committee be formed if boy-1 and girl-1 refuse to
attend the same committee. [AUB-00]
Solution: 3 girls can be selected out of 8 girls in 8C3 ways and 4 boys can be selected out
of 7 boys in 7C4 ways.
So, the number of ways of choosing the committee is
8! 7!
8
C3  7C4 =   1960
3! 5! 4! 3!
If both boy-1 and girl-1 are members then there remain to be selected 2 girls out of 7 girls
and 3 boys from 6 boys. It can be done by
7! 6!
7
C2  6C3 =   420 ways.
2! 5! 3! 3!
Therefore, the number of ways of forming the committee in which boy-1 and girl-1 are not
present together is (1960 – 420) = 1540.

4.10 Exercise:
1. Define permutations and combinations. Illustrate with examples.
2. Distinguish between permutations and combinations.
3. What do you mean by circular permutation?
4. Find the values of (i) 10P4 (ii) 15P4 [Answer: (i) 5040 (ii) 32760]
5. Find the number of permutations of the word ACCOUNTANT. [Answer: 226800]
6. Prove that n
pn  n pn1 .
7. Find the value of r if 7Pr = 60.7Pr – 3. [Answer: 3]

50
 Permutation and Combination

8. Find the number of permutations of letters in the word ENGINEERING. [Answer:

1108800]
9. Find the numbers less than 1000 and divisible by 5 which can be formed with
digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 such that each digit does not occur more than once
in each number. [Answer: 154]
10. How many different numbers of 3 digits can be formed from the digits 1, 2, 3, 4, 5
and 6, if no digits may be repeated? If repetitions are allowed? [Answer: 120, 216]
11. Find how many words can be formed with the letters of the word ‘FAILURE’, the
four vowels always coming together. [Answer: 576]
12. A telegraph post has five arms and each arm is capable of 4 distinct positions
including the positions of rest. What is the total number of signals that can be
made? [Answer: 1023]
13. A library has 7 copies of one book, 4 copies of each of two books, 6 copies of each
of three books and single copies of 9 books. In how many ways can all the books
42!
be arranged? [Answer: ]
7!(4!) 2 (6!) 3
14. Find the values of (i) 7C2 (ii) 20C4 [Answer: (i) 21 (ii) 4845]
15. What is the value of r for 18Cr = 18Cr + 2 ?
16. Prove that the number of combinations of n different things taken r at a time is
equal to the number of combinations of n different things taken (n – r) at a time,
that is nCr = nCn – r , where 0 ≤ r ≤ n. (Complementary combination)
17. Prove that n
Cr + n – 1Cr – 1 + n – 1Cr – 2 = n +1Cr.
18. From 6 boys and 4 girls a committee of 6 is to formed. In how many ways can this
be done if the committee contains (i) exactly 2 girls (ii) at least 2 girls. [Answer: (i)
90 (ii) 185]
19. A cricket team consisting of 11 players is to be formed from 16 players of whom 6
persons are bowlers. In how many different ways can a team be formed so that the
teams contain at least 4 bowlers? [Answer: 3096]
20. A cricket team consisting of 11 players is to be formed from 16 players of whom 4
can be bowlers and 2 can keep wicket and the rest can neither bowler nor keep
wicket. In how many different ways can a team be formed so that the teams
contain (i) exactly 3 bowlers and 1 wicket keeper, (ii) at least 3 bowlers and at
least 1 wicket keeper. [Answer: (i) 960, (ii) 2472]
21. 36% marks have to be secured in each of the 11 subjects in order to pass the B.Sc
examination. In how many ways can a student fail? [Answer: 2047]
22. Let you have 5 Winner classic pens, 7 6B pencils, 4 Business Math. book and 1
calculator. Find the number of combinations in which at least one thing is present.
[Answer: 479]
23. A party of 6 is to be formed from 7 boys and 10 girls so as to include 3 boys and 3
girls. In how many ways can the party be formed if 2 particular boys refuse to join
the same party? [Answer: 3600]

51
 S. M. Shahidul Islam

05
Chapter
Determinant and Matrix

Highlights:
5.1 Introduction 5.9 Types of matrices
5.2 Definition of determinant 5.10 Matrices operations
5.3 Value of the determinant 5.11 Process of finding inverse matrix
5.4 Minors and co-factors 5.12 Rank of a matrix
5.5 Fundamental properties of determinant 5.13 Use of matrix to solve the system of
5.6 Multiplication of two determinants linear equations
5.7 Application of determinants 5.14 Some worked out examples
5.8 Definition of matrix 5.15 Exercise

5.1 Introduction: The working knowledge of determinants and matrices is a basic

necessity for the students of Mathematics, Business Mathematics, Physics, Statistics,
Economics and Engineering. In 1683 a Japanese mathematician Kiowa first devised the
idea as well as the notation of determinants and J. J. Sylvester was the first man who
introduced the word ‘matrix’ in 1850. Gabriel Cramer successfully applied determinants in
solving systems of linear equations in 1750. At present matrix is a powerful tool of
modern mathematics.

5.2 Definition of determinant: A determinant of order n is a square array of n 2 quantities

aij (i, j = 1, 2, 3, ..., n) enclosed between two vertical bars and is generally written in the
form given below:
a11 a12 ... a1n
a 21 a 22 ... a 2 n
D=
... ... ... ...
a n1 a n 2 ... a nn
which is an ordinary number.
The n 2 quantities aij (i, j = 1, 2, 3, ..., n) are called the elements of the determinant. The
vertical lines of elements are known as columns whereas the horizontal lines of elements
are known as rows. Here, aij represents the element of the i-th row and j-th column of the
determinant.
52
 Determinant and Matrix

5.3 Value of the determinant: The value of a 2  2 determinant

a1 b1
is a1b2 – a2b1 and its Sarrus diagram is a1 b1
a 2 b2
a2 b2
Figure 5.1
a1 b1 c1
The value of a 3  3 determinant a 2 b2 c 2 is
a3 b3 c3
a1(b2c3 – b3c2) – b1(a2c3 – a3c2) + c1(a2b3 – a3b2)
and its Sarrus diagram is as follows:
a1 b1 c1 a1 b1

a2 b2 c2 a2 b2

a3 b3 c3 a3 b3
Figure 5.2

To find the value of a determinant we use Sarrus diagram. We multiply the elements
joined by arrows. Arrows downwards denote positive sign with the corresponding
expression and arrows upwards denote negative sign with the corresponding expression.
So, from figure 5.2 we find the value of above 3  3 determinant as follows:
a1b2c3 + b1c2a3 + c1a2b3 – a3b2c1 – b3c2a1 – c3a2b1
Or, a1(b2c3 – b3c2) – b1(a2c3 – a3c2) + c1(a2b3 – a3b2) which is same as above.
1 2 3
Example: Using Sarrus diagram find the value of 4 5 6 .
7 8 9
Solution: Sarrus diagram of the given determinant as follows:
1 2 3 1 2

4 5 6 4 5

7 8 9 7 8
Figure 5.3
1 2 3
From the above Sarrus diagram, we get 4 5 6 = 1.5.9+2.6.7+3.4.8–7.5.3–8.6.1–9.4.2
7 8 9
= 45+84+96-105-48-72 = 0 (Answer)
53
 S. M. Shahidul Islam

5.4 Minors and co-factors: If the determinant

a11 a12 ... a1n
a 21 a 22 ... a 2 n
D=
... ... ... ...
a n1 a n 2 ... a nn
Deleting the i-th row and j-th column we will get a new determinant of (n – 1) rows and
(n – 1) columns. This new determinant is called the minor of the element aij and is denoted
by Mij. The minor Mij multiplied by (-1)i+j is called the co-factor of the element aij and is
denoted by Aij. So, Aij = (-1)i+j Mij.
The determinant D is equal to the sum of the products of the elements of any row or
column and their respective co-factors. That is,
D = a11A11 + a12A12 + ... + a1nA1n
As for example let us consider 3  3 determinant
a1 b1 c1
b c2 a c2
D = a 2 b2 c 2 . So the co-factor of a1 is 2 , the co-factor of b1 is (–) 2 and
b3 c3 a 3 c3
a3 b3 c3
a2 b2
the co-factor of c1 is
a3 b3
b2 c2 a2 c2 a2 b2
Thus, D = a1 + b1 (–) + c1
b3 c3 a 3 c3 a3 b3
= a1(b2c3 – b3c2) – b1(a2c3 – a3c2) + c1(a2b3 – a3b2)

5.5 Fundamental properties of determinants: The followings are fundamental

properties of determinants:
1. If all the elements in a row (or in a column) of a determinant are zero, the
determinant is equal to zero.
0 0 0
Let D = a 2 b2 c 2 = 0(b2c3 – b3c2) – 0(a2c3 – a3c2) + 0(a2b3 – a3b2) = 0
a3 b3 c3
2. The value of the determinant is not altered when the columns are changed to rows
(or the rows to columns).
a1 b1 c1 Interchanging the a1 a 2 a3
Let D = a 2 b2 c 2 columns to rows, we get D/ = b1 b2 b3
a3 b3 c3 c1 c2 c3
/
Expressing both determinant, we will have D = D .

54
 Determinant and Matrix

3. The interchange of any two rows (or any two columns) of a determinant changes
the sign of the determinant.
a1 b1 c1 Interchanging first and a 2 b2 c 2
Let D = a 2 b2 c2 second rows, we get D/ = a1 b1 c1
a3 b3 c3 a3 b3 c3
/
So, D = a2(b1c3 – b3c1) – b2(a1c3 – a3c1) + c2(a1b3 – a3b1)
= – [a1(b2c3 – b3c2) – b1(a2c3 – a3c2) + c1(a2b3 – a3b2)]
= – D.
4. If two rows (or two column) of a determinant are identical, the determinant is
equal to zero.
a1 b1 c1
Let D = a1 b1 c1 which first and second rows are identical.
a3 b3 c3
a1 b1 c1
Interchanging first and second rows, we get a1 b1 c1 and by property-3, its
a3 b3 c3
value is – D.
So, D = – D or, 2D = 0 or, D = 0.
5. If each element in a row (or in a column) is multiplied by any scalar,  , the
determinant is multiplied by that scalar  .
a1 b1 c1
Let D = a 2 b2 c 2 = a1(b2c3 – b3c2) – b1(a2c3 – a3c2) + c1(a2b3 – a3b2)
a3 b3 c3
Multiplying each element in first row by  , we get the following determinant:
 a1  b1  c1
D/ = a 2 b2 c2 =  a1(b2c3 – b3c2) –  b1(a2c3 – a3c2) +  c1(a2b3 – a3b2)
a3 b3 c3
=  {a1(b2c3 – b3c2) – b1(a2c3 – a3c2) + c1(a2b3 – a3b2)}
= D
a1 b1 c1  a1  b1  c1
So,  a 2 b2 c2 = a2 b2 c2
a3 b3 c3 a3 b3 c3

55
 S. M. Shahidul Islam

6. If every element in any column (or in any row) of a determinant is expressed as the
sum of two quantities, the determinant can be expressed as the sum of two
determinants of the same order.
If A1, A2 and A3 are co-factors of the elements (a1+  1), (a2+  2) and (a3+  3)
respectively of the following determinant, then
a1   1 b1 c1
D = a2   2 b2 c 2 = (a1 +  1)A1+ (a2 +  2)A2+ (a3 +  3)A3
a3   3 b3 c3
= (a1A1 + a2A2 + a3A3) + (  1A1 +  2A2 +  3A3)
a1 b1 c1  1 b1 c1
= a2 c 2 +  2 b2 c 2
b2
a3 b3 c3  3 b3 c3
7. If we change each element of any column (or any row) by adding to them any
constant multiple of the corresponding element of other columns (or other rows),
then the value of the determinant will not be altered.
a1 b1 c1
Let D = a 2 b2 c2
a3 b3 c3
Multiplying column-2 by m and column-3 by n and then adding to column-1, we
have the following determinant:
a1  mb1  nc1 b1 c1
D = a 2  mb2  nc 2 b2 c 2
/
Using property-6, we have
a3  mb3  nc3 b3 c3
a1 b1 c1 mb1 b1 c1 nc1 b1 c1
/
D = a2 b2 c 2 + mb2 b2 c 2 + nc 2 b2 c2
a3 b3 c3 mb3 b3 c3 nc3 b3 c3
a1 b1 c1 a1 b1 c1 a1 b1 c1
/
Or, D = a 2 b2 c 2 + m a 2 b2 c 2 + n a 2 b2 c 2 [Using property-5]
a3 b3 c3 a3 b3 c3 a3 b3 c3
/
Or, D = D + m.0 + n.0 [Using property-4]
So, D/ = D.

56
 Determinant and Matrix

8. The sum of the products of the elements of a column (or a row) and respective co-
factors of any other column (or any other row) is zero.
a1 b1 c1
b c2
Let D = a 2 b2 c 2 . So the co-factor of a1 is A1 = 2 ,
b3 c3
a3 b3 c3
b1 c1 b1 c1
the co-factor of a2 is A2 = (–) and the co-factor of a3 is A3 =
b3 c3 b2 c2
Thus, D = a1A1+ a2A2+ a3A3
b1 b1 c1
But, b1A1+ b2A2+ b3A3 represents b2 b2 c 2 = 0 [By property-4]
b3 b3 c3
So, b1A1+ b2A2+ b3A3 = 0.

5.6 Multiplication of two determinants: To multiply two determinants, we have to make

them same order firstly. Then we use one of the following four methods for the
multiplication of two determinants. Such as (i) Row by row method, (ii) Row by column
method, (iii) Column by row method and (iv) Column by column method. Here, we try to
make you understand the row by column method with the following example:
a b e f ae  bg af  bh
 =
c d g h ce  dg cf  dh
a b 0 0
a b 0
a b c d 0 0
Note: We can increase the order of a determinant as =c d 0=
c d 0 0 1 0
0 0 1
0 0 0 1
5.7 Application of determinants (Cramer’s rule): Let us consider the following system
of n non-homogeneous linear equations in n unknowns x1, x2, x3, ---, xn.
a11 x1  a12 x 2  ...  a1n x n  b1 
a 21 x1  a 22 x 2  ...  a 2 n x n  b2 

... ... ... ... ... 
a n1 x1  a n 2 x 2  ...  a nn x n  bn 
D D D
Then x1  1 , x 2  2 , ..., x n  n where
D D D

57
 S. M. Shahidul Islam

a11 a12 ... a1n b1 a12 ... a1n a11 a12 ... b1
a 21 a 22 ... a 2 n b2 a 22 ... a 2 n a 21 a 22 ... b2
D= ≠ 0, D1 = , ..., Dn =
... ... ... ... ... ... ... ... ... ... ... ...
a n1 a n 2 ... a nn bn a n 2 ... a nn a n1 an2 ... bn
Proof: Let us consider the relation
a11 x1  a12 x 2  ...  a1n x n a12 ... a1n b1 a12 ... a1n
a 21 x1  a 22 x 2  ...  a 2 n x n a 22 ... a 2 n b2 a 22 ... a 2 n
=
... ... ... ... ... ... ... ... ... ... ...
a n1 x1  a n 2 x 2  ...  a nn x n a n 2 ... a nn bn a n 2 ... a nn
The left hand side determinant can be expressed as the sum of n determinants of which all
the determinants are zero except first one, since at least two columns of those determinants
are equal. Therefore, from above relation we have
a11 a12 ... a1n b1 a12 ... a1n
a 21 a 22 ... a 2 n b a 22 ... a 2 n
x1 = 2
... ... ... ... ... ... ... ...
a n1 a n 2 ... a nn bn a n 2 ... a nn
Or, x1 D = D1
D
So, x1  1
D
Similarly, we can find the value of x2, x3, ..., xn as follows:
D D D
x 2  2 , x3  3 , ..., x n  n
D D D
D1 D2 D3 D
Thus, x1  , x2  , x3  , ..., x n  n
D D D D
This solution technique of a system of linear equations is called the Cramer’s rule.

Example: Using Cramer’s rule solve the following system of linear equations:
x  2y - 3z  0

2 x - y  4 z  10 [AUB-02 MBA]
4x  3y - 4z  6

Solution: Forming determinant with coefficients of x, y and z, we get
1 2 3
D= 2 1 4 = 1(4 – 12) – 2(– 8 – 16) + (–3)(6 + 4) = – 8 + 48 – 30 = 10
4 3 4

58
 Determinant and Matrix

Forming determinant with constant terms and the coefficients of y and z, we get
0 2 3
Dx = 10 1 4 = 0(4 – 12) – 2(– 40 – 24) – 3(30 + 6) = 0 + 128 – 108 = 20
6 3 4
Forming determinant with coefficients of x, constant terms and coefficients of z, we get
1 0 3
Dy = 2 10 4 = 1(– 40 – 24) – 0(– 8 – 16) – 3(12 – 40) = – 64 – 0 + 84 = 20
4 6 4
Forming determinant with coefficients of x, y and constant terms, we get
1 2 0
Dz = 2  1 10 = 1(– 6 – 30) – 2(12 – 40) + 0(6 + 4) = – 36 + 56 + 0 = 20
4 3 6
Dx 20
So, x =   2,
D 10
Dy 20
y =   2,
D 10
D 20
And z = z   2.
D 10
Therefore, the solution of the system, (x, y, z) = (2, 2, 2).

Example: Using Cramer’s rule solve the following system of linear equations
2x1 – x2 = 2
3x2 + 2x3 = 16 [DU-80 Eco.]
5x1 + 3x3 = 21
Solution: We can rewrite the equations as follows:
2x1 – x2 + 0x3 = 2
0x1 + 3x2 + 2x3 = 16
5x1 + 0x2 + 3x3 = 21
2 1 0
Here, D = 0 3 2 = 2(3.3 – 0.2) – (-1)(0.3 – 5.2) + 0(0.0 – 5.3)
5 0 3
= 2.9 +1.(-10) + 0 = 18 – 10 = 8
2 1 0
D1 = 16 3 2 = 2(3.3 – 0.2) – (-1)(16.3 – 21.2) + 0 = 18 + 6 = 24
21 0 3

59
 S. M. Shahidul Islam

2 2 0
D2 = 0 16 2 = 2(16.3 – 21.2) – 2(0.3 – 5.2) = 12 + 20 = 32
5 21 3
2 1 2
D3 = 0 3 16 = 2(3.21 – 0.16) –(-1)(0.21 – 5.16) + 2(0.0 – 5.3) = 126 – 80 – 30 = 16
5 0 21
D1 24 D 32 D 16
So, x1    3 , x2  2   4 and x3  3   2.
D 8 D 8 D 8
Therefore, the required solution is (x1, x2, x3) = (3, 4, 2) (Answer)

5.8 Definition of matrix: A rectangular array of m  n quantities aij (i = 1, 2, 3, ..., m;

j = 1, 2, ..., n) enclosed by a pair of brackets or double vertical bars is known as matrix and
is generally written as follows:
 a11 a12 ... a1n 
 
 a 21 a 22 ... a 2 n 
A= 
... ... ... ... 
 
a 
 m1 a m 2 ... a mn 
It is not an ordinary number like determinant. The number a11, a12, ..., amn are called the
entries or the elements of the matrix. The above matrix has m rows and n columns and
hence it is called an m  n matrix (read as m by n matrix). And we say the order of the
matrix is m  n or ‘m by n’.

5.9 Types of matrices: There are various types of matrices. Here, we shall discuss
important types of matrices with examples.
1. Square matrix: A matrix with same number of columns and rows is called a
 3 4 9
 2 6  
square matrix. For examples   and  6 5 1  are square matrices.
 4 1  9 7 3
 
2. Row matrix (or row vector): If a matrix consists of only one row, then it is called
a row matrix or row vector. For example (2 5 7) is a row matrix of order 1  3.
3. Column matrix (or column vector): If a matrix consists of only one column, then
3
 
it is called a column matrix or column vector. For example  5  is a column matrix
8
 
of order 3  1.

60
 Determinant and Matrix

4. Diagonal matrix: A square matrix whose elements aij = 0 when i ≠ j is called a

3 0 0 
 
diagonal matrix. For example  0 7 0  is a diagonal matrix of order 3  3.
 0 0 1
 
5. Scalar matrix: A diagonal matrix whose diagonal elements are all equal is called
 3 0 0
 
a scalar matrix. For example  0 3 0  is a scalar matrix of order 3  3.
 0 0 3
 
6. Unit matrix (or identity matrix): A square matrix whose elements aij = 0 when
i ≠ j and aij = 1 when i = j is called a unit matrix or identity matrix and is denoted
1 0 0
 
by I. For example, I =  0 1 0  is a unit matrix.
0 0 1
 
7. Null matrix (or zero matrix): The matrix whose each element is 0 (zero) is
0 0 0 0 0
   
known as null matrix or zero matrix. For examples  0 0  and  0 0 0  are
0 0 0 0 0
   
null matrices of order 3  2 and 3  3.
8. Triangular matrix: The square matrix in which elements aij = 0 when i > j is
called an upper triangular matrix. And the square matrix in which elements aij = 0
 1 7 3
 
for i < j is called a lower triangular matrix. For examples  0 5 8  and
 0 0 1
 
 3 0 0
 
 4 1 0  are upper and lower triangular matrices respectively of order 3  3.
8  2 1
 
9. Idempotent matrix: A square matrix A is said to be an idempotent matrix if
1 3 5 
 
A = A. For example  1  3  5  is an idempotent matrix.
2

1 3 5 

10. Involutory matrix: A square matrix A is said to be an involutory matrix if A2 = I,
 2 1 
where I is the unit matrix. For example, A =   is an involutory matrix.
  3  2

61
 S. M. Shahidul Islam

11. Nilpotent matrix: A square matrix A is said to be a nilpotent matrix of order n if

An = 0 and An–1 ≠ 0 where 0 is the null matrix and n is a positive integer. For
1 0
example, A =   is a nilpotent matrix of order 2.
0 0
12. Singular matrix: A square matrix A is said to be a singular matrix if the value of
the determinant formed by the matrix is 0 (zero). For example,
1 2 3 1 2 3
 
A =  2 3 4  is a singular matrix because of A = 2 3 4 = 0.
3 5 7
  3 5 7
13. Transpose of a matrix: The matrix, which is formed by interchanging the
columns to rows (or the rows to columns), is called the transpose matrix of that
matrix. Let A be an m  n order matrix, then the matrix of order n  m obtained
from the matrix A by interchanging its columns to rows (or the rows to columns) is
called the transpose of A and is denoted by the symbol AT. That is, if A = (aij) is an
m  n matrix then AT = (aji) is a matrix of order n  m. For example, let
1 6
 1 2 5  
A =   , then A =  2 7  .
T

 6 7 8 5 8
 
14. Symmetric matrix: A square matrix A is said to be symmetric matrix if AT = A,
1 2 3
 
that is, elements aij = aji for all i, j. For example  2 5 7  is a symmetric matrix.
 3 7 3
 
15. Skew-symmetric matrix: A square matrix A is said to be skew-symmetric matrix
if AT = – A, that is, elements aij = – aji for i ≠ j and aij = 0 for i = j. For example
 0 2 3
 
  2 0 7  is a skew-symmetric matrix.
  3  7 0
 
16. Inverse matrix: If A and B are two non-singular square matrices such that
AB = BA = I, where I is the unit matrix, then B is said to be the inverse matrix of
A as well as A is the inverse matrix of B. The inverse matrix of A is denoted by
 2  1 2 1
A-1. For examples, A-1 =   is the inverse matrix of A =   .
 3 2   3 2
17. Orthogonal matrix: A matrix A is called an orthogonal matrix if AAT = ATA = I,
where I is the unit matrix. That is, AT = A-1. For example,

62
 Determinant and Matrix

 1 2 2 
 
 3 3 3 
A=    is an orthogonal matrix.
2 1 2
 3 3 3
 2 2 1
  
 3 3 3
18. Complex conjugate of a matrix: The complex number z = x – iy is the conjugate
of the complex number z = x + iy. The conjugate of a matrix A is the matrix whose
elements are respectively the conjugates of the elements of A and is denoted by A .
 
That is, if A = (aij), then A = a ij . For example,
 2 3  i4   2 3  i4 
A    is the complex conjugate of matrix A =   .
 5  i 2 3   5  i 2 3 
19. Hermitian matrix: The square matrix A = (aij) of complex numbers is said to be
Hermitian matrix if A* = A T = A, that is, aij = a ji for all i, j. For example,
 2 3  i4 
A =   is a Hermitian matrix.
 3  i4 3 
20. Skew -Hermitian matrix: The square matrix A = (aij) of complex numbers is said
to be skew-Hermitian matrix if A* = A T = – A, that is, aij = – a ji for all i, j. For
 2i  3  i4 
example, A =   is a skew-Hermitian matrix.
 3  i4 0 
 a11 a12 ... a1n 
 
 a 21 a 22 ... a 2 n 
21. Adjoint or adjugate matrix: Let a square matrix A =  .
... ... ... ... 
 
 a a ... a 
 n1 n 2 nn 

If A11, A12, A13, ..., Ann are respective co-factors of elements a11, a12, a13, ..., ann of
a11 a12 ... a1n
a 21 a 22 ... a 2 n
the determinant A = then the adjoint matrix of A is
... ... ... ...
a n1 a n 2 ... a nn
T
 A11 A12 ... A1n   A11 A21 ... An1 
   
A A ... A2 n   A12 A22 ... An 2 
Adj A =  21 22 =  ... ...
... ... ... ...  ... ... 
   
A A ... Ann  A A ... Ann 
 n1 n 2  1n 2 n

63
 S. M. Shahidul Islam

 1  2  5   1 3  11
T
1 2 3 
   
For example, let A =  1 3  1 then Adj A =  3  6 3  =  2  6 4 
2 1 0    11 4 1    5 3 1 
    

22. Augmented matrix: Consider the system of linear equations A x = b , where A is

an m  n matrix, x and b are column vectors. Then the m  (n – 1) matrix (A b ),
obtained by adjoining the column vector b to the matrix A on the right, is known
as the augmented matrix of the system A x = b . For example, consider the system
x  2y - 3z  6 . . . (1)

2x - y  4 z  2 . . . (2)
4x  3y - 4z  14
 . . . (3)
We form the following augmented matrix with the coefficients of x, y, z and the
constant terms of the considered system.
1 2 3 6 1 2 3   x 6
     
(A| b ) = 2  1 4 2 [Here, A =  2  1 4  , x =  y  , b =  2  ]
3  4 14 4 3  4  z 14 
4      
23. Echelon matrix: A matrix A = (aij) is an echelon matrix or is said to be in echelon
form if the number of zeros preceding the first non-zero entry of a row increases
row by row until only zero rows remain. For example,
 2 3 0 4
 
0 6 5 1
A=  is an echelon matrix.
0 0 3 1
 
 0 0 0 2
 
In particular, an echelon matrix is called a row reduced echelon matrix if the
distinguished elements are
(i) the only nonzero entries in their respective columns;
(ii) each equal to 1.
1 3 0 4
 
0 1 5 1
For example, A =  is a row reduced echelon matrix.
0 0 1 1
 
0 0 0 1
 

64
 Determinant and Matrix

5.10 Matrices operations:

1. Addition of matrices: Addition of matrices is defined only for the matrices having
same order, that is, same number of rows and same number of columns. Let A and
B be the two matrices having m rows and n columns. That is,
 a11 a12 ... a1n   b11 b12 ... b1n 
   
 a 21 a 22 ... a 2 n   b21 b22 ... b2 n 
A=  and B = 
... ... ... ...  ... ... ... ... 
   
a  b 
 m1 a m2 ... a mn   m1 m 2 b ... bmn 

Then the sum of A and B is

 a11  b11 a12  b12 ... a1n  b1n 
 
 a 21  b21 a 22  b22 ... a 2 n  b2 n 
A+B=  
... ... ... ...
 
a  b   
 m1 m1 a m2 bm2 ... a mn b mn 

1 2 3  3 4 9
   
For example, let A =  2 5 7  and B =  6 5 1  , then
 3 7 3  9 7 3
   
 1  3 2  4 3  9   4 6 12 
   
A + B =  2  6 5  5 7  1  =  8 10 8 
 3  9 7  7 3  3  12 14 6 
   
2. Subtraction of matrices: Subtraction of matrices is defined only for the matrices
having same order, that is, same number of rows and same number of columns. Let
A and B be the two matrices having m rows and n columns. That is,
 a11 a12 ... a1n   b11 b12 ... b1n 
   
 a 21 a 22 ... a 2 n   b21 b22 ... b2 n 
A=  and B = 
... ... ... ...  ... ... ... ... 
   
a  b 
 m1 a m 2 ... a mn   m1 bm 2 ... bmn 
Then the sum of A and B is
 a11  b11 a12  b12 ... a1n  b1n 
 
 a 21  b21 a 22  b22 ... a 2 n  b2 n 
A–B=  
... ... ... ...
 
a  b 
 m1 m1 a m 2  bm 2 ... a mn  bmn 

65
 S. M. Shahidul Islam

1 2 3  3 4 9
   
For example, let A =  2 5 7  and B =  6 5 1  , then
 3 7 3  9 7 3
   
 1  3 2  4 3  9   2  2  6
   
A – B =  2  6 5  5 7  1 =   4 0 6 
 3  9 7  7 3  3   6 0 0 
  
3. Scalar multiplication of matrix: The scalar (number) multiplication of a matrix is
defined as follows: The product of an m  n matrix A by a scalar k is denoted by
kA or Ak and is the m  n matrix obtained by multiplying each element of A by k.
 a11 a12 ... a1n   ka11 ka12 ... ka1n 
   
 a 21 a 22 ... a 2 n   ka21 ka22 ... ka2 n 
Let A =  then kA =  . For example,
... ... ... ...  ... ... ... ... 
   
a   ka 
 m1 a m 2 ... a mn   m1 kam 2 ... kamn 
1 2 3  2.1 2.2 2.3   2 4 6 
     
let A =  2 5 7  and k = 2, so 2A =  2.2 2.5 2.7  =  4 10 14 
 3 7 3  2.3 2.7 2.3   6 14 6 
     
4. Multiplication of matrices: The multiplication of two matrices A and B will be
possible if the number of columns in the first matrix is equal to the number of rows
of the second matrix. We multiply the rows of first matrix by the columns of the
a b  e f 
second matrix. Let A =   and B =   . Then the product of the
c d  g h
matrices A and B is given by
 a b   e f   ae  bg af  bh   ea  fc eb  fd 
A  B =      =   and B  A =   .
 c d   g h   ce  dg cf  dh   ga  hc gb  hd 
1 2 3 3 4 9
   
For example, let A =  2 5 7  and B = 6 5 1  , then
3 7 3  9 7 3 
 
1 2 3  3 4 9   1.3  2.6  3.9 1.4  2.5  3.7 1.9  2.1  3.3 
     
AB = 2 5 7  6 5 1  =  2.3  5.6  7.9 2.4  5.5  7.7 2.9  5.1  7.3 
 3 7 3 9 7 3   3.3  7.6  3.9 3.4  7.5  3.7 3.9  7.1  3.3 
  
 42 35 20 
 
=  99 82 44 
 78 68 43 
 
66
 Determinant and Matrix

Note – 1: Algebraic associative law is applicable for matrix multiplications That is, if A,
B and C are three matrices then A  (B  C) = (A  B)  C.
Note – 2: Matrix multiplication may or may not satisfy the commutative law. That is, if A
and B are two matrices then A  B = B  A or A  B ≠ B  A.

 a11 a12 ... a1n 

 
5.11 Process of finding inverse matrix: Let the square matrix A =  21
a a 22 ... a 2 n 
 ... ... ... ... 
 
a a n 2 ... a nn 
 n1
a11 a12 ... a1n
a a ... a2n
and D be the determinant of the matrix A, that is D = A = 21 22 . Find the
... ... ... ...
a n1 a n 2 ... a nn
value of the determinant D. If D = 0 the matrix A is singular and it has no inverse, and if
D ≠ 0 the matrix A is non-singular and the inverse A-1 exists. To find the inverse matrix,
we have to find the adjoint matrix first. Let A11, A12, A13, ..., Ann are respective co-factors
of elements a11, a12, a13, ..., ann of the determinant D = A , then the adjoint matrix of A is
T
 A11 A12 ... A1n   A11 A21 ... An1 
   
Adj A =  A21 A22 ... A2 n  =  A12 A22 ... An 2  . Then the inverse matrix of A is
 ... ... ... ...   ... ... ... ... 
   
A A ... Ann  A A ... Ann 
 n1 n 2  1n 2 n
 A11 A21 An1 
 ... 
 A A A 
1 AdjA  A12 A22 A 
-1
A = AdjA  . Therefore, A =  A
-1 ... n 2  .
D A  A A 
 ... ... ... ... 
 A1n A2 n A 
 ... nn 
 A A A 
 4 0 4 0
For example, let A =   . So, the determinant D = A = = (4 – 0) = 4. Here,
 2 1 2 1
co-factors are: A11 = (-)2 1 = 1, A12 = (-)3 2 = - 2, A21 = (-)3 0 = 0, A22 = (-)4 4 = 4. So,
1  2
T
 1 0
Adj A =   =  
0 4    2 4

67
 S. M. Shahidul Islam

 1 0   1 4 0 
1
Therefore, inverse matrix, A =-1
 =
  2 4    1 2 1 
4
Note: If A and B be any two non-singular square matrices then (AB)-1 = B-1.A-1 and
A 
1 1
 A . These are the properties of the inverse matrix.

5.12 Rank of a matrix: The rank of a matrix can be defined in many equivalent ways.
Generally, we use the following two definitions.
(i) Let A be an arbitrary m  n matrix over a field. The rank is the largest value of r
for which there exists at least one r  r sub-matrix of A which forms a non-
vanishing determinant.
(ii) Let A be an m  n matrix and let Ae be the row reduced echelon form of A.
Then number of non-zero rows of Ae is the rank of the matrix A.
 4 5 4 5
For example let A =   . Then the rank of the matrix A is 2 because of A = =
 0 3 0 3
12 ≠ 0 or the matrix A contains 2 non-zero rows being itself an echelon matrix.

5.13 Use of matrix to solve the system of linear equations: We solve the system of
linear equations using matrix in two ways, such as (1) By finding inverse matrix, (2) By
making echelon matrix.
(1) By finding inverse matrix: In this method, we use inverse matrix to solve a system of
linear equations. Let A be the coefficient matrix, B be the constant term matrix and X be
the variable matrix of a system of linear equations, then
AX = B
So, the solution will be X = A-1B.
To illustrate this method, let us consider the following system of linear equations.
x  2y - 3z  6 . . . (1)

2x - y  4 z  2 . . . (2)
4x  3y - 4z  14
 . . . (3)
From the system, we find the following coefficient matrix, constant term matrix and
variable matrix.
1 2  3 x 6 
     
A = 2 1 4  , X =  y  and B = 2 
4  4  z  14 
 3    
-1
To find the inverse matrix of matrix A, A let us consider the following augmented
matrix:

68
 Determinant and Matrix

1 2 3 1 0 0
2 1 4 0 1 0
4 3 4 0 0 1
Subtracting two times of 1 row from 2nd row and four times of 1st row from 3rd row, we
st

get
1 2 3 1 0 0
0 5 10  2 1 0
0 5 8 4 0 1
Subtracting 2nd row from 3rd row, we get
1 2 3 1 0 0
0 5 10  2 1 0
0 0  2  2 1 1
nd rd
Dividing 2 row by –5 and 3 row by –2, we get
1 2 3 1 0 0
0 1  2 2 / 5  1/ 5 0
0 0 1 1 1/ 2  1/ 2
Subtracting 2 times of 2 row from 1st row and adding 2 times of 3rd row with 2nd row, we
nd

get
1 0 1 1/ 5 2/5 0
0 1 0 12 / 5 4 / 5 1
0 0 1 1 1/ 2  1/ 2
rd st
Subtracting 3 row from 1 row, we get
1 0 0  4 / 5  1 / 10 1/ 2
0 1 0 12 / 5 4 / 5 1
0 0 1 1 1/ 2  1/ 2
  4 / 5  1 / 10 1/ 2
 
-1
So, A =  12 / 5 4 / 5 1 
 1  1 / 2 
 1/ 2
Therefore, X = A-1B
 x    4 / 5  1 / 10 1/ 2  6 
    
Or,  y  =  12 / 5 4 / 5 1   2 
z   1  1 / 2  14 
   1/ 2

69
 S. M. Shahidul Islam

 24 1 
   7
x  5 5 
   72 8 
Or,  y  =    14 
z   5 5 
  6 1 7 
 
 
 x   2
   
Or,  y  =  2
 z  0
   
That is, x = 2, y = 2 and z = 0
So, the solution is (x, y, z) = (2, 2, 0).

(2) By making echelon matrix: In this method, forming the augmented matrix with the
coefficients of x, y, z and the constant terms of the system, we try to make it echelon form.
Then we check it consistency by the following rules:
If A|B is the row reduced echelon matrix of a system:
i) If rank of matrix A ≠ rank of augmented matrix A|B, then the system is
inconsistent.
ii) If rank of matrix A = rank of augmented matrix A|B = number of variables,
then the system is consistent and has a unique solution.
iii) If rank of matrix A = rank of augmented matrix A|B < number of variables,
then the system is consistent and has many solutions.
To explain the method, let us again consider the system of 3 linear equations with 3
variables x, y and z that we have solved by the previous method:
x  2y - 3z  6 . . . (1)

2x - y  4 z  2 . . . (2)
4x  3y - 4z  14
 . . . (3)
We form the following augmented matrix with the coefficients of x, y, z and the constant
terms of the system.
1 2 3 6
A|B = 2 1 4 2
4 3  4 14
Now, our goal is to reach the echelon form of the augmented matrix by row reduced
technique. Subtracting 2 times of 1st row from 2nd row and 4 times of 1st row from 3rd row,
we get

70
 Determinant and Matrix

1 2 3 6
≈ 0 5 10  10
0 5 8 10
Again subtracting 2 row from 3rd row, we get
nd

1 2 3 6
≈ 0 5 10  10
0 0 2 0
Now, dividing 2nd row by –5 and 3rd row by –2, we have
1 2 3 6
≈ 0 1 2 2 , which is the echelon form.
0 0 1 0
In the echelon form, we see that there are 3 non-zero rows both in matrix A and
augmented matrix A|B. So, the rank of A = the rank of A|B = 3 = number of variables.
Therefore, the considered system is consistent and has unique solution. To find the
solution, we form the following equations by the rows of the echelon matrix:

z=0
y – 2z = 2 => y=2
x + 2y –3z = 6 => x=2
Thus, the solution of the system is (x, y, z) = (2, 2, 0)

5.14 Some worked out examples:

3 5 6
Example (1): Evaluate the determinant D = 4 3  7
8 1 0
Solution: We know that the determinant D is equal to the sum of the products of the
elements of any row or column and their respective co-factors.
3 7 4 7 4 3
Thus, D = 3 –5 +6
1 0 8 0 8 1
= 3{3.0 – 1(-7)} – 5{4.0 – 8(-7)} + 6(4.1 – 8.3)
= 3(0 + 7) – 5(0 + 56) + 6(4 – 24)
= 3.7 – 5.56 + 6.(- 20)
= 21 – 280 – 120
= – 379 (Answer)

4 2 2 5
Example (2): If A = and B = then find A  B.
1 3 2 8

71
 S. M. Shahidul Islam

And also show that (value of A)(value of B) = value of A  B.

4 2 2 5
Solution: Given that, A = and B =
1 3 2 8
4 2 2 5 4.2  1.5 2.2  3.5 13 19
So, A  B =  = = (Answer)
1 3 2 8 4.2  1.8 2.2  3.8 16 28
Now, value of A = 4.3 – 1.2 = 10,
value of B = 2.8 – 2.5 = 6,
value of A  B = 13.28 – 16.19 = 364 – 304 = 60.
So, value of A  value of B = 10  6 = 60
Thus, (value of A)(value of B) = value of A  B (Shown)
1 1 1
Example (3): Prove that a 2 b2 c 2 = (a – b)(b – c)(c – a)(ab + bc + ca) [DU-75]
a3 b3 c3
1 1 1 [Subtracting 2nd column from 1st
Solution: L.H.S = a 2 b2 c2 column and 3rd column from the 2nd
a3 b3 c3 column.]
0 0 1
= a b
2 2
b c
2 2
c2
a3  b3 b3  c3 c3
a2  b2 b2  c2
=1
a 3  b3 b3  c3
(a  b)(a  b) (b  c)(b  c)
=
(a  b)(a  ab  b ) (b  c)(b 2  bc  c 2 )
2 2

( a  b) (b  c)
= (a – b)(b – c)
(a  ab  b ) (b  bc  c 2 )
2 2 2

ab ca [Subtracting 1st column

= (a – b)(b – c) 2 nd
a  ab  b 2 c 2  a 2  bc  ab from 2 column]
ab ca
= (a – b)(b – c)
a  ab  b 2
2
(c  a)(a  b  c)
ab 1
= (a – b)(b – c)(c – a)
a  ab  b (a  b  c)
2 2

= (a – b)(b – c)(c – a)(a2 + ab + ac + ab + b2 + bc – a2 – ab –b2)

= (a – b)(b – c)(c – a)(ab + bc + ca) = R.H.S (Proved)

72
 Determinant and Matrix

Example (4): Solve the following system of linear equations with the help of determinant
(Cramer’s rule)
x + 2y – z = 9
2x – y + 3z = –2 [RU-78, AUB-01]
3x + 2y + 3z = 9
Solution: Forming determinant with coefficients of x, y and z, we get
1 2 1
D = 2  1 3 = 1(–3 – 6) – 2(6 – 9) + (–1)(4 + 3) = – 9 + 6 – 7 = –10
3 2 3
Forming determinant with constant terms and the coefficients of y and z, we get
9 2 1
Dx =  2  1 3 = 9(–3 – 6) –2(– 6 – 27) –1(– 4 + 9) = – 81 + 66 – 5 = – 20
9 2 3
Forming determinant with coefficients of x, constant terms and coefficients of z, we get
1 9 1
Dy = 2  2 3 = 1(–6 – 27) – 9(6 – 9) + (–1)(18 + 6) = – 33 + 27 – 24 = –30
3 9 3
Forming determinant with coefficients of x, y and constant terms, we get
1 2 9
Dz = 2  1  2 = 1(– 9 + 4) – 2(18 + 6) + 9(4 + 3) = – 5 – 48 + 63 = 10
3 2 9
Dx  20
So, x =   2,
D  10
Dy  30
y =   3,
D  10
D 10
And z = z   1.
D  10
Therefore, the required solution (x, y, z) = (2, 3, -1)

 3 2
Example (5): Let A =   . Show that A2 – 3A + 2I = 0, where I is the unit matrix of
 1 0
order 2  2 and 0 is the null matrix of order 2  2.
 3 2
Solution: Given that A =  
 1 0

73
 S. M. Shahidul Islam

 3 2   3 2   3.3  2.1 3.2  2.0   7 6 

So, A2 = A.A =   .   =   =   ,
  1 0    1 0    1.3  0.1  1.2  0.0    3  2 
 3 2   3 .3 3 .2   9 6 
3A = 3.   =   =  
  1 0   3.  1 3 .0    3 0 
1 0  2 0
And 2I = 2.   =  
 0 1  0 2 
2
L.H.S = A – 3A + 2I
 7 6   9 6  2 0
=   –  +  
  3  2   3 0   0 2
 792 660 
=  
  3  3  0  2  0  2
0 0
=   = R.H.S (Proved)
0 0

0 1 0  i
Example (6): If A =   and B =   then prove that AB = – BA
1 0 i 0 
and A2 = B2 = I [RU-88]
0 1 0  i
Solution: Given that A =   and B =  
1 0 i 0 
Proof of AB = – BA:
 0 1   0  i   0.0  1.i 0.  i  1.0   i 0 
L.H.S = AB =   .   =   =  
 1 0   i 0  1.0  0.i 1.  i  0.0   0  i 
0  i  0 1  0.0  i.1 0.1  i.0    i 0 i 0 
R.H.S = – BA = –   .   = –   = –   =  
 i 0  1 0  i.0  0.1 i.1  0.0   0 i 0  i
So, AB = – BA (Proved)
Proof of A2 = B2 = I:
 0 1   0 1   0.0  1.1 0.1  1.0   1 0 
A2 = A.A =   .   =   =   = I
 1 0   1 0  1.0  0.1 1.1  0.0   0 1 
0  i 0  i  0.0  i.i 0.  i  i.0    i 2 0  1 0
B2 = B.B =   .   =   =   = = I
 i 0   i 0   i.0  0.i  i.i  0.0   0  i 2   0 1 
So, A2 = B2 = I (Proved)

74
 Determinant and Matrix

2 1 3 
 
Example (7): Find the inverse of the matrix A =  4 0  1 [CU-88]
3 3 2 
 
Solution: Let D be the determinant of the matrix, then
2 1 3
D= A = 4 0  1 = 2(0 + 3) + 1(8 + 3) + 3(12 – 0) = 6 + 11 + 36 = 53 ≠ 0.
3 3 2
So the matrix A is non-singular and A-1 exists. Now the co-factors of D are
0 1 4 1 4 0 1 3
A11 = = 3, A12 = (-1) = - 11, A13 = = 12, A21 = (-1) = 11,
3 2 3 2 3 3 3 2
2 3 2 1 1 3 2 3
A22 = = - 5, A23 = (-1) = - 9, A31 = = 1, A32 = (-1) = 14,
3 2 3 3 0 1 4 1
 11 12 
T
3  3 11 1 
2 1    
A33 = = 4. So, Adj A = 11  5  9  =   11  5 14 
4 0 
 1 14 4   12  9 4 
 
 3 11 1   3 53 11
53
1 
53 
1 1  
Therefore, A-1 = Adj A =   11  5 14  =   11  5 14 
D 53    53 53 53 
 12  9 4   12 53  9 53 4 53 
 
1 2  3
 
Example (8): Find the inverse of the matrix A =  2  1 4  by using row canonical
4  4 
 3
form. [JU-93]
Solution: To find the inverse matrix of matrix A-1 by using row canonical form, let us
consider the following augmented matrix:
1 2 3 1 0 0
(AI3) = 2  1 4 0 1 0 r2/ = r2 – 2r1
4 3 4 0 0 1 r3/ = r3 – 4r1
Subtracting two times of 1 row from 2nd row and four times of 1st row from 3rd row, we
st

get
1 2 3 1 0 0
≈ 0 5 10  2 1 0
0 5 8 4 0 1 r3// = r3/ – r2/
Subtracting 2nd row from 3rd row, we get

75
 S. M. Shahidul Islam

1 2 3 1 0 0
≈ 0 5 10  2 1 0 r2/// = r2///-5
0 0 2  2 1 1 r3/// = r3/// – 2
nd rd
Dividing 2 row by –5 and 3 row by –2, we get
1 2 3 1 0 0 r1iv = r1/// – 2r2///
≈ 0 1  2 2 / 5  1/ 5 0 r2iv = r2/// + 2r3///
0 0 1 1 1/ 2  1/ 2
Adding 2 times of 3 row and 2 row and subtracting 2 times of 2nd row from 1st row, we
rd nd

get
1 0 1 1/ 5 2/5 0 r1v = r1iv – r3iv
≈ 0 1 0 12 / 5 4 / 5 1
0 0 1 1 1/ 2  1/ 2
rd st
Subtracting 3 row from 1 row, we get
1 0 0  4 / 5  1 / 10 1/ 2
≈ 0 1 0 12 / 5 4 / 5 1
0 0 1 1 1/ 2  1/ 2
  4 / 5  1 / 10 1/ 2
-1  
Therefore, the inverse matrix, A =  12 / 5 4 / 5 1  (Answer)
 1  1 / 2 
 1/ 2
 4 0  2 0
Example (9): Let A =   and B =   , then show that (AB)-1 = B-1.A-1. [AUB-99]
 2 1   1 2 
 4 0  2 0
Solution: Given that A =   and B =  
 2 1 1 2
 4 0  2 0  8  0 0  0 8 0 
So, AB =      =   =  
 2 1  1 2  4  1 0  2 5 2
8 0
The determinant formed by the matrix AB is AB = = 16 – 0 = 16.
5 2
The co-factors of the determinant AB are: A11 = (-)2 2 = 2, A12 = (-)3 5 = - 5,
 2  5
T
 2 0
A21 = (-) 0 = 0, A22 = (-) 8 = 8. So, Adj AB = 
3 4
 =  
0 8    5 8
1  2 0   18 0 
Therefore, the inverse matrix, (AB)-1 =  =
16   5 8    5 1 
 16 2

76
 Determinant and Matrix

4 0
Now, the determinant formed by the matrix A is A = = (4 – 0) = 4.
2 1
The co-factors of the determinant A are: A11 = (-)2 1 = 1, A12 = (-)3 2 = - 2,
1  2
T
 1 0
A21 = (-) 0 = 0, A22 = (-) 4 = 4. So, Adj A = 
3 4
 =  
0 4    2 4
1  1 0   1 4 0 
Therefore, the inverse matrix, A-1 =  =
4   2 4    1 1
 2 
2 0
Again, the determinant formed by the matrix B is B = = (4 – 0) = 4.
1 2
The co-factors of the determinant B are: A11 = (-)2 2 = 2, A12 = (-)3 1 = - 1,
 2  1
T
 2 0
A21 = (-) 0 = 0, A22 = (-) 2 = 2. So, Adj B = 
3 4
 =  
 0 2    1 2 
1  2 0   2 1 0 
Therefore, the inverse matrix, B-1 =   =
4  1 2   1 1 
 4 2
 1 0   1 0   1 .1 0 0  0   18 0 
So, A-1.B-1 =  2 . 4 = 4 2 =
 1 1   1 1   1 . 1  1 . 1 0  1. 1    5 1 
 4 2  2   4 4 2 2 2   16 2
-1 -1 -1
Thus, (AB) = B .A (Proved)

Example (10): Solve the following system with the help of matrix:
x+y+z=6
5x – y + 2z = 9 [RU-91]
3x + 6y – 5z = 0
Solution: Expressing the system by matrix, we get
1 1 1   x  6
     
5 1 2  .  y  = 9
 3 6  5  z   0 
     
1 1 1   x 6
     
Or, A.X = B where A =  5  1 2  , X =  y  and B =  9 
 3 6  5 z 0
     
Or, X = A-1.B
Let D be the determinant of the matrix A, then

77
 S. M. Shahidul Islam

1 1 1
D = A = 5 1 2 = 1(5 – 12) – 1(–25 – 6) + 1(30 + 3) = – 7 + 31 + 33 = 57 ≠ 0.
3 6 5
So the matrix A is non-singular and A-1 exists. Now the co-factors of D are
1 2 5 2 5 1 1 1
A11 = = –7, A12 = (-1) = 31, A13 = = 33, A21 = (-1) = 11,
6 5 3 5 3 6 6 5
1 1 1 1 1 1 1 1
A22 = = – 8, A23 = (-1) = – 3, A31 = = 3, A32 = (-1) = 3,
3 5 3 6 1 2 5 2
  7 31 33 
T
  7 11 3 
1 1    
A33 = = – 6. So, Adj A =  11  8  3  =  31  8 3 
5 1  3
 3  6   33  3  6 
 
  7 11 3 
-1 1 1  
Therefore, A = Adj A =  31  8 3 
D 57  
 33  3  6 
Since, X = A-1.B
 x   7 11 3   6    42  99  0   57   1 
  1   1   1    
 y =  31  8 3  .  9  =  186  72  0  = 114  =  2 
 z  57  33  3  6   0  57  198  27  0  57  171   3 
           
So, x = 1, y = 2 and z = 3 (Answer)

Example (11): A manufacturer produces three products P, Q and R, which he sells in two
markets. Annual sales volumes are indicated as follows:
Products:
P Q R
1 10000 2000 18000 
Markets:   [DU-87, AUB-03]
2  6000 20000 8000 
If unit sale prices of P, Q and R are Tk.2.50, Tk.1.25 and Tk.1.50 respectively, find the
total revenue in each market with the help of matrix algebra. And if the unit costs of the
above three commodities are Tk.1.80, Tk.1.20 and Tk.0.80, find the gross profit.
Solution: Given that, Products:
P Q R
1 10000 2000 18000 
Markets:  
2  6000 20000 8000 
10000 2000 18000 
Let sales volume matrix, S =   ,
 6000 20000 8000 

78
 Determinant and Matrix

 2.50   1.80 
   
Prices (per unit) matrix, Pr =  1.25  and Cost (per unit) matrix, C =  1.20 
 1.50   0.80 
   
We know, Total revenue = Sales volume (S)  Price per unit (Pr)
 2.50 
10000 2000 18000   
So, Total revenue =     1.25 
 6000 20000 8000   1.50 
 
10000  2.50 2000  1.25 18000  1.50 
=  
 6000  2.50 20000  1.25 8000  1.50 
 25000 2500 27000 
=  
 15000 25000 12000 
Hence, total revenue in first market = Tk.(25000 + 2500 + 27000) = Tk.54500,
Total revenue in second market = Tk.(15000 + 25000 + 12000) = Tk.52000
And total revenue in both markets = Tk.(54500 + 52000) = Tk.106500 (Answer)

It is known that, Total cost = Sales volume (S)  Cost per unit (C)
 1.80 
10000 2000 18000   
So, Total cost =     1.20 
 6000 20000 8000   0.80 
 
10000  1.80 2000  1.20 18000  0.80 
=  
 6000  1.80 20000  1.20 8000  0.80 
18000 2400 14400 
=  
10800 24000 6400 
We also know that, Total profit = Total revenue – Total cost
 25000 2500 27000  18000 2400 14400 
So, Total profit =   –  
 15000 25000 12000  10800 24000 6400 
 25000  18000 2500  2400 27000  14400 
=  
 15000  10800 25000  24000 12000  6400 
 7000 100 12600 
=  
 4200 1000 5600 
Hence, total profit in first market = Tk.(7000 + 100 + 12600) = Tk.19700,
Total profit in second market = Tk.(4200 + 1000 + 5600) = Tk.10800
And total profit in both markets = Tk.(19700 + 10800) = Tk.30500 (Answer)

79
 S. M. Shahidul Islam

5.15 Exercise:
1. Define determinant with examples.
2. What is difference between minor and co-factor?
3. What is difference between matrix and determinant?
4. State and prove Cramer’s rule for solving a system of linear equations.
5. Define matrix with examples and discuss the uses of matrix.
6. Define with examples (i) Square matrix, (ii) Singular matrix, (iii), Inverse matrix,
(iv) Augmented matrix, (v) Echelon matrix, (vi) Hermitian matrix, (vii) Skew-
Hermitian matrix.
7. Discuss the difference between adjoint and transpose matrices.
8. Evaluate the following determinants:
1 2 1 2
1 0 3
(i) 6 5 0 [Answer: -226] (ii) 3 0 1 5 [Answer: 90] [AUB-02]
1 2 0 3
21 3 7
2 4 1 6
9. Using Sarrus diagram find the value of
0 1 2 3
1 2 3
1 0 1 2
(i) 4 5 0 [Answer: - 42] (ii) [Answer: 4]
 2 1 0 1
2 3 7
 3  2 1 0
4 2 3 7
10. If A = and B = then find A  B.
3 6 2 8
And also show that (value of A)(value of B) = value of A  B.
1 a a 2  bc
11. Show that 1 b b 2  ca  0 [DU-79]
1 c c 2  ab
12. Solve the following system of linear equations with the help of determinant
(Cramer’s rule)
(i) 2x + 5y = 24 (ii) x – 3y – 8z = - 10 (iii) – x + 3y + 2z = 24
3x + 8y = 38 3x + y – 4z = 0 x+z=6
[Answer: (2, 4)] 2x + 5y + 6z = 13 5y – z = 8
[Answer: (-10, 3, 0)] [Answer: (-1, 3, 7)]

(iv) 3x1 + x2 + 2x3 = 3 (v) x + 2y + 3z = 14 (vi) x – 2y + 3z = 11

2x1 – 3x2 – x3 = -3 2x + 3y + 4z = 20 2x + y + 2z = 10
x1 + 2x2 + x3 = 4 3x + 4y + 6z = 33 3x + 2y + z = 9
[Answer: (1, 2, -1)] [Answer: (5, - 6, 7)] [Answer: (2, 0, 3)]

80
 Determinant and Matrix

1  2 3  2 3 5 
13. If A =   and B =   then find the matrices 2A, A + B and
 5 1  4   1 4  2 
2  4 6  3 1 8   1  5  2
A – B. [Answer:   ,   and   ]
10 2  8   6 5  6   4  3  2 
T
14. Write a square matrix and then show that A + A is a symmetric matrix.
1 2 3 1 0 0
   
15. If A =  2 5 7  and B =  0 1 0  , then show that A  B = B  A.
 3 7 3 0 0 1
   
1 2 3 3 4 9
   
16. Let A =  2 5 7  and B = 6 5 1  , then prove that A  B  B  A.
 3 7 3 9 7 3 
  
3 2 1  2 3
 1 0 3    
17. If A =   , B =  1 1 1  and C =  1 2 
 2 1 5 2
 1 0   2 4
 
then show that (A  B)  C = A  (B  C).
 2  1 1 2 3 1 2 3
18. Find the inverse of the matrix (i) A =   (ii) A = 1  (iii)  2 3 2  .
 3 5
1 1     3 3 4
1 5 12   
 1 1 1  11  9 1  1   6 1 5 
 , (ii)   7 9  2  , (iii)  2 5  4  ]
1
[Answer: (i) 
3  1 2 3 
7  
 2 3 1   3 3 1 
2 1 3 
19. Find the inverse of the matrix A =  4 0  1 by row canonical form.
3 3 2 
 
 3 11 1 
 53 53 53 
[Answer:   11 5 14  ]
 53 53 53 
 12 9 4 
 53 53 53 
 2 2  3
20. If A =  2  3 2
1  6  then prove that A + A – 21A – 54I = 0, where I is the unit
 1  2 0 
 
matrix of order 3  3 and 0 is the null matrix of order 3  3.
 3 6
21. If A =   
1
 then show that A 1  A .
 2 1

81
 S. M. Shahidul Islam

5 0  2 1
22. Let A =   and B =   , then show that (AB)-1 = B-1.A-1.
3 1  3
23. Solve the following systems of linear equations with the help of matrices:
(i) 5x – 6y + 4z = 15 (ii) x + 2y + 3z = 14 (iii) 2x + 3y – z = 1
7x + 4y – 3z = 19 2x + 3y + 4z = 20 3x + 5y + 2z = 8
2x + y + 6z = 46 3x + 4y + 6z = 33 x – 2y – 3z = –1
[Answer: (3, 4, 6)] [Answer: (5, - 6, 7)] [Answer: (3, – 1, 2)]
24. Solve the following system of linear equations by using row equivalent canonical
matrix (by elementary row transformations):
(i) 2x + 3y – z = 1 (ii) 2x – y + z = 1 (iii) 2x + 3y – 2z = 5
3x + 5y + 2z = 8 x + 4y – 3z = –2 x – 2y + 3z = 2
x – 2y – 3z = –1 3x + 2y – z = 0 4x – y + 4z = 1
[Answer: (3, – 1, 2)] [Answer: (0, 1, 2)] [Answer: No solution]
25. A manufacturer produces three products P, Q and R, which he sells in two markets.
Annual sales volumes are indicated as follows:
Products:
P Q R
1  4000 3000 2000 
Markets:   [DU-87, AUB-03]
2  3000 2000 1000 
If unit sale prices of P, Q and R are Tk.2.50, Tk.2.00 and Tk.1.50 respectively, find
the total revenue in both markets with the help of matrix algebra. And if the unit
costs of the above three commodities are Tk.2.00, Tk.1.50 and Tk.1.00, find the
gross profit. [Answer: Tk.32000 and Tk.7500]
26. A televisions manufacturer produces three types of TVs A, B and C, which he sells
in two markets. Annual sales volumes are indicated as follows:
Products:
A B C
1 10000 2000 18000 
Markets:   [RU-88]
2  6000 20000 8000 
If unit sale prices of A, B and C types of TVs are Tk.1500, Tk.3000 and Tk.4000
respectively, find the total revenue in both markets with the help of matrix algebra.
And if the unit costs of the above three TVs are Tk.1000, Tk.2000 and Tk.3000,
find the gross profit. [Answer: Tk.37,50,000 and Tk.11,50,000]
27. The daily cost of operating a hospital(C) is a linear function of the number of
inpatients(I) and outpatients(O) plus a fixed cost(x), i.e., C = x + yO + zI. Given
the following information from 3 days, find the values of x, y and z by setting up a
linear system of equations and using the matrix inverse.

82
 Determinant and Matrix

Day Cost(C) in Tk. No. of inpatients(I) No. of outpatients(O)

1  7000 50 10 
 
2  6500 40 10 
3   7500 50 15  
[Answer: x = 3500, y = 100 and z = 50]

28. A special food for athletes is to be developed from two foods: Food X and Food Y.
The new food is to be designed so that it contains exactly 16 ounces of vitamin A,
exactly 44 ounces of vitamin B and exactly 12 ounces of vitamin C. Each pound of
Food X contains 1 ounce of vitamin A, 5 ounces of vitamin B and 1 ounce of
vitamin C. On the other hand, each pound of Food Y contains 2 ounces of A, 1
ounce of B and 1 ounce of C. Find the number of pounds of each food to be used in
the mixture in order to meet the above requirements. [Answer: 8 pounds of Food
X and 4 pounds of Food Y.]

83
 S. M. Shahidul Islam

06
Chapter
Functions and Equations

Highlights:

6.1 Introduction 6.11 Formation of quadratic equation

6.2 Formula 6.12 Identity
6.3 Relation 6.13 Linear equation
6.4 Function 6.14 System of linear equations
6.5 Types of functions 6.15 Solution methods of a system of
6.6 Polynomial linear equations
6.7 Inequality 6.16 Break-Even point
6.8 Equation 6.17 Break-Even interpretation
6.9 Degree of an equation 6.18 Exercise
6.10 Quadratic equation

6.1 Introduction: A function can be viewed as an input-output device. The significant

relationships in mathematical models typically are represented by functions. It is the
purpose of this chapter to introduce this important topic. In business applications, we
sometimes are interested in determining whether there are values of variables, which
satisfy several attributes. In this chapter, we will also be concerned with the process used
to determine whether there are values of variables, which jointly satisfy a set of equations.

6.2 Formula: Computational procedures are described efficiently by formulas employing

the symbolism of algebra. Thus, if x is the length and y the width of a rectangle, the area A
of the rectangle is expressed by the formula:
A = xy
If a rectangle has a length of 10 inches and a width of 5 inches, then we compute the area
of that rectangle using the above formula, as follows:
Area, A = (10) (5) square inches
= 50 square inches.

84
 Functions and Equations

6.3 Relation: If A and B be two sets then non empty subset of ordered pairs of Cartesian
product, A  B is called relation of A and B and is denoted by R. If we consider x  A and
y  B then we get (x, y)  R.

Example (i): If A = {1, 2, 3} and B = {3, 7} then A  B = {(1, 3), (1, 7), (2, 3), (2, 7),
(3, 3), (3, 7)}. So, the relation x < y where x  A and y  B is R = {(1, 3), (1, 7), (2, 3),
(2, 7), (3, 7)}.

Example (ii): If A = {$2, $7, $8} is a set of cost of per unit product and B = {$5, $8} is
the set of selling price of per unit product of a production firm. Find the profitable relation
between cost and selling price.
Solution: Here, A  B = {($2, $5), ($2, $8), ($7, $5), ($7, $8), ($8, $5), ($8, $8)}
A firm becomes profitable if its selling price of per unit product is greater than the cost of
per unit product. So, the profitable relation, R = {($2, $5), ($2, $8), ($7, $8)}. (Answer)

6.4 Function: If ‘f ’ is a rule which associates every element of set X with one and only
one element of set Y, then the rule ‘f ’ is said to be the function or mapping from the set X
to the set Y. This we write symbolically as
f : X Y
If y is the element of Y, that corresponds to an element x of X, given by the rule f, we
write this as follows:
y = f(x) ; Here x is independent variable and y is dependent variable.

The set X is known as ‘Domain’ and the set Y is known as ‘Co-domain’. The set formed
with those elements of the set Y that are in correspondence with at least one element of X
is called the range.

Example: f : R  R; f(x) = x 2
Here, the set of real numbers, R are simultaneously domain and co domain and the set of
positive real number, R+ is the range.

6.5 Types of functions: We shall now introduce some different types of functions, which
are particularly useful in different branches of Mathematics.

1. One-one (1-1) Function: If the function f corresponds to the different elements of the
set Y for the different elements of set X, then the function is known as one-one (1-1)
function. f
3 
Example: f(x) = 2x + 1 1  7 
   
X  
2  Y
3  5 
 

10

Figure 6.1

85
 S. M. Shahidul Islam

2. Onto function: If every element of the co-domain set Y is a correspondence (image) of

at least one element of the domain set X, then the function is called onto function.
Example: f: R → R+ : f(x) = x2 1  f
2  1 
  9 
 Y
X 3   
4  4 
  
16 
 2
Figure 6.2

3. One-one and onto function: If the function f satisfies the properties of one-one
function and onto function then it is known as one-one and into function.
Example: f: X  Y : f(x) = x + 1 f
1  2 
2  3 
   Y
X   
3  4 

4 
 5 
Figure 6.3
4. One valued function: When a function has only one value corresponding to each value
of the independent variable, the function is called a one valued function.
Example: If f(x) = x3, f(x) is a one or single valued function.

5. Many valued function: When a function has several values corresponding to each
value of the independent variable, it is called many-valued function or multiple valued
function.
Example: If y = f(x) =  x , y is a many valued function of x.

6. Explicit function: A function expressed directly in terms of the dependent variable is

said to be an explicit function.
Example: y = f(x) = x2 + 5x – 4.

7. Implicit function: The function which is not expressed directly in terms of the
dependent variable, there is a mutual relationship between the dependent and the
independent variables.
Example: x2 + y2 = 10 is an explicit function because
y =  10  x 2 .

8. Algebraic function: When the relation, which involves only a finite number of terms
and the variables, are affected only by the operations of addition, subtraction,
multiplication, division, powers and roots, the relation is said to be an algebraic function.
86
 Functions and Equations

Example: y = f(x) = 2x3 + 3x2 – 9 is an algebraic function.

9. Transcendental function: All the functions of x, which are not algebraic, are called
transcendental functions. Thus, f(x) = ex + 2x + 1 is a transcendental function.
We have the following subclasses of transcendental functions:
i) Exponential function: f(x) = ex+1
ii) Logarithmic function: f(x, y) = log (x + y)
iii) Trigonometric function: f(x) = sin x
iv) Inverse trigonometric function: f(x) = sin-1x

10. Rational function: Expressions involving x, which consist of a finite number of terms
of the form axn, in which ‘a’ is a constant and ‘n’ a positive integer is called a rational
function of x.
x2  6
Example: y = f(x) = 4x4 + 9x – 7 and y = f(x) = are rational functions.
3x 3  2

11. Irrational function: An expression involving x, which involves root extraction of

terms, is called an irrational function.
Example: y = f(x) = x 2  4 x  10  9 x  2 is an irrational functional function.

12. Monotone Function: When the dependent variable increases with an increase in the
independent variable, the function is called a monotonically increasing function. And
when the dependent variable decreases with an increase in the independent variable, the
function is called a monotonically decreasing function.
Example: y = f(x) = 2x is a monotonically increasing function
1
y = f(x) = is a monotonically decreasing function
2x

13. Even function: If a function f (x) is such that f (-x) = f(x), then it is called an even
function of x.
Example: y = f(x) = 2x2 is an even function.

14. Odd function: If a function f (x) is such that f (-x) = -f (x), then, it is said to be an odd
function of x.
Example: y = f(x) = 2x is an odd function of x.

15. Periodic Function: If f (x) = f (x + p) for all value of x, then f (x) is called a periodic
function with period p.
Example: y = f(x) = sin x is a periodic function with period 2π.

87
 S. M. Shahidul Islam

16. Linear function: The relationship between y and x expressed by

y = ax + b; a and b are constants.
is called a functional relationship because for each value of x, there is one, and only one
corresponding value for y. Notice that the expression states y is in terms of x and so we
say y is a function of x. This type of function is known as linear function because it
represents straight line in the graph. Here, x is called independent variable and y is called
dependent variable because, the value of y depends upon what value we assign to x.
Example: y = f(x) = 3x + 4 is a linear equation.

Note: In coordinate geometry a is called slope and b is called y-intercept of the straight
line that is represented by y = ax + b.
Example: The equation of the line that has a slope of 3.2 and y-intercept of 5 is y =3.2x +5

6.6 Polynomial: If n is a positive integer, the expression

f ( x)  a0 x n  a1 x n1  a2 x n2  . . .  an1 x  an , a0  0 is called a polynomial of degree
n. As for example f ( x)  5x 2  3x  1 and f ( x)  2 x 4  5x  10 are polynomial of degree
2 and 4 respectively. Every binomial can be expressed as a polynomial by the following
formula: (x + y)n = nC0xn + nC1xn-1y + nC2xn-2y2 + . . . + nCnyn.

6.7 Inequality: When two mathematical expressions become connected by the inequality
sign ( or  or  or  or  ) is known as an inequality.
Example: 3x2 + 4x + x > 2x + 1
3x + 1 > 0, are inequalities.
Example: In the graph, draw the solution space of the inequality 2x + y ≥ 11.
Solution: To draw the graph, firstly we consider the equation 2x + y = 11.
So, y = 11 – 2x - - - (i)
Substituting x = 1, 3 and 5 in equation (i) we get the following chart:
x 1 2 5
y 9 7 1
Plotting the point (1, 9), (2, 7) and (5, 1) in the following graph we get a straight line.
Y
_
10
9
_
_

_(1, 9)
8
7 
_(2, 7)
6
_
5
_
4
_
3
_
2
_ (5, 1) 
1
| | | | | | | | | | | X
0 1 2 3 4 5 6 7 8 9 10 11
Figure 6.4

88
 Functions and Equations

The origin (0, 0) lies on the left side of the straight line. If we put (0, 0) in the given
inequality we get 0 ≥ 11, which is absurd. So, the points of the opposite side that means all
the points of the right side of the line and on the line are the solutions of the given
inequality. The shadow region shows the solution space of the given inequality in the
graph.

Example: Solve the inequality: x – 9 > 3x + 1 and represent it in the number line.
Solution: Given that, x – 9 > 3x + 1
Or, x – 9 + 9 > 3x + 1 + 9 [Adding 9 to both sides]
Or, x > 3x + 10
Or, x – 3x > 3x + 10 – 3x [Subtracting 3x from both sides]
Or, – 2x > 10
[We know, 5 > 3 but –5 < –3. For this inequality sign (>, <) changes when multiplied or
divided by a negative number]
 2 x 10
Or,  [Dividing both sides by – 2]
2 2
Or, x<–5
Therefore, the required solution: x < – 5.
–7 –6–5 –4 –3 –2 –1 0 1 2 3 4 5 6 7

Number line
Figure 6.5

6.8 Equation: Equations signify relation between two mathematical (algebraic /

trigonometric / transcendental) expressions symbolized by the sign of equality (=).
However, the equality is true only for certain value or values of the variable or variables
symbolized generally by x, y, z, u, v, w etc. and these values are known as the solution of
the equation.
Example: 3x + 7 = 8x-3 [Linear equation]
9x + 12 = 0
2x2 + 3x + 2 = 0 [Quadratic equation]
x
3e + 4 cos x + 5x + 9 = 0 [Transcendental equation]

6.9 Degree of an equation: The highest power of variable or variables of an equation is

called the degree of that equation. Such as
2x + 5 = x
3x2 + 2y = 2x + 1
4x + y3 = 1
The highest power of variable or variables in the 1st, 2nd and 3rd equations are respectively
1, 2 and 3. So, they are respectively 1st degree, 2nd degree and 3rd degree equations.
Generally, 1st degree equations are known as linear equations. Of one variable, second

89
 S. M. Shahidul Islam

degree equation, ax2 + bx + c = 0 is known as quadratic equation, third degree equation,

ax3+ bx2+ cx + d = 0 is known as cubic equation and fourth degree equation, ax 4 + bx3 +
cx2 + dx + e = 0 is known as bi-quadratic equation.

6.10 Quadratic equation: The second degree equations of one variable are known as
quadratic equations. The standard form of quadratic equation is ax2 + bx + c = 0.
2x2 – 5x = 6x + 10, x2 – 8x + 15 = 0 are quadratic equations.
Every quadratic equation has two roots (solutions). One of the most common solution
techniques is as follows:
 b  b 2  4ac
The solution of standard form ax2 + bx + c = 0 is x  . Here b 2  4ac is
2a
called discriminant and (1) if b 2  4ac > 0, there will be two real roots; (2) if b 2  4ac = 0,
there will be one real root; & (3) if b 2  4ac < 0, there will be no real roots.

Example: Find the roots of the quadratic equation: 2x2 – 5x = 6x + 10 [AUB-02]

Solution: Converting the given equation in the standard, we get
2x2 + x – 10 = 0
Comparing this equation to the standard form of quadratic equation, we get a = 2, b = 1
and c = –10.
 b  b 2  4ac
So, the value of x 
2a
 1  (1) 2  4.2.(10)
Or, x
2.2
1 9
Or, x
4
1 9 1 9
So, x = 2 and x  = – 2.5
4 4
Therefore, the roots of the given equation are 2 and – 2.5.

x  3 x  3 2x  3
Example: Solve:   [RU-92 A/C]
x2 x2 x 1
x  3 x  3 2x  3
Solution: Given that  
x2 x2 x 1
( x  3)( x  2)  ( x  3)( x  2) 2 x  3
Or, 
( x  2)( x  2) x 1
x 2  x  6  x 2  x  6 2x  3
Or, 
x2  4 x 1

90
 Functions and Equations

2 x 2  12 2 x  3
Or, 
x2  4 x 1
Or, 2 x  2 x  12 x  12  2 x 3  3x 2  8x  12
3 2

Or, x 2  4x  0
Or, x(x – 4) = 0
So, x = 0, x – 4 = 0 or x = 4
Therefore, the required solution x = 0 or 4 (Answer)

6.11 Formation of quadratic equation: If given that  and  are two roots of a
quadratic equation then we can form the equation as x2 – (  +  )x +   = 0,
That is, x2 – (sum of the roots) x + product of the roots = 0
3 and 5 are two roots of the quadratic equation which is x2 – (3 + 5)x + 3.5 = 0;
Or, x2 – 8x + 15 = 0

Example: If a and b be two roots of 2x2 – 4x + 1 = 0, then find the quadratic equation
whose roots are a2 + b and a + b2. [RU-93 A/C]
1
Solution: Given that a and b are roots of 2x2 – 4x + 1 = 0 or x2 – 2x + = 0,
2
1
So, a + b = 2 and ab =
2
Now, sum of the roots of the required equation is {(a2 + b) + (a + b2)}
= (a2 + b2) + (a + b)
= {(a + b)2 – 2ab}+ (a + b)
1
= 22 –2. +2
2
=5
And the product of the roots of the required equation is (a2 + b)(a + b2)
= ab + a2 b2+ a3 + b3
= ab + (ab)2 + (a + b)3 – 3ab(a + b)
1 1 1
= + ( )2 + 23 – 3. .2
2 2 2
23
=
4
We know that, the required equation is
x2 – (sum of the roots) x + product of the roots = 0
23
x2 – 5x + =0
4
4x2 – 20x + 23 = 0 (Answer)

91
 S. M. Shahidul Islam

6.12 Identity: When an equation holds true whatever be the value of its variables, it is
called an identity.
Example: (x + 2 )2 = x2 + 4x + 4.
2 (x2 + y) = 2x2 + 2y
x2 + 2x = x(x + 2), are identities
N.B: The following identities are used as formula in various problems. So, students should
memorize these identities.
1. (x + y)2 = x2 + 2xy + y2
= (x – y)2 + 4xy
2. (x – y)2 = x2 – 2xy + y2
= (x + y)2 – 4xy
3. x2 – y2 = (x + y)(x – y)
4. x2 + y2 = (x + y)2 – 2xy
= (x – y)2 + 2xy
5. 2(x2 + y2) = (x + y)2 + (x – y)2
6. 4xy = (x + y)2 – (x – y)2
x y x y
2 2

7. xy =    
 2   2 
8. (x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx)
9. (x + y)3 = x3 + 3x2y + 3xy2 + y3
= x3 + y3 + 3xy(x + y)
10. (x – y) = x3 – 3x2y + 3xy2 – y3
3

= x3 – y3 – 3xy(x – y)
11. x + y = (x + y)( x2 – xy + y2)
3 3

= (x + y)3 – 3xy(x + y)
12. x3 – y3 = (x – y)( x2 + xy + y2)
= (x – y)3 + 3xy(x – y)

6.13 Linear equation: a1x1 + a2x2 + ... + anxn = b; ai (i = 1, 2, 3, ..., n), b R

is a linear equation that means, the equation, which does not contain higher degree or
multiple terms of variables, is called linear equation. A linear equation always represents a
straight line in the graph.
For example: 3x + 4y + 5z = 1 is a linear equation. Here x, y, z are variables; 3, 4, 5 are
coefficients of x, y & z; 1 is constant term.

6.14 System of linear equations: The m linear equations in the n unknowns- x1, x2, ..., xn:
a11 x1 + a12 x2 + . . . + a1n xn = b1
a12 x1 + a22 x2 + . . . + a2n xn = b2
... ... ...
am1 x1 + am2 x2 + . . . + amn xn = bm

92
 Functions and Equations

where aij, bi  R; is a system of linear equations or simultaneous linear equation. That is, if
more than one linear equation makes a problem together, this problem is called a system
of linear equations.
If at least one bi (1, 2, 3, ..., m) is non - zero, the system is non- homogeneous and if
otherwise, the system is homogeneous. That is
a11 x1 + a12 x2 + . . . + a1n xn = 0
a12 x1 + a22 x2 + . . . + a2n xn = 0
... ... ...
am1 x1 + am2 x2 + . . . + amn xn = 0
is a system of homogeneous equations.
A homogeneous system always has a solution namely the zero n-tuple, 0 = (0, 0, . . ., 0)
called the zero or trivial solution. Any other solution if it exists, is called a non-zero or non
- trivial solution.
A non-homogeneous system may or may not have solution.
Example: x + 2y – 3z = 6
2x – y + 4 z = 2
4x + 3y – 4z = 14
is example of non-homogeneous system.
And x + 2y – 3z = 0
2x – y + 4 z = 0
4x + 3y – 4z = 0
is example of homogeneous system.
The following chart shows the types solutions of a system of linear equation.

System of linear
equations
Consistent
Inconsiste
nt
Unique solution More than one solution
No
solution Figure 6.6
solution
6.15 Solution methods of a system of linear equations: There are many solution
techniques of a system of linear equations. Such as
1. Substitution method.
2. Gauss elimination method.
3. Cramer's rule (Using determinant)
4. Matrix method: i) Finding inverse matrix
ii) Making echelon matrix.

93
 S. M. Shahidul Islam

1. Substitution method: In this method, finding the value of a variable in terms of other
variables and substitute it in other equations. To illustrate this method let us consider the
following problem:
x  2y - 3z  6 . . . (1)

2x - y  4 z  2 . . . (2)
4x  3y - 4z  14
 . . . (3)
From equation (1), we have value of x in terms of variables y and z as follows:
x = 6 - 2y + 3 z . . . (4)
Substituting the value of x in equation (2), we get
2(6 – 2y + 3z) – y + 4z = 2
Or, 12 – 4y + 6z – y + 4z = 2
Or, –5y + 10z = -10
Or, y – 2z = 2 [Dividing by –5]
Or, y = 2z + 2 . . . (5)
Again substituting the value of x in equation (3), we get
4(6 – 2y + 3z) + 3y – 4z = 14
Or, 24 – 8y + 12z +3y – 4z = 14
Or, – 5y + 8z = 14 – 24
Or, –5(2z + 2) + 8z = – 10 [Substituting y = 2z + 2]
Or, –10z –10 + 8z = – 10
Or, – 2z = 0
So, z=0
Substituting z = 0 in equation (5), we get y=2
Again substituting z = 0 and y = 2 in equation (4), we get x = 2
Therefore, the solution of the system is (x, y, z) = (2, 2, 0)

2. Gauss elimination method: In this method, we always try to eliminate one or more
variables from the equations to solve the system. This method will be clear by the
following solution.
Let us consider the problem:
x  2y - 3z  6 . . . (1)

2x - y  4 z  2 . . . (2)
4x  3y - 4z  14
 . . . (3)
To eliminate variable x from equation (1) and (2), we multiply equation (1) by 2 and then
subtract from equation (2), we get
– 5y +10 z = – 10
Or, y – 2z = 2 . . . (4)
Again to eliminate variable x from equation (2) and (3), we multiply equation (2) by 2 and
then subtract from equation (3), we get

94
 Functions and Equations

5y – 12z = 10 . . . (5)
Again to eliminate variable y from equation (4) and (5), we multiply equation (4) by 5 and
then subtract from equation (5), we get
– 2z = 0
So, z=0
And again to eliminate variable z from equation (4) and (5), we multiply equation (4) by 6
and then subtract from equation (5), we get
–y=–2
So, y=2
Putting the value of y & z in equation (1), we get x = 2.
So, the solution is (x, y, z) = (2, 2, 0)

3. Cramer's rule: In this method, we use determinant to solve a system of linear

equations. To make the method understand we use the same problem as follows:
x  2y - 3z  6 . . . (1)

2x - y  4 z  2 . . . (2)
4x  3y - 4z  14
 . . . (3)
Forming determinant with coefficients of x, y and z, we get
1 2 3
D= 2 1 4 = 1(4 – 12) – 2(– 8 – 16) – 3(6 + 4) = – 8 + 48 – 30 = 10
4 3 4
Forming determinant with constant terms and the coefficients of y and z, we get
6 2 3
Dx = 2  1 4 = 6(4 – 12) – 2(– 8 – 56) – 3(6 + 14) = – 48 + 128 – 60 = 20
14 3  4
Forming determinant with coefficients of x, constant terms and coefficients of z, we get
1 6 3
Dy = 2 2 4 = 1(– 8 – 56) – 6(– 8 – 16) – 3(28 – 8) = – 64 + 144 – 60 = 20
4 14  4
Forming determinant with coefficients of x, y and constant terms, we get
1 2 6
Dz = 2 1 2 = 1(– 14 – 6) – 2(– 28 – 8) + 6(6 + 4) = – 20 – 40 + 60 = 0
4 3 14
Dx 20
So, x =   2,
D 10

95
 S. M. Shahidul Islam

Dy 20
y =   2,
D 10
D 0
And z = z   0.
D 10
Therefore, the solution of the system is (x, y, z) = (2, 2, 0).

4. Matrix method finding inverse matrix: In this method, we use inverse matrix to solve
a system of linear equations. Let A be the coefficient matrix, B be the constant term matrix
and X be the variable matrix of a system of linear equations, then
AX = B
So, the solution will be X = A-1B.
To illustrate this method, let us consider the following system of linear equations.
x  2y - 3z  6 . . . (1)

2x - y  4 z  2 . . . (2)
4x  3y - 4z  14
 . . . (3)
From the system, we find the following coefficient matrix, constant term matrix and
variable matrix.
1 2  3 x 6 
     
A = 2 1 4  , X =  y  and B = 2 
4  4  z  14 
 3    
-1
To find the inverse matrix of matrix A, A let us consider the following augmented
matrix:
1 2 3 1 0 0
2 1 4 0 1 0
4 3 4 0 0 1
Subtracting two times of 1 row from 2nd row and four times of 1st row from 3rd row, we
st

get
1 2 3 1 0 0
0 5 10  2 1 0
0 5 8 4 0 1
nd rd
Subtracting 2 row from 3 row, we get
1 2 3 1 0 0
0 5 10  2 1 0
0 0  2  2 1 1
nd rd
Dividing 2 row by –5 and 3 row by –2, we get

96
 Functions and Equations

1 2 3 1 0 0
0 1  2 2/5  1/ 5 0
0 0 1 1 1/ 2  1/ 2
Adding 2 times of 3 row from 2nd row and subtracting 2 times of 2nd row from 1st row,
rd

we get
1 0 1 1/ 5 2/5 0
0 1 0 12 / 5 4 / 5 1
0 0 1 1 1/ 2  1/ 2
Subtracting 3rd row from 1st row, we get
1 0 0  4 / 5  1 / 10 1/ 2
0 1 0 12 / 5 4 / 5 1
0 0 1 1 1/ 2  1/ 2
  4 / 5  1 / 10 1/ 2
 
-1
So, A =  12 / 5 4 / 5 1 
 1  1 / 2 
 1/ 2
Therefore, X = A-1B
 x    4 / 5  1 / 10 1/ 2  6 
    
Or,  y  =  12 / 5 4 / 5 1   2 
z   1  1 / 2  14 
   1/ 2
 24 1 
   7
x  5 5 
   72 8 
Or,  y  =    14 
z   5 5 
  6 1 7 
 
 
 x   2
   
Or,  y  =  2
 z  0
   
That is, x = 2, y = 2 and z = 0
So, the solution is (x, y, z) = (2, 2, 0).

5. Matrix method making echelon matrix: In this method, forming the augmented
matrix with the coefficients of x, y, z and the constant terms of the system, we try to make
it echelon form. Then we check it consistency by the following rules:

97
 S. M. Shahidul Islam

If A|B is the row reduced echelon matrix of a system:

iv) If rank of matrix A ≠ rank of augmented matrix A|B, then the system is
inconsistent.
v) If rank of matrix A = rank of augmented matrix A|B = number of variables,
then the system is consistent and has a unique solution.
vi) If rank of matrix A = rank of augmented matrix A|B < number of variables,
then the system is consistent and has many solutions.
To explain the method, let us again consider the system of 3 linear equations with 3
variables x, y and z that we have solved by the previous four methods:
x  2y - 3z  6 . . . (1)

2x - y  4 z  2 . . . (2)
4x  3y - 4z  14
 . . . (3)
We form the following augmented matrix with the coefficients of x, y, z and the constant
terms of the system.
1 2 3 6 1 2 3 
 
A|B = 2  1 4 2 [Here, matrix A means  2  1 4 ]
3  4 14 4 3  4 
4 
Now, our goal is to reach the echelon form of the augmented matrix by row reduced
technique. Subtracting 2 times of 1st row from 2nd row and 4 times of 1st row from 3rd row,
we get
1 2 3 6
≈ 0 5 10  10
0 5 8 10
Again subtracting 2 row from 3rd row, we get
nd

1 2 3 6
≈ 0 5 10  10
0 0 2 0
Now, dividing 2nd row by –5 and 3rd row by –2, we have
1 2 3 6
≈ 0 1 2 2 , which is the echelon form.
0 0 1 0
In the echelon form, we see that there are 3 non-zero rows both in matrix A and
augmented matrix A|B. So, the rank of A = the rank of A|B = 3 = number of variables.
Therefore, the considered system is consistent and has unique solution. To find the
solution, we form the following equations by the rows of the echelon matrix:

98
 Functions and Equations

z=0
y – 2z = 2 => y=2
x + 2y –3z = 6 => x=2
Thus, the solution of the system is (x, y, z) = (2, 2, 0)

Example: Solve the following system of two linear equations and express it by a graph.
2x + y – 11 = 0 [RU-81 A/C]
3x + 5y –27 = 0
Solution: We solve it by substitution method and try to express it by graphical method.
Given that, 2x + y – 11 = 0 --- (1)
3x + 5y –27 = 0 --- (2)
From equation (1) we get, y = 11 – 2x --- (3)
Using equation (3) in equation (2) we get, 3x + 5(11 – 2x) – 27 = 0
Or, 3x + 55 – 10x – 27 = 0
Or, – 7x = – 28
Or, x = 4
Substituting x = 4 in equation (3) we get, y = 11 – 2  4 = 3
So, the required solution: x = 4, y = 3.
Substituting x = 1 and 4 in equation (1) we get y = 9 and 3. That is, from equation (1) we
get the following coordinates (1, 9) and (4, 3).
Substituting x = –1 and 4 in equation (2) we get y = 6 and 3. That is, from equation (2) we
get the following coordinates (–1, 6) and (4, 3).
Plotting the coordinates (1, 9), (4, 3) and (–1, 6), (4, 3) in the graph we get two straight
lines AB and CD respectively. They intersect each other at the point (4, 3). It means the
solution of the system: x = 4 and y = 3.
Y
_
10
9
_ A
_
8
C 7
_
_
6
_
5
_
4
3
_
_
 (4, 3)
2 D
_
1 B
| | | | | | | | | | | X
0 1 2 3 4 5 6 7 8 9 10 11
Figure 6.7

Example: Solve the system: 27 x  9 y and 81 = 243(3x) [NU-98 A/C]

Solution: We solve the system by substitution method.
Given that, 27 x  9 y - - - (1)
81 = 243(3x) - - - (2)

99
 S. M. Shahidul Islam

From equation (2) we get, 81 = 279x

81
Or, x =
279
1
So, x =
9
   
x
Now from equation (1) we get, 33  32
y

Or, 33 x  32 y
Or, 3x = 2y
1 1
Or, 3  = 2y [Substituting the x = ]
9 9
1
Or, 2y =
3
1
So, y =
6
1 1
Therefore, the required solution: x = and y = (Answer)
9 6

Example: Solve the system: x + y = 5 [RU-95 A/C]

x2 + y2 = 25
Solution: Given that, x + y = 5 - - - (i)
x2 + y2 = 25 - - - (ii)
From equation (i) we get, y = 5 – x - - - (iii)
Using equation (iii) in equation (ii) we get, x2 + (5 – x)2 = 25
Or, x2 + 52 – 2.5.x + x2 = 25
Or, 2x2 + 25 – 10x – 25 = 0
Or, 2x2 – 10x = 0
Or, 2x(x – 5) = 0
Or, x(x – 5) = 0
So, x = 0, 5
From equation (iii), y = 5 when x = 0
y = 0 when x = 5
Therefore, the solution: (x, y) = (0, 5), (5, 0) (Answer)

Example: An investor plan to invest 10000 taka in the bonds of two companies to receive
820 taka interest. First company pay 7% interest while second company pay 10% interest.
How much should be invested in each company? [JU-95]

100
 Functions and Equations

Solution: Let the investor invest Tk. x in the first company and Tk. y in the second
company in order to receive total of Tk. 820 as interest. So, we find the following system
of linear equations:
 x  y 10000

 x (7%)  y (10%)  820
 x  y 10000 . . . (i)
Or, 
0.07 x  0.1y  820 . . . (ii )
Multiplying equation (i) by 0.1 and then subtracting from equation (ii), we get
– 0.03x = –180
So, x = 6000
Substituting x = 6000 in equation (i), we get
6000 + y = 10000
So, y = 4000
Therefore, the investor should invest Tk. 6000 in the first company and Tk. 4000 in the
second company.

Example: (Market equilibrium for two competing products) Given supply and demand
functions for a product, market equilibrium exists if there is a price at which the quantity
demanded equals the quantity supplied. Suppose that the following demand and supply
functions have been estimated for two competing products.
q d 1  f1 ( p1 , p 2 )  100  2 p1  3 p 2

q s1  h1 ( p1 )  2 p1  4
q d 2  f 2 ( p1 , p 2 )  150  4 p1  p 2

q s 2  h2 ( p 2 )  3 p 2  6
where q d 1  quantity demanded of product 1
q s1  quantity supplied of product 1
q d 2  quantity demanded of product 2
q s 2  quantity supplied of product 2
p1  price per unit of product 1
p 2  price per unit of product 2
q1  equilibrium quantity of product 1
q 2  equilibrium quantity of product 2
Notice that the quantity demanded of a given product depends not only on the price of the
product but also on the price of the competing product. The quantity supplied of a product
depends only upon the price of that product.
Market equilibrium would exist in this two-product marketplace if prices were offered
such that

101
 S. M. Shahidul Islam

q d 1  q s1
and qd 2  q s 2
Supply and demand are equal for product-1 when
100 – 2 p1 + 3 p 2 = 2 p1 – 4
Or, 4 p1 – 3 p 2 = 104 . . . (i)
Supply and demand are equal for product-2 when
150 + 4 p1 – p 2 = 3 p 2 – 6
Or, – 4 p1 + 4 p 2 = 156 . . . (ii)
Solving equations (i) and (ii) simultaneously, we get the equilibrium prices
p1 = 221 and p 2 = 260
This result suggests that if the products are priced accordingly, the quantities demanded
and supplied will be equal for each product. And the equilibrium quantities will be
q1  438 and q 2  774

Example: Demand and supply equations are (q + 20)(p + 10) = 400 and q = 2p – 7
respectively, where p stands for price and q for quantity. Find the equilibrium price and
quantity and show it by a graph. [NU-2000 A/C]
Solution: Given that, Demand: (q + 20)(p + 10) = 400 --- (1)
Supply: q = 2p – 7 --- (2)
Using equation (2) in equation (1), we get
(2p – 7 + 20)(p + 10) = 400
Or, (2p + 13)(p + 10) = 400
Or, 2p2 + 33p + 130 – 400 = 0
Or, 2p2 + 33p – 270 = 0
Or, 2p2 + 45p – 12p – 270 = 0
Or, p(2p + 45) – 6(2p + 45) = 0
Or, (2p + 45)(p – 6) = 0
 45
Or, p=6&p= [Not acceptable]
2
So, p=6
Replacing p = 6 in equation (2), we get
q=5
Therefore, the equilibrium price is 6 and quantity is 5. This is shown in following graph:

102
 Functions and Equations

10
_ Supply
_
9
_
8
_
Price 7 _ (q =5, p =6)
6
_
5
_ Demand
4
_
3
_
2
_
1
| | | | | | | | | | |
0 1 2 3 4 5 6 7 8 9 10 11
Quantity
Figure 6.8

6.16 Break-Even point: Of specific concern in break-even analysis is identifying the

level of operation or level of output of a business that would result in a zero profit. This
level of operations or output is called the break-even point. The break-even point is a
useful reference point in the sense that it represents the level of operation at which total
revenue equals total cost. Any changes from this level of operation will result in either a
profit or a loss. Break-even analysis is valuable particularly as a planning tool when firms
are contemplating expansions such as offering new products or services. Similarly, it is
useful in evaluating the pros and cons of beginning a new business venture. In each
instance the analysis allows for a projection of profitability.

6.17 Break-Even interpretation: In this section we shall consider the case of a

manufacturer who produces q units of a product and sells the product at a price of Tk. p
per unit. The symbols to be used are:
C = Total cost of producing and selling q units
q = Number of units produced and sold.
v = Variable cost per unit made, assumed to be constant.
F = Fixed cost, a constant
p = Selling price per unit.
R = Total revenue received, which is the same as the volume of sales.
The cost function then is given by,
C = vq + F . . . (i)
And Revenue = (Price per unit)  (Number of units sold)
R = pq . . . (ii)
If the manufacturer is to achieve break-even on operations, neither incurring a loss nor
earning a profit, revenue (ii) must be equal to cost (i). That is, at beak-even,
pq = vq + F . . . (iii)
We now may solve (iii) for the production volume, q
pq – vq = F
Or, q (p – v) = F

103
 S. M. Shahidul Islam

F
 q= , this is the break-even quantity.
pv
_ Revenue
10 Profit
_
9 Break-even
_
8
7
_ Point
_ Cost
6 Variable cost
_
5
_
4
3
_ Loss
_
2
_ Fixed cost
1
| | | | | | | | | | |
0 1 2 3 4 5 6 7 8 9 10 11
Figure 6.9

Example: A manufacturer of cassette tapes has a fixed cost of Tk. 60,000 and variable
cost is Tk. 4 per cassette produced. Selling price is Tk. 7 per cassette.
a) Write the revenue and cost equations.
b) At what number of units will break-even occur?
c) At what sales (revenue) volume will break-even occur?
d) How much profit will be earned if 25000 cassettes are produced? [AUB-02]
Solution: Here, Fixed cost, F = Tk. 60,000,
Selling price per cassette, p = Tk. 7
Variable cost per cassette, v = Tk. 4
Let R means total revenue, C means total cost and q means the number of cassettes
produced and sold.
a) The equations are
Revenue: R = 7q
Cost: C = 4q + 60,000
b) We know that break-even quantity:
F
q= [Here, F = Tk. 60000, P = Tk. 7 and V = Tk. 4]
pv
60000
=
74
= 20,000 units
c) At break-even 20,000 cassettes would be produced and sold at Tk. 7 each. So, the
break-even sales volume would be: R = Tk. (7) (20,000)
= Tk. 1,40,000
d) If 25,000 cassettes produced and sold, then profit = Tk. {7  25000 – (4  25000 +
60000)}
= Tk. 15000

104
 Functions and Equations

6.17.1 Another interpretation of break-even: If cost = $ c, retail price = $ p,

pc
Then, mark-up = $ (p – c) and Margin, M =
p
pc  p pc c
So, cost = (1 – M)p = (1 – )p =   p = p = c
p  p  p
Generally, we let, fixed expense = $ F
Now, if total sales of a firm be $ x, then
Total cost, y = (1 – M)x + F
When break-even occurs, total cost must be equal to the total selling price.
So, from (1), we get.
x = (1 – M)x + F
Or, x – (1 – M)x = F
Or, x – x + Mx = F
Or, Mx = F
F
Or, x=
M
F
Thus, break-even occurs at sales, x =
M
Example: Margin is to be 33% of retail price and other variable cost is estimated at $ 0.13
per dollar of sales. Fixed cost is estimated at $ 4000.
a) What is the break-event point?
b) Estimate profit of sales for $ 50000.
Solution: Here, net margin, M = 33% - 0.13
= 0.33 - 0.13
= 0.20
And fixed cost F = $ 4000
Total cost equation: y = (1 – M)x + F
Or, y = (1 – 0.20)x + 4000
Or, y = 0.80x + 4000
F $4000
a) Break-even point is = = $ 20000
M 0.20
b) Here, total sales, x = $ 50000
Hence, total cost, y = (0.80)(50000) + 4000 dollars
= 40000 + 4000 dollars
= 44000 dollars
Therefore, estimated profit = (50000 - 440000) dollars
= 6000 dollars

105
 S. M. Shahidul Islam

6.18 Exercises:
1. Define identity and inequality with example.
2. What is difference between equation and identity?
3. Define linear equation and system of linear equations. Give a example of a system
of linear equation and solve it.
4. Define degree of an equation. Write an equation of 5 degree.
5. What do you mean by break-even quantity? Interpret the break-even quantity.
6. Write the equation of the line that has a slope of –3.1 and a y-intercept of 10.
[Answer: y = -3.1x + 10]
7. If A = {2, 3, 5} and B = {3, 7} then find the relation, R as x < y where x  A and
y  B, [Answer: {(2, 3), (2, 7), (3, 7), (5, 7)}]
8. If A = {$3, $7, $8} is a set of cost of per unit product and B = {$5, $8} is the set of
selling price of per unit product of a production firm. Find the profitable relation
between cost and selling price. [Answer: {($3, $5), ($3, $8), ($7, $8)}]
9. Write two polynomials of degree 6 and 10 respectively. [Answer: Many answers,
in particular, f ( x)  3x 6  7 x 2  x  1 , f ( x)  5  6 x  4 x 7  2 x10 ]
4
10. Solve: (i) 4(3x + 2) ≥ 2x, (ii) 5(3 – 2x) < 3(4 – 3x) [Answer: (i) x ≥  , (ii) x > 3]
5
11. A student bought x pencils per pencil $5 and bought (x + 4) pens per pen $8. At
best how many pencils did he buy, if his total spent not more than $97? [Answer:
At best 5 pencils.]
12. Solve the quadratic equation: (i) x2 – 8x + 15 = 0 [Answer: 3, 5]
(ii) x2 – 10x – 9 = 0 [Answer: –1, 9]
13. Kazi bought a bi-cycle with Rs. 6000 and sold it to Faruk earning profit x%. After
then Faruk sold it to Mizan earning profit x%. If Mizan’s cost price is greater than
Kazi’s cost price by Rs. 2640, what is the value of x? [Answer: x = 20.]
14. In the memorable cricket match between Bangladesh and Pakistan in the World
Cup’99 that Bangladesh won, one of Bangladeshi player scored 42 runs. As a
result his average run increased by 2. Before this match he played x number of
matches and his total score was 198. Find the value of x. [Answer: 11]
15. If a and b be two roots of x2 – 3x – 10 = 0, then find the quadratic equation whose
roots are a2 and b2. [Answer: x2 – 29x +100 = 0]
16. Solve the following systems of linear equations by Cramer’s rule:
i) x + 2y + z = 0 ii) 3x + 2y –5z = -8
x – 2y – 2z = 0 2x + 3y + 6z = 37
3x + y + 2z = 11 x – y + 6z = 7
[Answer: (x, y, z) = (2, -3, 4)] [Answer: x =2, y = 3, z = 4]
iii) 2x + y +2z = 10 x y
 2
x – y + 2z = 5 iv) a b
x+y+z=6 ax  by  a 2  b 2
[Answer: (x, y, z) = (1, 2, 3)]
[Answer: (x, y) = (a, b)]
106
 Functions and Equations

17. Solve the following systems of linear equations by any method:

i) x + 2y – z = 2 ii) x + 2y – 3z = -1
2x + y + z = 1 5x + 3y – 4z = 2
x + 5y – 4z = 5 3x – y + 2z = 7
[Answer: It has many solutions. [Answer: No solution.]
(-1, 2, 1) is a particular solution.]

iii) x + 2y – 3z = 4 iv) x+y+z=4

x + 3y + z = 11 2x – y + z = 3
2x + 5y – 4z = 13 x – 2y + 3z = 5
2x + 6y + 2z = 22 [Answer: (1, 1, 2)]
[Answer: (1, 3, 1)]

18. The sum of the digits of a two-digit number is 12 and if the digits are reversed the
number is decreased by 18. Find the number? [Answer: 75]
19. Solve the following simultaneous linear equations using determinant:
2x – 3y = 3
4x – y = 11 [Answer: x = 3, y = 1] [CMA-93]
20. Solve the system:
2x + 3y = 5 [NU-97]
xy = 1
21. Solve the following simultaneous equations:
3x2 – 5x – 3y = – 4 [NU-2000]
2x + 3y = 10
22. It takes 20 minutes and costs $2 to make one chair, whereas it takes 30 minutes
and costs $1 to make one table. If 600 minutes and $40 are available, how many
chairs and tables can be made? [Answer: 15 chairs and 10 tables]
23. Market Equilibrium problem: Given the following demand and supply functions
for two competing products,
q d 1  82  3 p1  p 2
q s1 15 p1  5
q d 2  92  2 p1  4 p 2
q s 2  32 p 2  6
Determine whether there are prices, which bring the supply and demand levels into
equilibrium for the two products. If so, what are the equilibrium quantities?
[Answer: p1  5, p2  3, q1  70, q2  90]
24. Demand (D) and supply (S) functions of a product, D = 17 – 4p – p2 and S = 4p –
3; where p means the price of the product per unit. Determine equilibrium price
and quantity. [Answer: p = 2, D = S = 5]

107
 S. M. Shahidul Islam

25. Demand and supply equations are 2p2 +q2 = 11 and p + 2q = 7 respectively, where
p stands for price and q for quantity. Find the equilibrium price and quantity.
5 29
[Answer: (p, q) = (1, 3) or ( , )] [RU-80 A/C, CMA-93]
9 9
26. A manufacturer of keyboards has a fixed cost of $10000 and variable cost per
keyboard made is $5. Selling price per unit is $10.
(a) Write the revenue and cost equations. [Answer: R = 10q, C = 5q + 10000]
(b) At what number of units will break-even occur? [Answer: 2000 units]
(c) At what sales volume (revenue) will break-even occur? [Answer: $20,000]
(d) Compute the loss if 1000 keyboards are produced and sold. [Answer: $5000]
27. A manufacturer has a fixed cost of Tk.7500 and variable cost per unit made is
Tk.7. Selling price per unit is Tk.10.
(a) Write the revenue and cost equations. [Answer: R = 10q, C = 7q + 7500]
(b) At what number of units will break-even occur? [Answer: 2500 units]
(c) At what sales volume (revenue) will break-even occur? [Answer: Tk.25000]
(d) Draw the break-even chart. [RU-95 A/C]
28. A manufacturer of handbag has a fixed cost of Tk.1,20,000 and a variable cost of
Tk.20 per unit made and sold. Selling price is Tk.50 per unit.
(a) Write the revenue and cost equations, using C for cost and q for number of units.
[Answer: R = 50q, C = 20q + 1,20,000]
(b) Find the break-even quantity. [Answer: 4,000 units]
(c) Find the break-even sales volume. [Answer: Tk. 2,00,000]
(d) Compute profit if 10,000 handbags are made and sold. [Answer: Tk. 1,80,000]
(e) Compute loss if 1,000 handbags are made and sold. [Answer: Tk. 90,000]
29. Compute total cost and profit/loss if sales are Tk. 40000, fixed expense is Tk.
12000 and margin is 25%. [Answer: Tk. 42000 and loss = Tk. 2000]

108
 Exponential and Logarithmic Functions

07
Chapter
Exponential and Logarithmic Functions
Highlights:
7.1 Introduction 7.4 Logarithmic function
7.2 Exponential function 7.4.1 Laws of logarithmic operations
7.2.1 Properties of Exponents 7.4.2 Relation between natural and
7.2.2 Graph of exponential function common logarithms
7.2.3 Applications of exponential functions 7.4.3 Graph of logarithmic function
7.3 Surds 7.5 Some worked out Examples
7.3.1 Formulae of surd 7.6 Exercise
7.3.2 Rationalization of surd

7.1 Introduction: The main object of this chapter is to review the nature and properties of
exponents, exponential functions, logarithms and logarithmic functions. The concept of
exponential and logarithmic functions is very useful in various parts of mathematics. We
shall look at some very important applications of these functions here and in the chapter of
mathematics of finance. Until twenty years ago, students labored with extensive tables of
logarithms and exponential values, but today we are fortunate to have these numerical
values at our fingertips via the scientific calculator and computer.

7.2 Exponential function: If a, x R, a > 0 and a ≠ 1 then the function f(x) = ax is called
an exponential function. Here, “a” is called ‘base’ and “x” is called the ‘exponent’.
Example: f(x) = 5ex
f(x) = 10x
f(x) = (2.5) x + 1 are exponential function
But f(x) = x10 is not an exponential function because the exponent “10” is not variable.

7.2.1 Properties of Exponents:

1. Any number to the power 0 is 1
i.e., a0 = 1
2. Always add exponents when multiplying two powers of the same base
i.e., ax . ay = a x +y
3. When dividing ax by a y , Subtract exponents

109
 S. M. Shahidul Islam

ax
i.e., y = a x – y
a
4. The quantity ax to the power y is equal to axy
  y
i.e., a x = axy
5. The base ab to the power x is equal to ax times ay
i.e., (ab)x = ax. by
6. The base a/b to the power x is equal to ax over bx
x
a ax
i.e.,   = x
b b
7. A base to the power –x is equivalent to one over that base to the power x
1
i.e., a-x = x
a

Example: Apply the law of exponents to simplify the following expression and write the
result with positive exponents.
3 x ( x 2 ) y 5
4 x4 y2
Solution: Given that,
3 x ( x 2 ) y 5 3 x12 y 5
=
4 x4 y2 4 x4 y2
3 x 1 y 5
=
4 x4 y2
3 1 4 52
= x .y
4
3 5 3
= x .y
4
3y3
= (Answer)
4x5

1 1 1
Example: Show that a b a c
 b c ba
 1
1 x  x 1 x  x 1  x  x c b
c a

1 1 1
Solution: L.H.S = a b a c
 b c ba

1 x  x 1 x  x 1  x  x c b
c a

1 1 1
= b c
 c a

1  x .x  x .x
a a
1  x .x  x .x
b b
1  x .x  x c .x b
c a

110
 Exponential and Logarithmic Functions

x  a .1 x b .1 x  c .1
= a  
x (1  x a .x b  x a .x c ) x b (1  x b .x c  x b .x a ) x c (1  x c .x a  x c .x b )
[Multiplying both the numerator & the denominator of first, second and third terms by
x  a , x b and x  c respectively.]
x a x b x c
= a  
x  x b  x  c x b  x  c  x  a x  c  x  a  x b
( x  a  x b  x  c )
= a
( x  x b  x  c )
= 1 = R.H.S (Proved)

8. Graph of exponential function: The graph of an exponential function is easy to

sketch. Consider the function f(x) = 2 x . To sketch the graph we determine the
following table of ordered pairs, which satisfy the function.

x 0 1 2 3 -1 -2 -3
f(x) = 2 x 1 2 4 8 0.5 0.25 0.125

Using the above ordered pairs we get the following graph:

f(x) = 2 x
6
5
4
3
2
1
-4 –3 –2 –1 0 1 2 3 4 5 6

Figure 7.1

From the graph we see that as x becomes large positively, 2 x increases rapidly; and as x
takes on values more and more negative, 2 x seems to decrease to zero. That is, the graph
of exponential function 2 x is asymptotic to the negative x-axis.

111
 S. M. Shahidul Islam

x
1
Graph of exponential function: f(x) =  
x
2
1
f(x) =  
2 6
5
4
3
2
1
-4 –3 –2 –1 0 1 2 3 4 5 6

Figure 7.2
From the above two graph we can see that exponential functions comes in two forms;
those with a > 1 increase to the right and those with 0< a <1 decrease to the right. All
exponential functions of the form ax
1. pass through the point (0, 1);
2. are positive for all values of x; and
3. tend to infinity in one direction and zero in the other.

7.2.3 Applications of exponential functions:

Example 1: (Compound Interest) The equation
F = P(1+i)n
can be used to determine the amount F that an investment of P taka will grow to if it
receives interest of i percent per compounding period for n compounding periods,
assuming reinvestment of any accrued interest. F is referred to as the compound amount
and P as the principal. If F is considered to be a function of n, the above equation can be
viewed as having the form of exponential function. That is,
F = f(n)
Or, F = Pan ; where a = 1+i
Assume that P = Tk. 1000 and i = 0.08 per period, so a = 1.08. Then we find
F = f(n) = (1000)(1.08)n
If we want to know to what sum Tk. 1000 will grow after 25 periods, we must evaluate
(1.08)25. And we find the compound amount
F = f(25) = Tk. 6848.50

Example 2: (Advertising Response) A large recording company sells tapes and CDs by
direct mail only. Advertising is done through network television. Much experience with
response to an advertising approach has allowed analysts to determine the expected

112
 Exponential and Logarithmic Functions

response to an advertising program. Specifically, the response function for classical music
CDs and tapes is R = f(t) = 1 – e-0.05t, where R is the percentage of customers in the target
market actually purchasing the CD or tape and t is the number of times an advertisement is
run on national TV.
a) What percentage of the target market is expected to buy a classical music offering
if advertisements are run one time on TV? 5 times? 10 times? 15 times? 20 times?
b) Sketch the response function R = f(t).
Solution: a) Given that the expected response to the advertising program,
R = f(t) = 1 – e-0.05t ; where t is the number of times to play the advertisement.
If the advertisement plays 1 time then the expected percentage of response,
R = f(1) = 1 – e(-0.05)(1) = 1 – 0.9512294 = 0.0488 = 0.0488  100% = 4.88%
If the advertisement plays 5 time then the expected percentage of response,
R = f(5) = 1 – e(-0.05)(5) = 1 – 0.7788007 = 0.2212 = 0.2212  100% = 22.12%
If the advertisement plays 10 time then the expected percentage of response,
R = f(10) = 1 – e(-0.05)(10) = 1 – 0.6065306 = 0.3935 = 0.3935  100% = 39.35%
If the advertisement plays 15 time then the expected percentage of response,
R = f(15) = 1 – e(-0.05)(15) = 1 – 0.4723665 = 0.5276 = 0.5276  100% = 52.76%
If the advertisement plays 20 time then the expected percentage of response,
R = f(20) = 1 – e(-0.05)(20) = 1 – 0.3678794 = 0.6321 = 0.6321  100% = 63.21%
b) The graph is as follows
0.6 f(t) = 1 – e-0.05t
0.5
0.4
0.3
0.2
0.1
-10 -5 0 5 10 15 20

Figure 7.3

7.3 Surds: A surd is defined as the irrational root of a rational number of the type n a ,
where it is not possible to extract exactly the nth root of “a”. In other words, a real number
n
a is called a surd, if and only if
i) it is an irrational number, and
ii) it is a root of a rational number.
In the surd n a , the index “n” is called the order of the surd and “a” the radicand. A surd
can always be expressed with fractional indices.
1
n n
i.e., a = a

113
 S. M. Shahidul Islam

1
10 10
And 5 = 5 etc.
Illustration: 2 , 3 and 3 7 are surds, since 2, 3 and 3
7 are the irrational roots
of the rational numbers 2, 3 and 7 respectively.
 
1
But 4 16 is not a surd, because 4 16 = 2 4 4 = 2 is not an irrational number.

7.3.1 Formulae of surd: 1. n

a .n b  n ab ; p n a  q n b  pq n ab
n
a a pn a pn a
2. n
n ; n

b b q b q b
3. m n
a  mn a  n m
a
am 
pn
4. n
a pm
5.  a
n
m
 n am
1 1
6. n
a an ; a a2
7. p n a  q n a  ( p  q)n a

7.3.2 Rationalization of surd: If multiplication of two surds makes a rational number

then this multiplication is known as rationalization of surds. One of these surds is called
the rationalizing factor of other. 5 is the rationalizing factor of 5 because 5  5  5 ,
rational number.
7 3 5 2
Example: Rationalize the denominator of the irrational fraction:
48  18
7 3  5 2 (7 3  5 2) ( 48  18 )
Solution: Given that 
48  18 ( 48  18 )( 48  18 )
[Multiplying numerator and denominator by ( 48  18 ) ]
(7 3  5 2 )(4 3  3 2 )
=
48  18
84  21 6  20 6  30
=
30
114  41 6
= [Answer]
30

114
 Exponential and Logarithmic Functions

1
 n  14 n
 9 . 3.3 
n
Example: Show that  n
 = 27 [AUB-2002]
 3 3 
 
1
 n
1
 n
 9 . 3.3 
n 4
Solution: L.H.S =  
n
 3 3 
 
1
 2 n  14 n 1 n
   
1
3 .3 2 
=  
 
1
 3. 3  n 2 
1
 2n
1 n 1
 n
 3 .3 2 2 
=  n 
 31 2 
 
1
 2 n  12  n21 n
3 
=  2 n 
 3 2 
 
1
 2 n  1  n 1 2 n  n
=  3 2 2 2 
 
1
 4 n 1 n21 2 n  n
=  3 

 
1
 6n  n
=  3 2 
 
6n 1
.
2 n
=3
= 33
= 27 = R.H.S
So, L.H.S = R.H.S (Proved)

7.4 Logarithmic function: If ax = n; a >0 and a  1 then “x” is said to be the logarithm of
the number n to the base “a”. Symbolically it can be expressed as follows:
Log a n = x

115
 S. M. Shahidul Islam

We read it as the logarithm of n to the base a is x

Note: 1) The logarithm of a number to the base “e” (e = 2.718282 approximately) is called
“Natural logarithm” or “The Napierian logarithm” and is denoted by ln N.
That is, ln N = loge N
2. The logarithm of a number to the base 10 is called “Common logarithm” or “The
Briggsian logarithm” and is denoted by log N.
That is, log N = log 10 N
When no base is mentioned, it is understood to be 10, that is, the common logarithm.
Logarithmic functions are inverses of exponential functions. Ln x is inverse of ex and
log 10 x is inverse of 10 x . That is,
Antilog10 x = 10 x so, antilog (log10 x) = x = 10log10 x
Antiln x = e x so, antiln (ln x) = x = eln x

7.4.1 Laws of logarithmic operations:

1. The logarithm of 1 to any base is 0,
i.e., log a1 = 0 because of a0 = 1
2. The logarithm of any quantity to the same base is unity,
i.e., loga a = 1 because of a1 = a
3. The logarithm of the product of two numbers is equal to the sum of the logarithms
of that numbers to the same base,
i.e., loga (mn) = loga m + loga n
Proof: Let log a m = x so that ax = m - - - (i)
and log a n = y so that ay = n - - - (ii)
Multiplying (i) and (ii) we get
ax  ay = m  n
=> ax +y = mn
Then by the definition of logarithm, we have
log a (mn) = x + y
=> log a (mn) = log a m + log an (Proved)
Remark: This formula can be extended in a similar way to the product of any number of
quantities, that is,
log a (mnpq...) = log a m + log a n + log a p + log a q + ...
We should remember that log a (m + n)  log a m + log a n
4. The logarithm of the quotient of two numbers is equal to the difference of their
logarithms to the same base,
m
i.e., log a  log a m  log a n
n
5. The logarithm of the number raised to a power is equal to the index of the power
multiplied by the logarithm of the number to the same base,
i.e., loga mn = n log a m

116
 Exponential and Logarithmic Functions

Proof: Let log a m = x so that ax = m - - - (i)

Raising the power “n” on both sides of equation (i), we get
a 
x n
 mn
=> a xn  m n
Then by the definition of logarithm, we get
log a m n  nx
So, log a m n  n log a m (Proved)
6. The logarithm of any number “b” to the base “a” is equal to the reciprocal to the
logarithm of “a” to the base “b”
1
i.e., log a b =
log b a
Or, log a b  log b a = 1
Proof: Let log a b = x so that ax = b - - - (i)
1
y y
and log b a = y so that b = a => b = a - - - (ii)
From equation (i) we have
ax = b
1

Or, ax = a y [Using equation (ii)]

1
Or, x=
y
1
i.e., log a b =
log b a
So, log a b  log b a = 1 (Proved)

7.4.2 Relation between natural and common logarithms: The natural logarithm of a
number is equal to the quotient of the common logarithm of that number and the common
logarithm of e
log m log 10 m
that is, ln m = Or, log e m 
log e log 10 e
Proof: Let log e m  x so that e = m x
- - - (1)
log 10 m = y so that 10 y = m - - - (2)
1
z z
and log 10 e = z so that 10 = e => 10 = e - - - (3)
From equations (1) and (2), we get
ex = 10 y

117
 S. M. Shahidul Islam

y
 1z 
=> e =  e 
x
[Using equation (3)]
 
y
z
=> ex = e
y
=> x =
z
log 10 m
i.e., log e m 
log 10 e
log m
Or, ln m = (Proved)
log e

7.4.3 Graph of logarithmic function: The graph of a logarithmic function is easy to

sketch. Consider the function f(x) = ln x. To sketch the graph we determine the following
table of ordered pairs, which satisfy the function.
x 0 1 2 3 4 5 6
f(x) = ln x - ∞ 0 0.69 1.99 1.39 1.61 1.79
Using the above ordered pairs we get the following graph:

2.0
1.5
1.0
0.5
-3 -2 -1 0 1 3 4 5 6 7 8
-0.5
-1.0

Figure 7.4

From the graph we see that the curve rises slowly, but always rises, to the right of x = 1.
Recalling that x can not be zero or negative, we conclude that to the left of x = 1 the curve
falls indefinitely as x gets closer to zero, but never touches the y-axis.

7.5 Some worked out Examples:

Example (1): Find the value of log 6 6 6 .
Solution: Let x = log 6 6 6
Then by the definition of logarithm, we get

118
 Exponential and Logarithmic Functions

6x  6 6
1 1 3
1
Or, 6 x  6.6 2 = 6 2 = 6 2
3
So, x = (Answer)
2

Example (2): Find the value of x: log x 25  2 .

Solution: Given that, log x 25  2
By the definition of logarithm, we get
x 2  25
Or, x 2  52
So, x = 5 (Answer)

Example (3): Using logarithm find the value of 1000(1.06)110. 5

Solution: Let x =1000 (1.06)110. 5
Taking common logarithm on both sides, we have
log 10 x = log 10 {1000 (1.06)110. 5}
= log 10 1000+ log 10 (1.06)110. 5
= log 10 (10)3 + 110.5 log 10 1.06
= 3 log 10 10 +110.5 log 10 1.06
= 3  1 + 110.5  0.0253
= 3 + 2.79565
= 5.79565
Taking anti logarithm, we have,
anti log( log 10 x ) = anti log 5.79565
x = 105.79565
= 624669.07 (Answer)

Example (4): Assume that, for some base, log x = 0.5, log y = 1.5 and log z = 3. Compute
 
 xy 
the value of log  1  . [RU-91]
 3
z 
Solution: Given that, log x = 0.5, log y = 1.5 and log z = 3
  1
 xy 
Now, log  1  = log (xy) – log z 3
 3
z 

119
 S. M. Shahidul Islam

1
= log x + log y – log z
3
= 0.5 + 1.5 –
1
3
3
= 2.0 – 1
=1 (Answer)

Example (5): Solve for x: 3x = 8

Solution: Given that, 3x = 8
Taking common logarithm on both sides, we have
log 3x = log 8
Or, x log 3 = log8
log 8
Or, x =
log 3
0.903090
Or, x =
0.477121
So, x = 1.893 (approximately) (Answer)

Example (6): Solve for x: 9x = 2

Solution: Given that, 9x = 2
Taking common logarithm of both sides, we have
log 9x = log 2
Or, x log 9 = log 2
log 2
Or, x =
log 9
0.3010
Or, x = [Using calculator]
0.9542
So, x = 0.31546 (approximately)

Example (7): Find the value of i when 10000 (1 + i)10 = 30000 [AUB-03,]
Solution: Given that, 10000(1 + i)10 =30000
30000
Or, (1+ i)10 =
10000
Or, (1+ i)10 = 3
Or, ln (1+ i)10 = ln 3 [Taking natural logarithm of both sides]
Or, 10 ln (1+ i) = 1.0986 [Using calculator]
Or, ln (1+ i) = 0.10986
Or, antiln{ln(1+ i)} = antiln 0.10986
Or, 1+ i = 1.11612 [Using calculator]

120
 Exponential and Logarithmic Functions

Or, i = 1.11612 – 1
So, i = 0.11612 (Answer)

Example (8): Solve the equation for x: ln (1 + x) = – 0.5

Solution: Given that, ln (1 + x) = – 0.5
Or, Antiln ln (1+ x) = Antiln (–0.5) [Taking Antiln of both sides]
Or, 1 + x = 0.60653
Or, x = 0.60653 – 1
So, x = – 0.39347 (Answer)

Example (9): If log 10 2 = 0.3010, then find the value of log 8 25 [AUB-02]
Solution: Given that, log 10 2 = 0.3010
log 25 log 10 m
Now, log 8 25  10 [We know, log e m  ]
log 10 8 log 10 e
 100 
log 10  
=  4 
log 10 2 3
log 10 100  log 10 4
=
3 log 10 2
log 10 (10) 2  log 10 (2) 2
=
3 log 10 2
2 log 10 10  2 log 10 2
=
3 log 10 2
2  1  2  0.3010
=
3  0.3010
= 1.548 (Answer)

16 25 81
Example (10): Show that log 2 + 16 log + 12 log + 7 log =1
15 24 80
[CMA-94, NU-94]
16 25 81
Proof: L.H.S = log 2 + 16 log + 12 log + 7 log
15 24 80
16 12 7
 16   25   81 
= log 2 + log   + log   + log  
 15   24   80 

121
 S. M. Shahidul Islam


 2 
4 16
 52 
12
 34  
7

= log 2      3    4  

 
3.5   2 .3   2 .5  
 2 64 5 24 328 
= log  2  16 16  36 12  28 7 
 3 .5 2 .3 2 .5 

= log 21643628  3281216  524167 

= log 21  30  51 
= log 2  1 5
= log 10
= 1 [We know that log 10 10 1 ]
= R.H.S (Proved)

1 3 log 1728
Example (11): Find the value of . [CMA-93]
6 1 1
1  log 0.36  log 8
2 3
1 3 log 1728 1 3  3.2375437
Solution: . = .
6 1 1 6 1 1
1  log 0.36  log 8 1   (0.4436975)   0.9030899
2 3 2 3
[Using calculator]
1 9.7126311
= .
6 1  0.2218487  0.3010299
1 3.1165094
= .
6 1.0791812
3.1165094
=
6.4750872
= 0.4813077 (Answer)

1 1 1
Example (12): Show that    1 [NU-95]
log a bc  1 log b ca  1 log c ab  1
1 1 1
Solution: L.H.S =  
log a bc  1 log b ca  1 log c ab  1
1 1 1
=  
log a bc  log a a log b ca  log b b log c ab  log c c
1 1 1
=  
log a abc log b abc log c abc

122
 Exponential and Logarithmic Functions

= log abc a  log abc b  log abc c

= log abc (abc)
= 1 = R.H.S (Proved)

  
Example (13): Find the value of log 2 log 2 log 3 log 3 27 3  [AUB-02 MBA]

Solution: log 2 log 2 log 3 log 3 27  = log log log log 3 
3
2 2 3 3
9

= log 2 log 2 log 3 9 log 3 3

= log 2 log 2 log 3 9  1
= log 2 log 2 log 3 9

= log 2 log 2 log 3 32  
= log 2 log 2 2 log 3 3
= log 2 log 2 2
= log 2 1
=0 (Answer)

7.6 Exercise:
1. Define exponential function. And discuss the properties of exponents.
2. What do you mean by surd?
3. Define natural and common logarithms. What is difference between natural and
common logarithms?
4. State and prove relation between natural and common logarithms.
5. (Credit Card Collections) A major bank offers a credit card, which can be used
internationally. Data gathered over time indicate that the collection percentage for
credit issued in any month is an exponential function of the time since the credit
was issued. Specifically, the function approximating this relationship is
C = f(t) = 0.92(1- e-0.1t); t ≥ 0
where C equals the percentage of accounts receivable (in taka) collected t months after the
credit is granted.
(a) What percentage is expected to be collected after 1 month?
[Answer:8.75%]
(b) What percentage is expected after 3 months? [Answer: 23.84%]
(c) What value does C approach as t increases without limit (t → ∞)?
[Answer:92%]
3

6. Find the value of (i) (23)2 (ii) (121)0.5 (iii) 16 4
[Answer: (i) 64 (ii) 11 (iii) 1/8]

  
1


7. Find the value of 1  1  1  x 3 1 1 3
[Answer:
1
]
 x

123
 S. M. Shahidul Islam

1 1 1
 rqpr  p  q  rppq  q r  qp rq  r  p
8. Show that  x  x  x  1
     
     
1632  2 4  55 1
m 3m  2 m 1 m 1
9. Show that [AUB-02]
152 16
m 1 m
52m
  
10. Find the value of log 5 3 5 5 [Answer: ]
5
6
11. Find the value of x: (i) log x 81  4 (ii) log 2 5 400  x [Answer: (i) 3, (ii) 4]
12. Using the definition of logarithm find the value of x: log x (4 x  3)  log x 4  2
3 1
[Answer: , ] [NU-2000]
2 2
13. Solve for x: (i) 10x = 8. (ii) 2x.32x +1 = 74x + 3 [Answer: (i) 0.90309 (ii) – 0.9685]
14. Solve for x: log x 3  log x 9  log x 729  9 [Answer: 3] [AUB-01]
15. Assume that for some base, log a = 0.3, log b = 2.8642 and log c = 1.7642. Find
 3 12 
a b 
the value of log   [Answer: 0.5676] [NU-01]
 c 
 
16. Compute the value of x from 10 log5 = x [Answer: 5]
1

17. Using logarithm find the value of 789.45 8 [Answer: 2.3023]

5
10  3 10
18. Find the value of [Answer: 158.49] [RU-84]
3
10 2  3 10 3
33 33

[Hints: After simplification, we get 10 15 . Let x = 10 15 and then use log and antilog
respectively to find the value of x.]
16 25 81
19. Simplify: 7 log  5 log  3 log [Answer: log 2 ] [NU-99]
15 24 80
20. Find the value of log 2 6  log 2 2 [Answer: 1]
3
 b bb 
a

21. Show that log b log b log b

  = a.
3 5 1
22. Show that log 2  log 2  log 2 5  
2 3 2
1 1 1 1
23. Show that (i)    1 [Hints:  log abe a ]
log a (abc) log b (abc) log c (abc) log a (abc)

124
 Exponential and Logarithmic Functions

1 1 1
(ii)   2
log pq ( pqr ) log qr ( pqr ) log rp ( pqr )
1 1 1
(iii)   2 [CMA-95]
log 6 24 log 8 24 log 12 24
24. Using the definition of logarithm find the value of x from log 2 log 3 log 2 x 1.
[Answer: 512] [AUB-02]
log a log b log c
25. If   , prove that
x y yz zx
(i) abc = 1
log a log b log c log a
(ii) a x y .b y  z .c z  x  1 [Hints: let   = M, So, M
x y yz zx x y
Or, log a = M(x – y) Or, (x + y)log a = M(x – y)(x + y) Or, log a(x + y) = M(x2 – y2).
Similarly, log a(y + z) = M(y2 – z2) and log a(z + x) = M(z2 – x2). Now add.]

125
 S. M. Shahidul Islam

08
Chapter
Mathematics of Finance
Highlights:

8.1 Introduction 8.5.1 Effective interest rate

8.2 Simple interest and the future value 8.5.2 Future value with continuous
8.2.1 Definition of simple interest compounding
8.3 The yield on the common stock of a 8.6 Ordinary annuity
company 8.6.1 Present value of ordinary annuity.
8.4 Bank discount 8.7 Exercise
8.5 Compound Interest and the future
value

8.1 Introduction: This chapter is concerned with interest rates and their effects on the
value of money. We shall discuss the nature of interest and its computational processes.
Interest rates have widespread influence over decisions made by businesses and by us in
our personal lives. Corporations pay millions of dollars in interest each year for the use of
money they have borrowed. We earn money on sums we have invested in savings
accounts, certificates of deposit, and money market funds. We also pay for the use of
money, which we have borrowed for school loans, credit card purchases or mortgages.
The interest concept also has applications that are not related to money such as population
growth.

8.2 Simple interest and the future value: Interest rates are generally quoted in
percentage form and, for use in calculations, must be converted to the equivalent decimal
value by dividing the percentage by 100; that is, by moving the decimal point in the
percentage two places to the left. For example,
1
i = 5 % = 5.25% = 0.0525.
4
Unless otherwise stated, a quoted rate is a rate per year. Thus, Tk.1 at 5 percent means that
interest of Tk. 0.05 will be earned in a year, and Tk. 100 at this rate provides.
Tk. 100(0.05) = Tk. 5
of interest in one year. Interest on Tk. 100 at 5 percent for 9 months is interest for 9/12
year; that is,

126
 Mathematics of Finance

Interest = Tk.100 (0.05) (9/12) = Tk. 3.75.

Interest = (Principal) (Rate of interest) (Time in year)

The last line introduces the following definitions of simple interest, which apply in simple
interest calculation:

8.2.1 Definition of simple interest: In this case, we do not calculate interest of interest;
here we calculate only interest of capital or principal. The simple interest formula is as
follow:
I = Pin
And F = P + I = P + Pin = P (1 + in)
Where I = Total interest
P = Capital, the sum of money on which interest is being earned
i = Rate of interest, interest of per unit capital for a year
n = Number of years
F = Future value, the sum of capital and total interest
When time is given in days, there are two ways of computing the interest: the exact
method and the ordinary method (often called the Banker’s Rule). If the exact method is
used, then the time is
Number of days
n =
365
but if the ordinary method is used, then the time is
Number of days
n =
360
Bank, for convenience, often count a year as twelve 30-day months, 360 days for a year.

1
Example: Compute the interest on Tk. 480 at 6 % for 9 months.
4
Solution: Here,
Capital, P = 480 Tk.
1 25 25
Rate of interest, i = 6 % = %= = 0.0625
4 4 4 x100
9
Number of years, n = year
12
We know, simple interest, I = Pin
9
= 480  0.0625  taka
12
= 22.50 taka (Answer)

127
 S. M. Shahidul Islam

Example: Find the interest rate if Tk.1000 earns Tk.45 interest in 6 months.
Solution: Here,
Capital, P = Tk. 1000
Interest, I = Tk. 45
6
Number of years, n =
12
= 0.5
Rate of interest, i = ?
We know,
I = Pin
I
Or, i =
Pn
45
=
1000 X 0.5
= 0.09
100
= 0.09  100% [We know, 100% = = 1]
100
= 9%
Hence, the required interest rate is 9%.

Example: Find the exact and ordinary interest on Rs. 1460 for 72 days at 10 percent
interest. [AUB-02]
Solution: In both case:
Capital, P = Rs. 1460
Interest rate, i = 10% = 0.1
For exact interest,
72
Number of years, n =
365
For ordinary interest,
72
Number of years, n =
360
So, exact interest, I = Pin
72
= Rs. 1460  0.1 
365
= Rs. 28.80 (Answer)
And ordinary interest, I = Pin
= Rs 1460 0.1 
72
360
= Rs 29.20 (Answer)

128
 Mathematics of Finance

Example: Find the future value of $ 5000 at 10 percent for 9 months.

Solution: Here,
Capital = $ 5000
10
Interest rate, i = 10% =  0.1
100
9 3
Number of years, n = =
12 4
We know that,
Future value, F = P(1+in)
3
= $ 5000 (1 + 0.1  )
4
= $ 5000 (1+ 0.075)
= $ 5375 (Answer)

Example: Lovlu has placed Tk. 500 in a savings account that pays 8% simple interest.
How long will it be, in months, until the investment amounts to Tk. 530?
Solution: Here,
Capital, P = Tk. 500
Future value, F = Tk. 530
8
Interest rate, i = 8% = = 0.08
100
Number of years, n =?
We know that,
F = P(1 + in)
F
Or, 1 + in =
P
F
Or, in = -1
P
FP
Or, n=
Pi
530  500
=
500  0.08
30
=
500  0.08
3
=
4
3 3
So, the required time = of a year =  12 months = 9 months (Answer)
4 4

129
 S. M. Shahidul Islam

8.3 The yield on the common stock of a company: The yield on the common stock of a
company is a percent obtained by dividing the amount (called the dividend) that a
shareholder receives per share of stock held by the price of a share of the stock. A stock
market report showing
Shain Pu 3.50 87 ½
Means that at the time of the quotation, a share of Shain Pukur sold for Tk. 87.50 and the
annual dividend was estimated to be Tk. 3.50 per share.

Example: Compute the yield for Shain Pukur from the market report Shain Pu 3.50 87 ½.
Solution: Given that,
The annual dividend = Tk. 3.50
The value of a share = Tk. 87 ½ = Tk. 87.50
3.50
So, the yield for general electric =  100%
87.50
= 4% (Answer)

8.4 Bank discount: In many loans, the interest charge is computed not on the amount the
borrower receives, but on the amount that is repaid later. A charge for a loan computed in
this manner is called the bank discount, and the amount borrower receives is called the
proceeds of a loan and is denoted by ‘P’. The future amount to be paid back is ‘F’, now
called the maturity value of the loan. The bank discount rate is denoted by ‘d’ and the loan
time is denoted by ‘n’, which is in years.
The formula is to calculate the proceeds: P = F (1- dn)

Example: (a) A borrower signs a deed promising to pay a bank $ 5000 in ten months from
now. How much will the borrower receive if the discount rate is 8.4%? (b) How much
would the borrower have to repay in order to receive $ 5000 now?
Solution: (a) Here,
The maturity value, F = $ 5000
The bank discount rate, d = 8.4% = 0.084
10
The loan time, n = 10 months = of a year.
12
We know that,
The proceeds, P = F (1- dn)
10
= 5000 (1- 0.084  ) dollar
12
= 5000(1- 0.07) dollar
= 4650 dollar
So, the borrower will receive $ 4650 (Answer)

(b) Here, The proceed value, P = $ 5000

130
 Mathematics of Finance

The bank discount rate, d =8.4% = 0.084

10
The loan time, n = 10 months = of a year
12
The maturity value, F =?
We know that,
P = F (1- dn)
P
Or, F =
1  dn
5000
= dollar
10
1  0.084 
12
5000
= dollar
1  0.07
5000
= dollar
0.93
= 5376. 34 dollar

So, the borrower will have to repay $ 5376.34 (Answer)

8.5 Compound interest and the future value: To see how compound interest works and
develop a formula for computing the future value, suppose Tk. 4000 in invested at 10%
interest compounded each year. The amount at the end of the first year would be
F1 = Tk. [4000+ 4000 (0.10) (1)]
= Tk. (4000+400)
= Tk. 4400
This Tk. 4400 becomes the principal at the beginning of the second year, and the amount
at the end of the second year is
F2 = Tk. [4400 + 4400 (0.10) 1]
= Tk. (4400+ 440)
= Tk. 4840
Thus, in the second year, interest is earned not only on Tk. 4000 invested, but also on
Tk.400 of interest earned in the first year. This common practice of computing interest on
interest is called compounding interest.
To formulate a formula for computing the future value, we will use ‘i’ as the interest rate
per period. Period means the time interval of two consecutive compounding; it may be a
year, a month, a semiannual etc.
Assuming, that the compounding period is 1 year, a capital of Tk. P will amount to

F1 = P(1 + i )

131
 S. M. Shahidul Islam

at the end of the first year. At the beginning of the second year, P(1+ i) becomes the new
principal, which is multiplied by (1+ i) to find the future value at the end of the second
year.
Thus,
F2 = P(1 + i)(1 + i) = P(1 + i)2

after two years. At the beginning of the third year, the new principal is P(1 + i) 2, and to
obtain the future value at the end of the third year, this must be multiplied by (1 + i). Thus,

F3 = P(1 + i)2(1 + i) = P(1+ i)3

after three years. Similarly, the future value at the end of 20 years would be

F20 = P(1 + i)20

and, in general, at the end of n years, the future value will be

Fn = P(1 + i)n .

Thus in this case of compound interest, we calculate the interest of interest and the capital
time to time. The compound interest formula is given by

F = P(1 + i )n

Where, F = Future value, the sum of the capital and total interest.
P = Capital or principal or present value
i = Interest per unit per period
n = Total number of periods

Example: Find the compound future value of Tk. 1000 at 7 % interest compounded
annually for 10 years.
Solution: Here,
Capital, P = Tk. 1000
Interest rate, i = 7% = 7/100 = 0.07
Total number of periods, n = 10
We know that,
Future value, F = P(1 + i )n
= Tk. 1000 (1 + 0.07)10
= Tk. 1000 (1.07)10
= Tk. 1000 (1.96715)
= Tk. 1967.15 (Answer)

132
 Mathematics of Finance

Example: If $500 is invested at 6 percent compounded annually, what will be the future
value 30 years later?
Solution: Here, Capital, P = $ 500
Interest rate, i = 6% = 6/100 = 0.06
Total number of periods, n = 30
Future value, F =?
We know that, F = P(1 + i )n
= $ 500 (1 + 0.06)30
= $ 500 (1.06)30
= $ 500 (5.7435)
= $ 2871.75
So, the future value = $ 2871.75 (Answer)

Example: If Tk. 800 is invested at 6% compounded semiannually, what will be the

amount in 5 years?
Solution: Here, Capital, P = 800 taka
6%
Interest rate per period, i = = 3% = 0.03
2
Total number of periods, n = (5 years)(2 periods per year)
= 10 periods
Future value, F =?
We know that,
F = P(1 + i )n
= 800 (1 + 0.03) 10 taka
= 800 ( 1.03 ) 10 taka
= 800 ( 1.3439 ) taka
= 1075.13 taka
So, after 5 years, total amount will be 1075.13 taka. (Answer)

Example: Find how many years it will take at 9% compounded annually for $1000 to
grow to $ 2000. [AUB-01]
Solution: Here, Capital, P = $ 1000
Future value, F = $ 2000
Interest rate, i = 9% = 0.09
Number of periods, n =?
We know that,
F = P(1 + i )n
Or, 2000 = 1000 (1 + 0.09) n
2000
Or, (1.09) n =
1000
n
Or, (1.09) = 2

133
 S. M. Shahidul Islam

Or, n Ln (1.09) = Ln 2 [Taking natural logarithm on both sides.]

Ln2
Or, n=
Ln1.09
0.6931471
Or, n=
0.0861776
Or, n = 8.0432317
So, n = 8.04 (Approximately)
Hence, the required time = 8.04 years.

Example: A man built up a scholarship fund to give prize of Tk.500.00 every year. If the
fund provides 10% interest compounded semiannually, what is the amount of the fund?
[DU-78]
Solution: Since the fund gives prize of Tk.500 every year, the amount of the interest of a
year must be equal to Tk.500.
10%
Given that, Interest rate, i = = 5% = 0.05
2
Number of periods (in a year) = 1  2
Let the amount of the fund, P = x taka
So, the future value, F = (x + 500) taka
We know that,
F = P(1 + i )n
Or, x + 500 = x(1 + 0.05) 2
Or, x + 500 = 1.1025x
Or, 0.1025x = 500
Or, x = 500
0.1025
So, x = 4878.05
Therefore, the amount of the fund = Tk.4878.05 (Answer)

Example: Find how many months it will take at 10% compounded quarterly of a year for
Tk. 5000 to grow to Tk. 20000. [DU-78]
Solution: Here, Capital, P = $ 5000
Future value, F = Tk.20000
10%
Interest rate, i = = 2.5% = 0.025
4
Number of periods (Quarter year), n =?
We know that,
F = P(1 + i )n
Or, 20000 = 5000 (1 + 0.025) n
20000
Or, (1.025) n =
5000

134
 Mathematics of Finance

Or, (1.025) n = 4
Or, n Ln (1.025) = Ln 4 [Taking natural logarithm on both sides.]
Ln4
Or, n=
Ln1.025
1.386294
Or, n=
0.024693
Or, n = 56.1411736
So, n = 56.14 (Approximately)
56.14  12
Hence, the required time = months.
4
= 168.42 months. (Answer)

Example: Find the rate of interest that compounded annually, will result in tripling a sum
of money in 10 years. [AUB-03, DU-77]
Solution: Here, let the capital = P taka
So, the future value, F = 3P taka
Total number of period, n = 10
The rate of interest, i =?
We know that,
F = P(1 + i )n
Or, 3P = P (1 + i ) 10
3P
Or, (1 + i ) 10 =
P
10
Or, (1 + i ) = 3
Or, Ln(1 + i ) 10 = Ln3
Or, 10 Ln (1 + i ) = Ln 3 [Taking natural logarithm on both sides.]
1.0986123
Or, Ln (1 + i ) =
10
Or, Ln (1 + i ) = 0.10986123
Or, 1 + i = Anti ln 0.10986123
Or, i = 1.1161231-1
Or, i = 0.1161231
Or, i = 0.1161231  100%
So, i = 11.61%
Hence, the required rate of interest = 11.61%. (Answer)

Example: What is the present value of $ 2500 payable 4 years from now at 10%
compounded quarterly of a year?
Solution: Here, the future value, F = $ 2500

135
 S. M. Shahidul Islam

10%
Interest rate per period, i =
4
= 2.5%
= 0.025
Number of periods, n = 4  4 = 16
Present value (Capital), P =?
We know that, F = P( 1 + i )n
F
Or, P=
(1  i ) n
2500
Or, P=
(1  0.025)16
2500
Or, P=
1.4845056
So, P = 1684.0624
Thus, the present value = $ 1684.0624.

Example: To buy a car Mr. Amin pays Tk.500000 in cash and promises to pay Tk.300000
(including interest) in 3 years later. Find the present value of the car if he pays 12% annual
interest compounded semiannually. [NU-96]
Solution: The present value of the car = Tk.500000 + present value of Tk.300000
12%
Here, F = Tk.300000, i =  6%  0.06 , n = 3  2 = 6, P = ?
2
We know that, F = P(1 + i )n
Or, 300000 = P (1 + 0.06) 6
Or, 300000 = 1.4185191P
300000
Or, P=
1.4185191
Or, P = 211488.16
Therefore, the present value of the car = Tk.(500000 + 211488.16)
= Tk. 711488.16 (Answer)

8.5.1 Effective interest rate: Because of lack of comparability, it is hard to judge whether
interest quoted at 18% compounded semiannually result in more or less interest than
would be the case if the rate were 17% compounded monthly. To make the comparison
possible, we change both to their equivalent annual rates; these equivalents are called
effective rates. To find out effective rate formula let us consider annual interest rate, j
compounded m times in a year, Tk.1 grows to
j
F = (1)(1+i)m [Interest rate per period, i = ]
m

136
 Mathematics of Finance

at the end of a year. At the effective rate, re, Tk. 1 grows to

F = 1+ re
in a year. So,
1 + re = (1)(1+i)m
Hence, the effective rate, re = (1+i)m - 1 , where m = number of periods in a year and
i = interest rate per period.

Example: Find the effective rate of 16% compounded quarterly of a year.

16%
Solution: Here, interest rate per period, i = = 4% = 0.04
4
Number of period in a year, m = 4
We know that,
Effective rate, re = (1+i)m - 1
= (1 + 0.04)4 – 1
= (1.04)4 – 1
= 0.1699
= 0.1699  100%
= 16.99% (Answer)

8.5.2 Future value with continuous compounding: In this case we calculate future value
at every time (moment) for compound interest. To formulate a formula, let us consider
compound interest formula:
F = P(1 + i )n
Where, F = Future value, the sum of the capital and total interest.
P = Capital or principal
j
i = = Interest per unit per period ( j = Interest rate per year, m = Periods in a year)
m
n = mt = Total number of periods (t = Time in years, m = Periods in a year)
For continuous compounding a period is very very small, so m is very very large that
means m tends to infinity.
So, we can write the formula as follows:
j mt m
F = P(1 + ) [Let, x = , so m = xj and when m → ∞ , x → ∞ ]
m j
1
Or, F = P(1 + )xjt [x → ∞]
x
1 x jt
Or, F = P{(1 + )} [x → ∞]
x
1
So, F = Pejt [In the chapter of limit, we knew, Lt x→ ∞ (1 + )x = e ]
x
Hence, the formula for the future value with continuous compounding is

137
 S. M. Shahidul Islam

F = Pejt.
Where, F = Future Value
P = The present value
j = Interest rate per year
t = Time in years.

Example: Find the future value of $ 500 at 8 percent compounded continuously for 9
years and 3 months.
Solution: Here, The present value, P= $ 500
Interest rate per year, j = 8% = 0.08
3
Number of years, t = (9+ ) years
12
= 9.25.
We know that, Future value, F = Pejt
= $ 500 e(0.08) (9.25)
= $ 500 e 0.74
= $ 1047.97. (Answer)

8.6 Ordinary annuity: An ordinary annuity is a series of equal periodic payments in

which each payment is made at the end of the period. To find out a formula, we shall use
the following symbols:
n = Number of periods
i = Interest rate per period
R = Payment per period
F = Future value of the annuity
Since, ordinary annuity is a series of equal periodic payments, the first payment of Tk. R
accumulates interest for n-1 periods, the second payment of Tk. R for n-2 periods, etc. The
next-to-last payment R accumulates one period of interest, and the last payment R
accumulates no interest. So using the future value formula for compound interest, we set
the future value of the annuity:
F = R(1+i)n-1 + R(1+i)n-2 + ------ + R(1+i)1 + R
= R1 + R(1+i)1 + - - - + R(1+i)n-2 + R(1+i)n-1 [This is a geometric progression of first
term R and common ratio 1+i > 1. ]
(1  i ) n  1
=R[ ]
(1  i )  1
(1  i ) n  1
=R[ ]
i
(1  i ) n  1
That is, F = R [ ]
i

138
 Mathematics of Finance

Example: If $100 is deposited in an account at the end of every quarter for the next 10
years, how much will be in the account at the time of the final deposit if interest is 8%
compounded quarterly?
Solution: Here, Payment per period, R = $100
Number of periods, n = (10 years) (4 quarters per year)
= 40 periods
8%
Interest rate per period, I = = 2% = 0.02
4
The future value, F =?
(1  i ) n  1
We know that, F = R [ ]
i
(1  0.02) 40  1
= R[ ] dollar
0.02
= 100 (60.401983) dollar
= 6040.20 dollar (Answer)

Example: A company issues $ 1 million of bonds and sets up a sinking fund at 8 percent
compounded quarterly to accumulate $ 1 million 15 years hence to redeem the bonds. Find
the quarterly payment to the sinking fund. [AUB-02, CMA-96]
Solution: Here, The future value, F = $ 1 million = $ 10,00,000
8%
The rate of interest, i = = 2% = 0.02
4
Number of periods, n = (15 years) (4 quarters per year)= 60 periods
Periodic payment, R =?
We Know that,
(1  i ) n  1
F=R[ ]
i
(1  0.02) 60  1
10,00,000 = R [ ]
0.02
10,00,000 = R (114.0515)
10,00,000
R=
114.0515
R = 8767.97.
So, the payment of every period = $ 8767.97 (Answer)

8.6.1 Present value of ordinary annuity: Present value annuity calculations arise when
we wish to determine what lump sum must be deposited in an account now if this some
and the interest it earns are to provides equal payment for a stated number of periods, with
the last payment making the account balance zero.

139
 S. M. Shahidul Islam

Form the compound interest formula we found that

Present value, P = F (1 + i )-n [F is future value.]
So the present value of any amount due n periods from now is found by multiplying the
amount by the compound discount factor,
(1 + i )-n
(1  i ) n  1
If we multiply the future amount of an annuity of R per period, R [ ], by the
i
compound discount factor we have the present value of an annuity of R per period. Hence
(1  i ) n  1
P=R[ ](1 + i )-n
i
1  (1  i )  n
So, P = R [ ];
i
where, P = Present value
R = Payment of every period
i = Interest rate for a period
n = Total numbers of periods.
The process of repaying a loan by installment payments is referred to as amortizing a
loan. To find the amortization payment we use the above formula. In the formula R is
referred to as amortization payment or installment payment per period.

Example: a) A sum of money invested now at 10 percent compounded semiannually is to

provide payments of $ 1500 every 6 months for 8 years, the first payment due 6 months
from now. How much should be invested? b) How much interest will the investment earn?
Solution: Here, number of periods, n = (8 years) (2 periods per year)
= 16 periods
10%
Interest rate, i = = 5% = 0.05
2
Payment for every period, R = $ 1500.
Present Value, P =?
1  (1  i )  n
We know that, P = R[ ]
i
1  (1  0.05) 16
= $ 1500 [ ]
0.05
= $ 1500 (10.83777)
= $ 16256.65
So, $ 16256.65 should be invested.
b) The amount of interest = (16  1500 – 16256.65) dollar
= 7743.35 Dollar.

140
 Mathematics of Finance

Example: To buy a computer Nahar borrowed Tk.50000.00 at 10% interest per year,
compounded quarterly. She will amortize the debt by equal payments each quarter over 15
years. a) Find the quarterly payment. b) How mach interest will be paid?
Solution: a) Here, number of periods, n = (15 years) (4 periods per year)
= 60 periods
10%
Interest rate, i = = 2.5% = 0.025
4
Present Value, P = Tk.50000.00
Payment for every period, R = ?
1  (1  i )  n
We know that, P = R[ ]
i
1  (1  0.025) 60
Or, 50000 = R[ ]
0.025
Or, 50000 = R(30.908656)
50000
Or, R=
30.908656
So, R = 1617.67
Therefore, the quarterly payment = Tk.1617.67
b) Payment for 60 quarters will be Tk.1617.67  60 = Tk.97060.20
So, interest paid will be (Tk.97060.20 – Tk.50000) = Tk.47060.20

8.7 Exercises:
1. Discuss simple and compound interest formulae.
2. What is difference between simple and compound interest?
3. What do you mean by period in compound interest? Discuss that the interest will
increase as the duration of a period decreases.
4. What do you understand by amortization of loan?
1
5. Compute the interest on Tk.5480 at 9 % for 9 months. (Answer: Tk.380.18)
4
6. Find the interest rate if Tk.5250 earns Tk.55 interest in 6 months. (Answer: 2.1%)
7. Find the exact and ordinary interest on $ 2190 for 75 days at 12 percent interest.
(Answer: $54.00, $54.75)
8. Find the future value if Tk.1000 is borrowed at 10 percent for 5 years. (Answer:
Tk.1250)
9. Find the future value if Tk.20000 is invested at 6 percent for 3 months. (Answer:
Tk.20300)
1
10. Compute the future value of Tk.480 at 6 % interest for 1 year and 6 months.
4
(Answer: Tk.525)

141
 S. M. Shahidul Islam

11. Liton has invested $5350 in a savings account that pays 12% simple interest. How
long will it be, in years, until the investment amount to Tk.10165?
(Answer: 7.5 years)
12. Lalu received Rs.50 for a diamond at a pawnshop and a month later paid Rs.53.50
to get the diamond back. Find the present interest rate. (Answer: 84%)
13. (a) A borrower signs a deed promising to pay a bank Tk.15000 in 5 years from
now. How much will the borrower receive if the bank discount rate is 10%? (b)
How much would the borrower have to repay in order to receive Tk.15000 now?
(Answer: (a) Tk.7500, (b) Tk.30000)
14. Find the Future value of Tk.500 at 8% compounded quarterly for 10 Years.
(Answer: Tk.1104.02)
15. Find the future value of Rs.10000 at 12% compounded monthly for 3 years and 4
months. (Answer: Rs.14888.64)
16. Compute the future value of Tk.5000 at 9 percent compounded monthly for 10
years. (Answer: Tk.12256.79)
17. How many years it will take for $5630 to grow to $12657.45 where interest rate is
10% and compounded yearly? (Answer: 8.5 years)
18. Find how many months it will take at 11% compounded quarterly of a year for
Tk.550 to grow to Tk.946.24. (Answer: 60 months)
19. Find the rate of interest that compounded yearly for $1200 to grow to $1636.40 in
5 years. (Answer: 6.4%)
20. Find the rate of interest that compounded semiannually for $1550 to grow to
$3144.32 in 10 years. (Answer: 7.2%)
21. A man built up a scholarship fund to give prize of $2562.50 every year. If the fund
provides 10% interest compounded semiannually, what is the amount of the fund?
(Answer: $25000.00)
22. Find the effective rate of 24 percent compounded monthly. (Answer: 26.824%)
23. How much will a deposit of 5000 taka grow to in 20 years at 6% interest
compounded continuously? (Answer: 16600.59 taka)
24. Sums of Tk.1000 are deposited in an account at the end of each 6 months period
for 9 years. Find the amount in the account after the last deposit has been made if
interest is earned at the rate of 10% compounded semiannually. (Answer:
Tk.28132.39)
25. How much should be deposited in a sinking fund at the end of each quarter for 8
years to accumulate Tk.15000 if the fund earns 10 percent compounded quarterly?
(Answer: Tk.311.52)
26. a) A sum of money invested now at 10 percent compounded quarterly is to provide
payments of $ 1000 every 3 months for 12 years, the first payment due 3 months
from now. How much should be invested? b) How much interest will the
investment earn? (Answer: (a) $27773.15, (b) $20226.85)
27. A real estate developer borrows Tk.100000 at 12% compounded monthly. The
debt is to be discharged by monthly payments for the next 6 years. (a) Find the

142
 Mathematics of Finance

monthly payment. (b) How much interest will be paid? (Answer: (a) Tk.1955.02,
(b) Tk.40761.44)
28. (a) Determine the annual payment necessary to repay a Tk.350000 loan if interest
is computed at 9% per year, compounded annually. Assume the period of the loan
is 6 years. (b) How much interest will be paid over the 6-year period? (Answer: (a)
Tk.78021.92, (b) Tk.118131.52)
29. (a) Determine the quarterly payment necessary to repay a Tk.25000 loan if interest
is computed at the rate of 14% per year, compounded quarterly. Assume the loan is
to be repaid in 10 years. (b) How much interest will be paid over the 10-year
period? (Answer: (a) Tk.1170.68, (b) Tk.21827.20)

143
 S. M. Shahidul Islam

09
Chapter
Limit and Continuity

Highlights:
9.1 Introduction 9.6 Left hand side and right hand side
9.2 Limit limits
9.3 Difference between Lim f (x) and f (a) 9.7 Continuity
xa 9.8 Some solved problems.
9.4 Methods of evaluating limit of a function 9.9 Exercise
9.5 Some important limits

9.1 Introduction: Limit and continuity are the core concept in the development of
calculus. In the calculus there is often an interest in the limiting value of a function as the
independent variable approaches some specific real number. There are different
procedures for finding the limit of a function. One procedure is simply to substitute the
value x = a in the function. Another one procedure is to substitute values of the
independent variable into the function while observing the behavior of f(x) as the value of
x comes closer and closer to a from both sides. In an informal sense, a function is said to
be continuous if it can be sketched without lifting our pen or pencil from the paper (that is,
it has no jumps, no breaks and no gaps). A function, which is not continuous, is termed
discontinuous.

9.2 Limit: If corresponding to a positive number , however small, we are able to find a
number δ such that f ( x)  l  for all values of x satisfying x  a   then we say that
f (x)  l as x  a; and write this symbolically as
Lim f ( x)  l , this is the limiting value as xa.
x a
Note: It should be remembered that the function may not actually reach the limit l but it
may get closes and closer to l as x approaches a so that f ( x)  l is less than any given
value.

144
 Limit and Continuity

Another definition of limit: The limit of f (x) as x approaches a is l,

Lim f ( x)  l ,
xa
if and only if f (x) approaches ‘l’ as x approaches ‘a’ along any sequence of values.

Example: Let us have a function

f ( x)  x 2  2
The function approaches the limit 7 as x approaches 3, we can express it is
Lim( x 2  2)  7 . This can be shown below first with x approaching closer and closer to 3
x3
from the lower side:
When x  2.99, f (x)  6.9401
” x  2.999, f (x)  6.994001
” x  2.9999, f (x)  6.99940001
” x = 2.99999, f (x)  6.9999400001
So, when x approaching closer and closer to 3 from the lower side, then f(x) tends to 7.
Now when x approaches 3 from the higher side, we have
When x  3.01, f (x)  7.0601
” x  3.001, f (x)  7.006001
” x  3.0001, f (x)  7.00060001
” x  3.00001, f (x)  7.0000600001
Thus, when x approaching closer and closer to 3 from the higher side, then f(x) tends to 7.
It is evident from the above that as x is taken closer and closer to 3 from any side, f (x)
moves closer and closer to 7.

9.3 Difference between Lim f (x) and f (a) : The statement Lim f (x) is a statement
xa xa
about the value of f(x) when x has any value arbitrarily near to a, except a. In this case, we
do not care to know what happens when x is put equal to a. But f(a) stands for the value of
f(x) when x is exactly equal to a, obtained either by the definition of the function at a, or
else by substitution of a for x in the expression f(x), when it exists.

9.4 Methods of evaluating limit of a function: The following are some theorems on the
limits, which are often used for evaluating the limits of a function.
If Lim f ( x)  l and Lim Q( x)  m , then
x a x a

1) Lim f ( x)  Q( x)  Lim f (x)  Lim Q( x)  l  m,

x a x a xa

2) Lim f ( x).Q( x)  Lim f (x). Lim Q( x)  lm,

xa xa xa

3) Lim kf ( x)  k Lim f ( x)  k.l

x a xa

145
 S. M. Shahidul Islam

f ( x) Lim f ( x) l
4) Lim  x a 
x a Q( x) Lim Q( x) m
xa

5) Lim log f ( x)  log {Lim f ( x)}  log l.

xa

 
x a

6) Lim f ( x)  Lim f ( x)

n n
x a x a
The proofs of these formulae are beyond the scope of this book.

9.5 Some important limits: 1) Lim(c)  c; c is cons tan t.

x a

xn  an
2) Lim  na n 1
x a xa
x
1 1
3) Lim(1  n)  Lim1    e n
n 0 x 
 x
sin x tan x
4) Lim  Lim  Lim cos x =1
x 0 x x  0 x x0

The proofs of these limits are beyond the scope of this book.

9.6 Left hand side and right hand side limits:

L.H.S Lim f ( x)  Lim f (a  h)  limit of f (x) ;
xa h0
when x approaches "a" from the L.H.S
R.H.S Lim f ( x)  Lim f (a  h)  Limit of f (x) ;
x a h0
when x approaches "a" from the R.H.S
Note: If Lim f ( x)  l  Lim f ( x) , then we say limit of f(x) exists when xa and is l, that
x a x a

is Lim f ( x)  l ; otherwise, we say the limit does not exist.

xa

 x, if x  0
Example: If f ( x)  
 1, if x  0
Find Lim f ( x) .
x0

Solution: L.H.S Lim f ( x)  Lim (1)  1 [For x ≤ 0, f(x) = -1]

x0 x0

And R.H.S Lim f ( x)  Lim ( x) = 0 [For x > 0, f(x) = x]

x 0 x 0
Since, L.H.S limit = -1  0 = R.H.S limit, the limit
Lim f ( x) does not exist.
x0
The graph of the function is as follows:

146
 Limit and Continuity

f(x)
4
3
2
1 x
-2 -1 0 1 2 3 4 5
-1
-2
-3

Figure 9.1

Example: Find f(5) and Lim f ( x) where f(x) = 3x2 + 2.

x5
Solution: Given that, f(x) = 3x2 + 2
So, f(5) = 3(5)2 + 2
= 77. (Answer)
And Lim f ( x) = Lim (3x  2)
2
x5 x 5
2
= Lim (3x ) + Lim (2)
x5 x 5

[Since, Lim f ( x)  Q( x)  Lim f (x)  Lim Q(x) ]

x a x a xa
2
= 3 Lim ( x ) + 2
x5

= 3  25 + 2
= 77. (Answer)

Example: Evaluate Lim f ( x) where f ( x)  5x 3  3x  1.

x 1

Solution: Given that, f ( x)  5x 3  3x  1.

So, Lim f ( x) = Lim(5 x 3  3x  1)
x1 x 1

= 5 1 + 3 1 +1
=9 (Answer)

(2 x  1)( x  3)
Example: Find the values of f(3) and Lim f ( x) where f(x) =
x3 (5 x  2)( x  3)
(2 x  1)( x  3)
Solution: Given that, f(x) =
(5 x  2)( x  3)
(2  3  1)(3  3) 7  0 0
So, f(3) =   , which is indeterminate form.
(5  3  2)(3  3) 13  0 0

147
 S. M. Shahidul Islam

Therefore, f(3) does not exist.

(2 x  1)( x  3)
Now, Lim f ( x) = Lim
x3 x 3 (5 x  2)( x  3)

(2 x  1)
= Lim
x 3 (5 x  2)

Lim(2 x  1) f ( x) Lim f ( x)
= x 3 [Since, Lim  x a ]
Lim(5 x  2) x a Q( x) Lim Q( x)
x 3 x a
7
= (Answer)
13

a  x2  a  x2 1
Example: Prove that Lim 2
 [AUB-02]
x0 x a
0
Solution: If we put x = 0 in given function, we get , which is indeterminate form. In
0
such cases rationalizing the numerator, we have
a  x2  a  x2 
 a  x2  a  x2 a  x2  a  x2   
Lim
x 0 x2
 Lim 
x 0
 
x2 a  x2  a  x2

 
 a x   ax 2
2
2
2

x  ax  ax 
= Lim
x0 2 2 2

a  x 2  a  x 2
= Lim
x 0
x2  ax 2
 a  x2 
2 x 2
= Lim
x 0
x 2  ax 2
 a  x2 
2
= Lim
x0
a  x  a  x2
2

2 2 1
=   (Proved)
a  a 2 a a

x 2  2 x  15
Example: Evaluate Lim
x3 x2  9
0
Solution: Replacing x by 3 in the expression, we get , which is indeterminate form,
0

148
 Limit and Continuity

(x – 3) must therefore be a factor of the numerator as well as of the denominator.

Factorizing, we get
2
x 2  2 x  15 x 2  5 x  3x  15 1
Lim = Lim
x3 x2  9 x 3 x 2  32 -6 –5 –4 -3 –2 –1 0 1 2 3
xx  5  3( x  5) -1
= Lim -2= x 2  2 x  15
x3 x  3x  3 Graph of f(x)
x2  9
= Lim
x  5x  3 Figure 9.2
x 3 ( x  3)( x  3)

= Lim
x  5 [For all x ≠ 3] 2
x 3 ( x  3) 1
-6 -5 -4 -3 –2 –1 0 1 2 3
8 -1
=
6 -2
Graph of f(x) = x  5
4 x3
= (Answer)
3 Figure 9.3
x
 a
Example: Find the limiting value of Lim1   [NU-99 Mgt.]
x 
 x
x
 a
Solution: Given that Lim1  
x 
 x
= Lim1  h  h
a a
[Let = h, so h → 0 as x →  ]
h 0 x
a
 
= Lim 1  h  h 
1

h 0  

= Lim1 h 
h 0
1
h 
a

x a
n

[Since Lim f ( x)  Lim f ( x) ]
x a

n

1
=e a
[We know, Lim(1  n) n
 e]
n 0

9.7 Continuity: A function f (x) is said to be continuous at a point ‘a’ if the following
three conditions are met
1. f (a) is defined
2. Lim f (x) exists
xa

3. Lim f (x) = f (a)

xa
If a function is not continuous at a point ‘a’, we say it is discontinuous at ‘a’ or has a
discontinuity at ‘a’.

149
 S. M. Shahidul Islam

x2
Example: Discuss the continuity of f (x) =
x  5x  6
2

Solution: First, we factor the denominator

x2
f ( x)  2
x  3x  2 x  6
x2
=
x( x  3)  2( x  3)
x2
=
x  3x  2
Condition 1: f (x) is defined for all x except x = 2 and x = 3.
Condition 2: By definition of limit, Lim f (x) exists for all x except x = 2 and x = 3.
xa

Condition 3: Again by definition of limit, we find that Lim f (x) = f (a) for all x except
xa
x = 2 and x = 3.
Hence, f (x) is continuous at all x except x = 2 and x = 3, where it has discontinuities.

1
Example: Show that the function f (x) as defined below, is discontinuous at x  .
2
x , for 0  x  1 / 2

f ( x)  1 , for x  1 / 2 [AUB-03, DU-88]
1  x , for 1 / 2  x  1
 f(x)
Solution: We are given that 2
1 1
f (x) =1 when x  , which means that f ( )  1
2 2 1 -
 the first condition is satisfied.
Now let us find Lim f (x) and Lim f (x) x
1 1 -2 -1 0 1/2 1 2
x x
2 2
-1
1
 L.H.S Lim f (x) = Lim (x) 
x
1
x
1 2
2 2 -2
1
And R.H.S Lim f (x) = Lim (1  x) 
x
1
x
1 2 Figure 9.4
2 2

1
Since, L.H.S limit = = R.H.S limit,
2
1
Lim f (x) = , exist
x
1 2
2

150
 Limit and Continuity

 the second condition is satisfied.

1 1
But Lim f (x) =  1  f ( )
x
1 2 2
2
So, the third condition is failed.
1
Therefore, the given function f (x) is discontinuous at x  . (Proved)
2

9.8 Some solved problems:

Problem (1): If f ( x)  10 x 4  5x 3  x 2  9 , then find the value of f (3) .
Solution: Given that f ( x)  10 x 4  5x 3  x 2  9
So, f (3)  10(3) 4  5(3)3  (3) 2  9
= 10  81 + 5  27 + 9 + 9
= 963 (Answer)

Problem (2): Find g(a) – g(x – a) if g(x) = x2 +10 [AUB-01]

Solution: Given that g(x) = x2 + 10
 g(a) = a2 + 10
 g(x – a) = (x – a)2 + 10
Now g(a) – g(x – a) = a2 + 10 – {(x-a)2 + 10}
= a2 + 10 – {x2 – 2ax + a2 + 10}
= a2 + 10 – x2 + 2ax – a2 – 10}
= – x2 + 2ax
= x(2a – x) (Answer)

x2  9
Problem (3): Evaluate the limit: Lim [NU-98]
x 3 x  3

x 9
2
x 2  32
Solution: Given that, Lim = Lim
x 3 x  3 x 3 x3
(2 – 1) xn  an
= 2.3 [Since Lim  na n 1 ]
x a xa
= 2.3
=6 (Answer)

1  2 x  1  3x
Problem (4): Evaluate the limit: Lim [RU-90]
x0 x
0
Solution: If we put x = 0 in given function, we get , which is indeterminate form. In
0
such cases rationalizing the numerator, we have

151
 S. M. Shahidul Islam

1  2 x  1  3x  1  2x  1  3x  1  2x  1  3x 
x 1  3x  1  3x 
Lim = Lim
x0 x x0

 1  2 x    1  3x 
2 2

x 1  2 x  1  3x 
= Lim
x0

1  2 x  1  3x
= Lim

x0 x 1  2 x  1  3x 
5 x
= Lim

x 0 x 1  2 x  1  3 x
 
5
= Lim
x0 1  2 x  1  3x
5
=
11
5
= (Answer)
2

Problem (5): Show that Lim f (x) exists and is equal to f (1) ,
x 1

x  1 for x  1
where f ( x)   [NU-96]
3  x for x  1
2

Solution: Given that f(x) = x + 1 for x  1

So, f(1) = 1 + 1
=2
To find Lim f (x) , we have to find Lim f (x) and Lim f (x)
x 1 x 1 x 1

 L.H.S Lim f (x) = Lim (x + 1) = Lim {(1 – h) + 1} = 2

x 1 x 1 h 0

And R.H.S Lim f (x) = Lim (3  x ) = Lim {3 – (1 + h)2} = 2

2
x 1 x 1 h 0
Since, L.H.S limit = 2 = R.H.S limit,
Lim f (x) = 2, exist
x 1

Therefore, Lim f (x) = f(1) = 2 (Shown)

x 1
Note: This function satisfies all three conditions of continuity. So it is continuous at x = 1.

x2
Problem (6): Discuss the continuity of f ( x) 
x  3x  2
2

x2
Solution: Given that, f ( x) 
x  3x  2
2

152
 Limit and Continuity

x2
=
x  2x  x  2
2

x2
=
xx  2  1x  2
x2
=
( x  2)( x  1)
Condition 1: f (x) is defined for all x except x = 1 and x = 2.
Condition 2: By definition of limit, Lim f (x) exists for all x except x = 1 and x = 2.
xa

Condition 3: Again by definition of limit, we find that Lim f (x) = f (a) for all x except
xa
x = 1 and x = 2.
Hence, f (x) is continuous at all x except x = 1 and x = 2, where it has discontinuities.

9.9 Exercise:
1. Define limit and continuity.
2. What is difference between limit and continuity?
3. When does the limit of a function exist?
4. If f ( x)  10 x 4  5x 3  x 2  9 , then find the value of f (0) . [Answer: 9]
5. If f ( x)  x 4  2 x 2  5 , then find f(0), f(-2) and f(3). [Answer: 5, 29 and 104]
6. Evaluate Lim f ( x) where f ( x)  3x 3  2 x 2  6. [Answer: 38]
x 2

7. Find f(5) and Lim f ( x) where f(x) = 3x2 + 10x + 5 [Answer: 130, 130]
x5

8. Show that f(2) and Lim f ( x) are equal where f(x) = 3x2 + 2x – 1.
x2
9. For the following exercises, find the indicated limit.
x 2  8 x  14 x 2  81
(i) Lim(3x 2  5 x  3) ; (ii) Lim ; (iii) Lim ;
x 0 x 4 2x  7 x 9 x9
[Answer: 3] [Answer: 14] [Answer: 18]
x 2  3x 1 x2  1 x2 x10  1 2x
(iv) Lim ; (v) Lim ; (vi) Lim ; (vii) Lim
x 3 x3 x  0 x x  1 x 1 x   x3
[Answer: 3] [Answer: 0] [Answer: -10] [Answer: 2]
1 1 1
( x  1) (  2)
x 1 2 h  h2
(viii) Lim 2 [Hints: x = x x = ; h→0 as x →∞]
x  x  1 1 2 1 1  h 2
( x  1) 1 2
x2 x
[Answer: 0]

153
 S. M. Shahidul Islam

10. A function f(x) is defined as follows:

 x when x  0

f(x) = 0 when x  0 [Answer: 0]
 x when x  0

Find the value of Lim f ( x) .
x0

2 x for x  5
11. Find Lim f ( x) where f(x) = 
x5
20  2 x for x  5
[Answer: 10]
x5
12. Discuss the continuity of 2 .
x  7 x  12
13. A function f(x) is defined as follows:
1 1
 2  x, when 0  x  2

 1
f(x) = 0, when x 
 2
3 1
 2  x, when 2  x  1

Show that f(x) is continuous at x = ½ .
14. Show that the function defined below is discontinuous at x = 1
 x  1, for x  1
f(x) = 
 x, for x  1

154
 Differentiation

10
C h a p te r
Differentiation and its applications
Highlights:
10.1 Introduction 10.7 Maxima, minima and point of
10.2 Differential coefficient inflection
10.3 Fundamental theorem on 10.8 Determination of maxima & minima
differentiation 10.9 Calculus of multivariate functions
10.4 Meaning of derivatives and 10.10 Business application of differential
differentials calculus
10.5 Some standard derivatives 10.11 Some worked out examples
10.6 Successive differentiation 10.12 Exercise

10.1 Introduction: Differential calculus is the most important part of mathematics. The
word differentiation means the rate of change in one variable with reference to an
infinitesimal variation in the other variables. There is then a dependent variable which gets
an impulse for changing in the dependent variables. Differential calculus is concerned
with the average rate of changes, whereas Integral calculus, by it nature, considered the
total rate of changes in variables. It has a large use in business problems. For example with
a given cost function it would be possible to find average change of cost, i.e., marginal
cost with reference to a small change in product or in other related factors and also be
found out the minimum value of the function. In this chapter we discuss nature of
differentiation, how to find the differential coefficient of various functions and the use in
business problems.

10.2 Differential coefficient (or Derivation): Let y = f(x) be a finite and single valued
function defined in any interval of x and assume x to have any particular value in the
interval. Let ∆x (or h) be the increment of x, and let ∆y (or k) = f ( x  x)  f ( x) be the
corresponding increment of y. If the ratio ∆y/∆x of these increments tends to a definite
finite limit as ∆x tends to zero, then this limit is called the differential coefficient (or
derivative) of f ( x) (or y) for the particular value of x and is denoted by f1(x) or f/(x) or

{ f ( x)} or D f (x) or
d dy d
(here, is called differential operator with respect to x).
dx dx dx
Thus, symbolically, the differential coefficient of y [= f (x) ] with respect to x [for any
particular value of x] is

155
 S. M. Shahidul Islam

dy y f ( x  x)  f ( x)
f / ( x) or  lim  lim
dx x  0 x x  0 x
f ( x  h)  f ( x )
= lim , provided this limit exists.
h0 h
Note 1: The process of finding the differential coefficients is called differentiation, and we
are said to differentiate f (x) with respect to x.
f ( x  h)  f ( x )
Note 2: The right-hand limit lim for any particular value of x, when it
h0 h
exists, is called the right-hand derivative of f(x) at that point and is denoted by Rf/(x).
f ( x  h)  f ( x )
Similarly, the left-hand limit lim  , when it exists is called the left-hand
h0 h
derivative of f(x) at x, denoted by Lf/(x). When these two derivatives both exist and are
equal, it is then only that the derivative of f(x) exists at x. When, however, the left-hand
and right-hand derivatives of f(x) at x are unequal, or one or both are non-existent then
f(x) is said to have no proper derivative at x.
Note 3: If f/(a) is finite, the function f(x) must be continuous at x = a.

Example: If f(x) = x2 + 3x then find f(x + ∆x) – f(x). What is the change of f(x) if x
changes from 5 to 5.5?
Solution: Given that f(x) = x2 + 3x
Then f(x + ∆x) – f(x) = {(x + ∆x)2 + 3(x + ∆x)} – (x2 + 3x)
= x2 + 2x(∆x) + (∆x)2 + 3x + 3(∆x) – x2 – 3x
= ∆x(2x + ∆x + 3) (Answer)
Here, ∆x = 5.5 – 5 = 0.5
So, the change of f(x) is f(5 + 0.5) – f(5) = 0.5{2(5) + 0.5 + 3}
= 0.5(13.5)
= 6.75 (Answer)

Example: Find the differential coefficient of f(x) = x2 using first principle.

Solution: Given that f(x) = x2 .
 f(x + h) = (x + h)2 = x2 + 2xh + h2.
dy f ( x  h)  f ( x )
We know that f / ( x) or = lim
dx h0 h
x  2 xh  h 2  x 2
2
= lim
h0 h
2 xh  h 2
= lim
h0 h
h( 2 x  h)
= lim
h0 h

156
 Differentiation

= lim ( 2 x  h)
h0

= 2x
d 2
Therefore, ( x )  2 x (Answer)
dx

Example: A function f(x) is defined as follows:

 x when x  0

f(x) = 0 when x  0
 x when x  0

Show that f/(0) does not exist.
f ( x  h)  f ( x )
Solution: We know that f/(x) = lim
h0 h
f ( 0  h )  f ( 0) f ( h)
 f/(0) = lim = lim
h0 h h0 h
f ( h) h
Now, Rf/(0) = hlim0 
= lim 
h0
= lim  1 = 1
h0
h h
f ( h) h
And Lf/(0) = lim  = lim  = lim   1 = –1
h0 h h  0 h h0

Since the right-hand derivative is not equal to the left-hand derivative, the derivative at
x = 0 does not exist.

10.3 Fundamental theorem on differentiation:

d
1. (c)  0 ; c  Cons tan t with respect to x
dx
2.
d
c f ( x)  c d  f ( x) c f / ( x)
dx dx
3.
d
 f ( x)  g ( x) d f ( x)  d g ( x)  f / ( x)  g / ( x)
dx dx dx
4.
d
 f ( x)  g ( x) f ( x) g ( x)  g ( x) d f ( x)  f ( x) g / ( x)  g ( x) f / ( x)
d
dx dx dx
5.
d
 f1 ( x). f 2 ( x)... f n ( x) = f1/ ( x) f 2 ( x). f 3 ( x)... f n ( x) + f 2/ ( x) f1 ( x). f 3 ( x)... f n ( x)
dx
+ . . . + f n/ ( x) f1 ( x). f 2 ( x)... f n1 ( x)
d d
f ( x)  f ( x) g ( x)
g x  f / ( x)  f x g / x 
g ( x)
d  f ( x) 
6.   = dx dx = ; g x   o
dx  g ( x)  g ( x)2 g ( x)2

157
 S. M. Shahidul Islam

dy dy du
7. If y = f(u), u = f(x) then  .
dx du dx
dy d d
That is, if y = f(g(x)) then  f ( g ( x)). g ( x)
dx dg ( x) dx
dy 1 dy dx
8.  , that is,  1
dx dx dx dy
dy
dy
dy dt
9. If y = f(t) and x = g(t) then 
dx dx
dt
dy d
10. If y  { f ( x)}g ( x ) than  { f ( x)}g ( x ) {g ( x) . log e f ( x)}
dx dx
The proofs of these formulae are beyond the scope of this book.

10.4 Meaning of the derivatives and differentials:

dy
  tan , when  is the angle which the tangent at any point to the curve
dx
y = f x  makes with the positive direction of the x -axis , that means ,
dy
is the
dx
slope of the tangent of the curve y = f x  at the point (x, f x  ).
dy
 =Rate of change of dependent variable y with respect to independent variable x
dx
 dy  f / x dx , if y = f x  .

10.5 Some standard derivatives:

1)
d n
dx
 
x  nx n 1 ; 2)
d
dx
[ f ( x)]n  n[ f ( x)]n 1
d
dx
f x  ;
d x d mx
3) (e ) = ex; 4) (e ) = memx;
dx dx
d x d mx
5) (a ) = ax loge a; 6) (a ) = m amx logea;
dx dx
d d 1 d 1
7) (log ex) = (ln x) = ; 8) ( (log a x) = log a e
dx dx x dx x
Note: We may use these standard derivatives as formula where necessary.

158
 Differentiation

Example: Find f x  if (a) f x  = x ; (b) f x  = 3 x and (c) f x  =

/ 10 3
1
x3
Solution: (a) Given that f x  = x10
Differentiating with respect to x, we get
d d
[ f ( x)] = ( x10 )
dx dx
Or, f / ( x)  10 x101 [We know
d n
dx
 
x  nx n 1 ]

So, f / ( x) = 10 x9 (Answer)
2

(b) Given that f x  = 3 x 3

Differentiating with respect to x, we get,
2
d d
[ f ( x)] = (3 x 3 )
dx dx
2
/ d 3
Or, f ( x) = 3 ( x ) [We know
d
c f ( x)  c d  f ( x) ]
dx dx dx

 
2
2 1 d n
Or, f / ( x) = 3. x 3 [We know x  nx n 1 ]
3 dx
1

3
= 2x (Answer)
(c) Given that f x  = 3
1
x
= x 3
Differentiating with respect to x, we get
d d
f (x) = ( x 3 )
dx dx
1
So, f / ( x) =  3x 31  3x 4  3 4 (Answer)
x

Example: Find h / ( x) if h(x) = (3x2 + 4)100

Solution: Given that h(x) = (3x2 + 4)100
Differentiating it we respect to x, we get
d
h / ( x) = {(3x2 + 4)100}
dx
f x  ]
d d d
=100(3x2 + 4)100 -1 (3x2 + 4) [ [ f ( x)]n  n[ f ( x)]n 1
dx dx dx
= 100(3x2 + 4)99(6x + 0)

159
 S. M. Shahidul Islam

= 100(3x2 + 4)99.6x
= 600x(3x2 + 4)99 (Answer)

Example: Find the differential coefficient of f(x) = 4x4 + (2x +1)3 – ¼ e4x.
Solution: Given that, f(x) = 4x4 + (2x +1)3 – ¼ e4x
Differentiating with respect to x, we get
d
f / ( x) = {4x4 + (2x +1)3 – ¼ e4x}
dx
d d d
= (4x4) + {(2x +1)3}– (¼ e4x) [Since d  f ( x)  g ( x) d f ( x)  d g ( x) ]
dx dx dx dx dx dx
d d d 1 d
= 4 ( x 4) + (u3) (2x +1) – (e 4 x ) [Let u = 2x +1]
dx du dx 4 dx
1
= 4.4x3 + 3u2.2 – .4e 4 x
4
= 16 x + 6(2x +1)2 – e 4 x
3
(Answer)

Example: Find the differential coefficient of f(x) = (ln x)2.

Solution: Given that f(x) = (ln x)2.
Differentiating with respect to x, we get
d
 f / ( x) = {(ln x)2} [Let u = ln x]
dx
d 2 d
= (u ). (ln x)
du dx
1
= 2u.
x
1
= 2.ln x . [Putting the value of u]
x
2 ln x
= (Answer)
x

Example: Find f / x  if f x  = (2x)(5x + 2).

Solution: Given that, f x  = (2x)(5x + 2).

f / x  =
d
 [(5x + 2) ln(2x)] [Using fundamental formula 10]
dx
d d
= (5x + 2) {ln(2x)} + ln(2x) (5x + 2)
dx dx
1 d
= (5x + 2). . (2x) + ln(2x).5
2 x dx

160
 Differentiation

1
= (5x + 2). .2 + 5 ln(2x)
2x
5x  2
= + 5 ln(2x) (Answer)
x

dy
Example: Find the differential coefficient of the following implicit function:
dx
x2 + y – 2x = 0
Solution: Given that x2 + y – 2x = 0
Differentiating with respect to x, we get
d 2 d
(x + y – 2x) = (0)
dx dx
d 2 dy d
Or, (x ) + – (2x) = 0
dx dx dx
dy
Or, 2x + –2=0
dx
dy
So, = 2 – 2x (Answer)
dx

dy
Example: Find the differential coefficient of the following implicit function:
dx
x2 – y2 + 3x = 5y [NU-99 A/C]
Solution: Given that x2 – y2 + 3x = 5y
Differentiating with respect to x, we get
d d
( x2 – y2 + 3x) = (5y)
dx dx
d 2 d 2 d dy
Or, (x ) – (y ) + (3x) = 5
dx dx dx dx
d 2 dy dy
Or, 2x – (y ). +3=5
dy dx dx
dy dy
Or, 2x – 2y +3=5
dx dx
dy dy
Or, – 2y –5 = – 2x – 3
dx dx
dy
Or, – (2y + 5) = – (2x + 3)
dx
dy 2x  3
So, = (Answer)
dx 2y  5

161
 S. M. Shahidul Islam

1 2
Example: Find the slope of f ( x)  x3  x  x  1, at x = – 1. [AUB-02]
2
Solution: Let us consider, y = f(x)
1
 y  x3  x2  x  1
2
Differentiating with respect to x, we get
dy d 1
= ( x3  x 2  x  1, )
dx dx 2
d d 1 2 d d
= ( x3 )  ( x ) + (x) + (1)
dx dx 2 dx dx
1 d
= 3x3-1 – . (x2) + 1.x1-1 + 0
2 dx
1
= 3x2 – .2 x2-1 +1.xo + 0
2
= 3x2 – x + 1.1
= 3x2 – x +1
So slope of the given curve at x = –1 is
 dy 
 dx  = 3(–1)2 – (–1) + 1
  x  1
=3+1+1
=5 (Answer)

10.6 Successive differentiation: We have seen that the first derivative (or first differential
coefficient) of a function y = f(x) of x is in general a function of x. This new function may
have a derivative, which is called the second derivative (or second differential coefficient)
// d2y
of f(x) and is denoted by f (x) or . Similarly, the derivative of the second derivative is
dx 2
d3y
called the third derivative of f(x) and is denoted by f///(x) or , and so on for the nth
dx 3
dny
derivative fn(x) or .
dx n
dy
Thus, if y = x5, = 5x4 is the first derivative of y with respect to x.
dx
d 2 y d  dy  d
Again,     (5 x 4 )  20 x 3 is the second derivative of y with respect to x.
dx 2
dx  dx  dx
d3y d d2y d
Again, 3
  2   (20 x 3 )  60 x 2 is the third derivative of y with respect to x.
dx dx  dx  dx

162
 Differentiation

Example: Find the third derivative of f(x) = 5x7 + 3x3 – 2x2 + 10.
Solution: Given that f(x) = 5x7 + 3x3 – 2x2 + 10.
f / x  =
d
 (5x7 + 3x3 – 2x2 + 10) = 35x6 + 9x2 – 4x
dx
f // x  =
d
Or, (35x6 + 9x2 – 4x) = 210x5 + 18x – 4
dx
So, third derivative f /// x  =
d
(210x5 + 18x – 4) = 1050x4 + 18 (Answer)
dx

10.7 Maxima, minima and point of inflection: A function f(x) is said to be maximum at
x = a if f(a) is greater than every other values assumed by f(x) in the immediate
neighbourhood ofYx = a.

Maximum point y = f(x)


Point of inflection  
Minimum point

O X
Figure 10.1

A function f(x) is said to be minimum at x = a if f(a) is less than every other values
assumed by f(x) in the immediate neighbourhood of x = a. And a function f(x) has a point
of inflection at x = a if f(a) is neither less nor greater than every other values assumed by
f(x) in the immediate neighbourhood of x = a.
dy
The gradient of the curve measured by = 0, at all turning points. At the maximum
dx
dy
point the sign of changes from positive to negative as x increases and at the minimum
dx
dy
point the sign of changes from negative to positive as x increases. At the point of
dx
dy
inflection does not change it sign as x increases or decreases.
dx

10.8 Determination of maximum and minimum:

(A) If c be a point in the interval in which the function f x  is defined, and if
f / c   0 and f // c   0 , then f c  is (i) a maximum if f // c  is negative and (ii) a
minimum if f // c  is positive.
(B) If f / c   f // c   . . .  f n1 c   0 and f n c   0 , than

163
 S. M. Shahidul Islam

(i) If n be even, f c  is a maximum or a minimum according as f // c  is negative or

positive.
(ii) If n be add, f c  is neither a maximum nor a minimum.

Example: Find for what values of x, the following expression is maximum and
minimum respectively: 15x4 + 8x3 – 18x2
Find also the maximum and minimum values of the expression. [AUB-03]
Solution: Let us consider,
f x   15x 4  8x 3  18x 2
 f / x   (15 x 4  8 x 3  18 x 2 ) = 60x3 + 24x2 – 36x
d
dx
And f // x  =
d
(60x3 + 24x2 – 36x) = 180x2 + 48x – 36
dx
Now, when f x  is a maximum or a minimum, f / x  = 0
Therefore, we should have, 60x3 + 24x2 – 36x = 0
Or, 12x(5x2 + 2x – 3) = 0
Or, x(5x2 + 2x – 3) = 0
Or, x(5x2 + 5x – 3x – 3) = 0
Or, x{5x(x +1) – 3(x +1)} = 0
Or, x(x +1)(5x – 3) = 0
3
 x = 0, x = –1 and x =
5
Now, when x = 0, f 0  36, which is negative
//

when x = –1, f // (1)  96, which is positive

3 //  3  288
And when x = f   , which is positive.
5, 5 5
3
Hence, the given expression is maximum at x = 0 and minimum at x = –1 and x =
5
Therefore, f(0) = 0
f(–1) = –11
3
And f   = – 2.808
5
Thus, the maximum value of the expression is 1 and the minimum values are –10 and
–1.808.

164
 Differentiation

10.9 Calculus of multivariate functions:

10.9.1 Definition of the function of two independent variables: If three variables x, y, z
are so related that for every pair of values of x and y within the defined domain, z has a
single definite value, z is said to be a function of the two independent variables x and y,
and is denoted by z = f(x, y).
z = f(x, y) = x2 + 2hxy + y2 is a function of two independent variables x and y.

10.9.2 Partial derivatives: The result of differentiating z = f(x, y), with respect to x,
treating y as a constant, is called the partial derivatives of z with respect to x, and is
 z f
denoted by one of the symbols , , z x , f x ( x, y) or briefly , f x , etc.
 x x
f f ( x  x, y)  f ( x, y )
Analytically, = lim
x x  0 x
f f ( x, y  y )  f ( x, y )
Similarly, = lim .
y y  0 y
f
  x   x ( f y  0)
dy f
And
dx f fy
y
Illustration: Let z = x2 + xy + y2 ; then
z z dy 2x  y
= 2x + y ; = x + 2y. So,  .
x y dx x  2y

dy
Example: Find the differential coefficient of the following implicit function:
dx
f(x, y) = x2 + y – 2x
Solution: Given that f(x, y) = x2 + y – 2x
Here, fx = 2x – 2 ; fy = 1
dy f 2x  2
So, =  x = = 2 – 2x (Answer)
dx fy 1

z
10.9.3 Successive partial derivatives: Since each of the first order partial derivatives ,
x
z
is in general, a function of x and y, each may possess partial derivatives with respect
y
to these two independent variables, and these are called the second order partial
derivatives of z. The usual notations for these second order partial derivatives are

165
 S. M. Shahidul Islam

  z  2z
  , i.e., or fxx .
x  x  x 2
  z  2z
 , i.e., or fxy .
x  y  xy
  z  2z
 , i.e., or fyx .
y  x  yx
  z  2z
 , i.e., or fyy .
y  y  y 2
2z 2z
And always = .
xy yx
z z
Illustration: Let z = x2 + xy + y2 ; then = 2x + y ; = x + 2y.
x y
2z  2z  2z 
So, = (2x + y) = 2; = (x + 2y) = 1; = (2x + y) = 1 and
x 2
x xy x yx y
2z 
= (x + 2y) = 2.
y 2
y

10.9.4 Determination of maximum and minimum of multivariate function: If (a, b) be

a point in the domain in which the function f(x, y) is defined, and if f x(a, b) = 0,
fy(a, b) = 0 and fxx(a, b).fyy(a, b) – [fxy(a, b)]2 > 0, then f(a, b) is a maximum or a minimum
according as fxx(a, b) < or > 0 (and consequently fyy(a, b) < or > 0).
But if fxx(a, b).fyy(a, b) – [fxy(a, b)]2 < 0, f(a, b) is neither a maximum nor a minimum and
if fxx(a, b).fyy(a, b) – [fxy(a, b)]2 = 0, further analysis is necessary.
In other word we can say that f(a, b) is a maximum (or a minimum) if the determinant
 2 f 2 f 
 2 
 x xy 
formed by the Hessain, Hf = is positive and the leading diagonal elements
 2 f 2 f 
 2 
 yx y 
are positive (or negative).

Example: Examine for extreme values of the function: f(x, y) = x2 + y2 + (x + y +1)2.

Solution: Given that f(x, y) = x2 + y2 + (x + y +1)2.
 fx = 2x + 2(x + y +1) = 4x + 2y + 2
 fy = 2y + 2(x + y +1) = 2x + 4y + 2
 fxx = 4, fyy = 4 and fxy = 2.
The equations fx = 0 and fy = 0 are equivalent to

166
 Differentiation

2x + y + 1 = 0
and x + 2y + 1 = 0
x y 1
These give  
1 2 1 2 4 1
x y 1
Or,  
1 1 3
1 1
Or, x =  and y = 
3 3
1 1
The function may have an extreme value at (  ,  )
3 3
Now, fxx.fyy – (fxy)2 = 4.4 – 22 = 12 > 0 ,
Also fxx > 0.
1 1 1
Therefore, f(x, y) is maximum at (  ,  ) and the maximum value is .
3 3 3

10.9.5 Constrained optimization with Lagrangian multipliers: Differential calculus is

also used to maximize or minimize a function subject to constraints. Consider a function
f(x, y) subject to a constraint h(x, y) = k, a new function F can be formed as follows:
F(x, y, λ) = f(x, y) – λ[h(x, y) – k]
Here, F(x, y, λ) is called Lagrangian function.
Since the constraint is always set equal to 0, the value of the new function f(x, y, λ) is
same as the original objective function f(x, y). Now solving simultaneously:
Fx(x, y, λ) = 0, Fy(x, y, λ) = 0 and Fλ(x, y, λ) = 0
we shall get some critical point. At these critical points the function will be maximum or
minimum according the maximum or minimum criteria of multivariate function.

Example: Determine the maximum value of the objective function f(x, y) = 4x – x3 + 2y

subject to x + y = 1.
Solution: Here, the Lagrangian function: F(x, y, λ) = 4x – x3 + 2y – λ(x + y – 1)
 Fx = 4 – 3x2 – λ
Fy = 2 – λ
and Fλ = – (x + y – 1)
The equations Fx = 0, Fy = 0 and Fλ = 0 are equivalent to
4 – 3x2 – λ = 0 - - - (1)
2–λ=0 - - - (2)
and x+y–1=0 - - - (3)
From equation (2), we get λ = 2
Substituting λ = 2 in equation (1), we get x = 0.8165
Now from equation (3), we get y = 0.1835

167
 S. M. Shahidul Islam

So, (0.8165, 0.1835, 2) is the critical point of F(x, y, λ) as well as (0.8165, 0.1835) is the
critical point of f(x, y).
Now, at the point fxx = – 4.899, fxy = 0, fyx = 0 and fyy = 0.
So, fxx.fyy – (fxy)2 = – 4.899  0 – 0 = 0, also fxx < 0.
Hence, (0.8165, 0.1835) maximizes the objective function and the maximum value is
3.0887 (Answer)

10.10 Business Applications of Differential Calculus:

Key Concepts: Profit: Profit is defined as the excess of total revenue over total cost. That
is, Profit = Total revenue – Total cost. If c(x) is the total cost and r(x) is the total
revenue, then profit, p(x) = r(x) – c(x).
Marginal cost: Marginal cost is the rate of change of total cost with respect to units of
production. If cx  is the total cost of producing x units of a products, c / x  is the point of
marginal cost.
Marginal production: Marginal production is the incremental production i.e., the
additional production added to the total production. If tp(x) is the total production,
d
[tp(x)] is the marginal production.
dx
Marginal revenue: Marginal revenue is the rate of change in revenue with respect to total
d
output. If tr(x) is the total revenue, [tr(x)] is the marginal revenue.
dx

Example: If the total cost of producing p units of pen is c(p) = 0.0015 p3 – 0.9p2 + 200p +
60000; compute the marginal cost at outputs of (a) 100 units, (b) 200 units, (c) 300 units.
Solution: Given that, cost: c(p) = 0.0015p3 – 0.9p2 + 200p + 60000
Differentiation with respect to p, we have,
c / x  
d
(0.0015p3 – 0.9p2 + 200p + 60000)
dx
= 0.0015  3p2 – 0.9  2p + 200 + 0
= 0.0045p2 – 1.8p + 200, this the marginal cost,
(a) c/ (100) = 0.0045(100)2 – 1.8(100) + 200
= 65 taka per unit pen
(b) c/ (200) = 0.0045(200)2 – 1.8(200) + 200
= 20 taka per unit pen
(c) c (300) = 0.0045(300)2 – 1.8(300) + 200
/

= 65 taka per unit pen

168
 Differentiation

5 2
Example: Total cost of producing x units is x + 175x + 125 and the price at which
4
5
each unit can be sold for 250 – x . What should be the output for a maximum profit?
2
Calculate the maximum profit.
5 2
Solution: Given that, cost: c(x) = x + 175x + 125
4
5 5
And revenue: tr(x) = (250 – x )  x = 250x – x 2
2 2
5 5
So, profit: p(x) = 250x – x 2 – ( x 2 + 175x + 125)
2 4
15 2
= – 125 + 75x – x
4
d d 15 2
Now, [p(x)] = (– 125 + 75x – x )
dx dx 4
30
= 75 – x
4
The necessary and sufficient conditions for maximization are that the first derivative of a
profit function is equal to zero and the second derivative is negative.
30
So, 75 – x =0
4
30
Or, x = 75
4
Or, x = 10
d2 30
And 2
[p(x)] = – , which is negative.
dx 4
Therefore, the profit is maximum at the output, x = 10 (Answer)
Putting x = 10 in the profit function, we get
15
The maximum profit = – 125 + 75  10 –  (10) 2 = 250
4
Hence, the maximum profit = Tk. 250 (Answer)

10.11 Some worked out examples:

Example (1): Find the differential coefficient of f(x) = x using first principle.
Solution: Given that f(x) = x
 f(x + h) = x  h

169
 S. M. Shahidul Islam

dy f ( x  h)  f ( x )
We know that f / ( x) or = lim
dx h0 h
xh  x
= lim
h0 h
( x  h  x )( x  h  x )
= lim
h0
h( x  h  x )
( x  h)2  ( x )2
= lim
h0
h( x  h  x )
xhx
= lim
h0
h( x  h  x )
h
= lim
h0
h( x  h  x )
1
= lim
h0
( x  h  x)
1
=
x x
1
=
2 x
d 1
Therefore, ( x)  (Answer)
dx 2 x

3
x15 .x 2
Example (2): Find f / ( x) for f ( x)  .
x 20
3
15 2
Solution: Given that, f x  
x .x
x 20
3
15
x 2
= 20
x
3
15  20
= x 2

7
= x2
Differentiating with respect to x, we get

170
 Differentiation

7
d 2
f x  
/
(x )
dx
7
7  1
= x 2
2.
9
7 2
= x (Answer)
2.

Example (3): Find the differential coefficient with respect to x of the following implicit
function: x2 + 2hxy + y2 = 0 ; h is a constant. [NU-99 Mgt.]
2 2
Solution: Given that x + 2hxy + y = 0
Differentiating with respect to x, we get
d d
( x2 + 2hxy + y2) = (0)
dx dx
d 2 d d 2
Or, (x ) + (2hxy) + (y ) = 0
dx dx dx
d dy
Or, 2x + 2h (xy) + 2y =0
dx dx
d d dy
Or, 2x + 2h{x (y) + y (x)} + 2y =0
dx dx dx
dy dy
Or, 2x + 2h{x + y.1} + 2y =0
dx dx
dy dy
Or, 2x + 2hx + 2hy + 2y =0
dx dx
dy
Or, (2hx + 2y) = – 2x – 2hy
dx
dy  2( x  hy )
Or, =
dx 2(hx  y )
dy ( x  hy )
So, =  (Answer)
dx (hx  y )

 
4
45
Example (4): Find f / ( x) if f x  = 2 x 3  3x 2  10 3 – 1
 e 9 x [AUB-02]
(5 x 2  4) 2

 
4
45
Solution: Given that, f x  = 2 x 3  3x 2  10 3 – 1
 e9x
(5 x 2  4) 2

Differentiating with respect to x, we get

171
 S. M. Shahidul Islam

 
d  3
  9x 
4
45
f x  =  2 x  3x  10  e 
/ 2 3

 
1
dx  
 5x  4
2 2

d   
   
4 1
d d 9x
2 x 3  3x 2  10 3 – 45 5 x  4 2  +
2
= (e )
dx dx   dx
       
4 1
4 1 d 1  1 d
= 2 x 3  3x 2  10 3 2 x 3  3x 2  10 – 45(– ) 5 x 2  4 2 5x 2  4
3 dx 2 dx
+ 9e9x
45  10 x
1
4
= 6x(x –1) (2 x 3  3x 2  10) 3 + 3
 9e 9 x
3
2(5 x 2  4) 2
1
225 x
= 8x(x –1)(2x 3 3x 2  10) 3  3
 9e 9 x (Answer)
(5 x 2  4) 2

Example (5): Find the differential coefficient of y = e ax bxc  (ln x) 5 [RU-96 A/C]
2

Solution: Given that y = e ax bxc  (ln x) 5

2

Differentiating with respect to x, we get

dy d
[ e ax bxc  (ln x) 5 ]
2
=
dx dx
dy d d
( e ax bxc ) +
2
Or, = [ (ln x) 5 ]
dx dx dx
dy d d
= e ax bxc . (ax2 + bx + c) + 5 (ln x) 4
2
Or, (lnx)
dx dx dx
dy 1
= e ax bxc .(2ax + b) + 5 (ln x) 4 .
2
Or,
dx x
dy 5
= (2ax + b) e ax bxc + (ln x) 4
2
So, (Answer)
dx x

Example (6): Find the differential coefficient of f(x), where

3x  2
1
f x   (2 x  1)(4 x  5) 2  1
+ e 2 x log e 2 x [AUB-03]
(6 x 2  5) 3
Solution: Given that,
3x  2
1
f x   (2 x  1)(4 x  5) 
2
1
+ e 2 x log e 2 x
(6 x 2  5) 3

172
 Differentiation

Differentiation with respect to x, we get

 
d  3x  2 
1
f x  
/
(2 x  1)(4 x  5) 
2
1
 e log e 2 x 
2x

dx  
 (6 x 2  5) 3 
 
d   d  3x  2  d 2 x
 
1
= (2 x  1)(4 x  5)   
2
1 
 e log e 2 x
dx   dx  (6 x 2  5) 3  dx
 
= 2 x  1 4 x  5 2  4 x  5 2 2 x  1
d 1 1 d

dx dx

1 1
d d
(6 x 2  5) 3 (3x  2)  (3x  2) (6 x 2  5) 3
dx dx d d
 2
 e2x (log e 2 x)  log e 2 x (e 2 x )
 2 1
 dx dx
( 6 x  5) 3

 
=
1 1
1 1
1
1
1 ( 6 x 2
 5) 3
.3  (3 x  2). ( 6 x 2
 5) 3
.12 x
1 3
(2 x  1). (4 x  5) .4  (4 x  5) .2 
2 2
2
2
(6 x  5) 3
2

1
+ e 2 x . .2  log e 2 x.2e 2 x
2x
1 2

3(6 x  5)  4 x(3x  2)(6 x  5)
1 1 2 2
 3 3
= 2 (2 x  1)(4 x  5) 2
 2(4 x  5) 
2
2

(6 x 2  5) 3
1
+ e 2 x  2e 2 x log e 2 x
x
1 2

2(2 x  1) 3(6 x 2  5) 3  4 x(3x  2)(6 x 2  5) 3
 24 x  5 
1
= 1
2
2
+
(4 x  5) 2
(6 x  5)
2 3

1 
e 2 x   log e 2 x 
x 
x
dy x
Example (7): Find of y = x . [RU-91 Mgt., RU-90 A/C]
dx x
x
Solution: Given that y = .x

173
 S. M. Shahidul Islam

Taking natural logarithm on both sides, we get

log y  x x log x
Again taking natural logarithm on both sides, we get
log(log y)  x log x  log(log x)
Differentiating both sides with respect to x, we get
d d
( log(log y) ) = ( x log x  log(log x) )
dx dx
1 d d d 1 d
Or, . (log y ) = x (log x)  log x. ( x)  (log x)
log y dx dx dx log x dx
1 1 dy 1 1 1
Or, . = x.  (log x).1  .
log y y dx x log x x
dy  1 
Or, = y log y1  log x  
dx  x log x 
x
dy x  1 
Or, = x . x x log x1  log x  
dx  x log x 
x
dy x x
Or, = x . [ x log x(1  log x)  x x1 ]
dx

Example (8): Find f / x  if f x  = (12x2 + 6) (5 x  2) [AUB-01]

Solution: Given that, f x  =(12x2 + 6) (5 x  2)
Taking natural logarithm of bath sides, we get
log e f x   log e (12 x 2  6) (5 x2)
= (5x + 2) loge (12x2 + 6)
Now differentiating with respect to x, we have
{log e f x }  {(5 x  2) log e (12 x 2  6)}
d d
dx dx
[ f x ]  (5 x  2) {log e (12 x 2  6)}  log e (12 x 2  6) (5 x  2)
1 d d d
Or,
f ( x) dx dx dx
[ f x ]  f x [(5 x  2).
d 1 d
Or, . (12 x 2  6)  log e (12 x 2  6).5]
dx 12 x  6 dx
2

5x  2
Or, f / x   (12 x 2  6) (5 x  2) [ (24 x)  5 log e (12 x 2  6)
6(2 x  1)
2

4 x(5 x  2)
 f / x   (12 x 2  6) (5 x  2) [  5 log e (12 x 2  6)] (Answer)
2x  1
2

Example (9): Find the maximum and minimum value of f(x) = x3 – 4x2 + 4x – 10.

174
 Differentiation

[RU-91 A/C]
Solution: Given that, f(x) = x3 – 4x2 + 4x – 10.
d 3
 f/(x) = (x – 4x2 + 4x – 10) = 3x2 – 8x + 4
dx
d
And f//(x) = (3x2 – 8x + 4) = 6x – 8
dx
Now, when f x  is a maximum or a minimum, f / x  = 0
Therefore, we should have, 3x2 – 8x + 4 = 0
Or, 3x2 – 6x – 2x + 4 = 0
Or, 3x(x – 2) –2(x – 2) = 0
Or, (x – 2)(3x – 2) = 0
2
 x = 2 and x =
3
Now, when x = 2, f 2  4, which is positive.
//

2 //  2 
And when x = f    4, which is negative.
3 3
2
Hence, the given function is maximum at x = and minimum at x = 2
3
Therefore, f(2) = – 10
2 238
And f  =– = – 8.815
3 27
Thus, the maximum value of the function is – 8.815 and the minimum values is –10

Example (10): Average cost of each radio for producing x radios is x2 – 10x + 30 (in
hundred taka). Find the marginal cost when x = 6. [RU-87 A/C & RU-95 BBA]
2
Solution: Given that, average cost = x – 10x + 30
So, the total cost: c(x) = (x2 – 10x + 30)x = x3 – 10x2 + 30x
d
We know that, marginal cost = [c(x)]
dx
d
= ( x3 – 10x2 + 30x)
dx
= 3x2 – 20x + 30
When x = 6, marginal cost = 3(6)2 – 20(6) + 30 hundred taka = Tk.1800 (Answer)

Example (11): A study has shown that the cost of producing pencils of a manufacturing
concern is given by c(x) = 30 + 1.5x + 0.0008x2. What is the marginal cost at x = 1000
units? If the pencils are sold for Tk.5 each for what values of x does marginal cost equal to
marginal revenue? [AUB-99]
Solution: Given that cost: c(x) = 30 + 1.5x + 0.0008x2.

175
 S. M. Shahidul Islam

d
We know that, marginal cost = [c(x)]
dx
d
= (30 + 1.5x + 0.0008x2)
dx
= 1.5 + 0.0016x
Putting the value x = 1000, we get
Marginal cost = 1.5 + 0.0016  1000 = 3.1 (Answer)
We know that, Total revenue [tr(x)] = Selling  price output.
So, tr(x) = 5x
d
We know that, Marginal revenue = [tr(x)]
dx
d
= (5x)
dx
= 5
When marginal cost = marginal revenue, then
1.5 + 0.0016x = 5
Or, 0.0016x = 3.5
Or, x = 2187.5
Therefore, marginal cost will equal to marginal revenue when x = 2187.5 (Answer)

Example (12): Production function of a firm is f(x, y) = 20x + 10y – 2x2, where x means
number of labours and y means number of units raw material. If the cost of per unit labour
is Tk. 4, the cost of per unit raw material is Tk. 5 and total spent is Tk. 24, find the
maximum production of the firm.
Solution: Here, the objective function is f(x, y) = 20x + 10y – 2x2,
And the constraint is 4x + 5y = 24.
So, the Lagrangian function: F(x, y, λ) = 20x + 10y – 2x2 – λ(4x + 5y – 24)
 Fx = 20 – 4x – 4λ
Fy = 10 – 5λ
and Fλ = – (4x + 5y – 24)
The equations Fx = 0, Fy = 0 and Fλ = 0 are equivalent to
20 – 4x – 4λ = 0 - - - (1)
10 – 5λ = 0 - - - (2)
and 4x + 5y – 24 = 0 - - - (3)
From equation (2), we get λ = 2. Then from equation (1), we get x = 3. And using x = 3
in equation (3), we have y = 2.4.
 Fxx Fxy Fx 
 
We know that Hessain, HF =  Fyx Fyy Fy 
F 
 x Fy F 

176
 Differentiation

Here, Fxx = – 4, Fxy = 0, Fxλ = – 4, Fyx = 0, Fyy = 0, Fyλ = – 5, Fλx = – 4, Fλy = – 5 and
Fλλ = 0.
4 0 4
So, the determinant formed by the Hessain is 0 0  5 = 100 > 0.
4 5 0
Therefore, the production will be maximum when x = 3, y = 2.4 and the value of the
maximum product = 20(3) + 10(2.4) – 2(3)2 = 66 units. (Answer)

10.12 Exercise:
1. Define differential coefficient and calculus of multivariate functions.
2. Why is differentiation necessary for business education?
3. If f(x) = x2 + 8x then find f(x + ∆x) – f(x). What the change of f(x) if x changes
from 6 to 5.5? [Answer: ∆x(2x + ∆x + 8), – 9.75]
4. Using first principle find the differential coefficient of (i) f(x) = x 3 (ii) f(x) = ax
[Answer: (i) 3x2 (ii) ax ]
2x
5. A function f(x) is defined as follows:
5 x when x  0

f(x) = 0 when x  0
 3x when x  0

Show that f/(0) does not exist.
6. Find the differential coefficient of the following function:
(i) y = x3 – 20x2 + 3x + 4 [Answer: 3x2 – 40x + 3]
(ii) y = ax4 + bx2 + c [Answer: 4ax3 + 2bx]
2
 1
(iii) y =  x 2   [Answer: 4x3 – 2x-3 + 2]
 x
7. Find the first derivatives with respect to x of the following functions:
(i) f(x) = (7x + 3)(4 – 3x) [Answer: 19 – 42x]
(ii) f(x) = x2/(3x + 2)½ [Answer: x(9x + 8)/2(3x + 2)3/2]
(iii) f(x) = x3(6x – 1)2/3 [Answer: x2(22x – 3)/(6x – 1)1/3]
(iv) f(x) = 5x/(3 – 4x) [Answer: 15/(3 – 4x)2]
(v) f(x) = (1 + x)2x [Answer: 2(1 + x)2x{x/(1 + x) + log(1 + x)]
logx
(vi) f(x) = x [Answer: 2xlogx–1.logx]
2 2
(vii)f(x) = (2x – 5x – 8) [Answer: 2(2x2 – 5x – 8)(4x – 5)]
(viii) f(x) = (1 – x)(1 – 2x)(1 – 3x)(1 – 4x) [Answer: 96x3 – 150x2 + 7x – 10]
3 1 2 1 1
(ix)f(x) =  2  [Answer:   3]
9x 2x 5 2 x 3 x

177
 S. M. Shahidul Islam

dy
8. Find the of the following functions:
dx
(i) 2x2 + 3x + y2 = 0 [Answer: –(4x + 3)/2y]
3 2 2
(ii) 5x – 3x + 3y – 6 = 7 [Answer: 2(2x + 1)/(2y – 5)]
(iii) x3 + 3bxy + y3 = b4 [Answer: - (bx + y2)/(x2 + by)]
2 2
(iv) ax + 2mxy + by = k [Answer: -(ax + my)/(mx + by)]
(v) log (xy) = x2 + y2 [Answer: y(2x2 – 1)/x(1 – 2y2)]
xy
(vi) e – 4xy = 2 [Answer: – y/x]
dy
9. Find the partial derivatives fx, fy and then find of the following functions:
dx
(i) f(x, y) = 2x2 + 3x + y2 [Answer: 4x + 3, 2y, –(4x + 3)/2y]
(ii) f(x, y) = ax2 + 2mxy + by2 [Answer: 2ax+2my, 2mx+2by, -(ax+my)/(mx+by)]
(ii) f(x, y) = – 3x5 + 4y3 + 6y [Answer: –15x4, 12y2 + 6, 15x4/(12y2 + 6)]
2z 2z
x2 y
10. If z = e , prove that  .
xy yx
1
11. Find the differential coefficient of y = log( x  1  x  1 ) [Answer: ]
2 x2 1
12. Find the differential coefficient of y = e ax bxc  ln x [NU-96 A/C]
2

[Answer: (2ax  b)e ax bxc  1 ]

2

x
 3

  x  1 2  x2  2
13. Differentiate the function f(x) = log e x    [Answer: ]
  x  1   x 2
 1
2
2 x 2
2 x
14. Find the differential coefficient of y = e x [Answer: (2x + 2) e x ]
5
15. Differentiate x with respect to x . 2
[Answer: 5 x3]
2
1
16. Differentiate ln x with respect to x2. [Answer: ]
2x 2
4 d2y dy
17. If y = 2x + , then prove that 2
x2
+x – y = 0.
x dx dx
 
m d2y
18. If y  x  1 x 2 , then show that (1 + x2) 2 + x
dx
dy
dx
– m2y = 0.
19. Find for what values of x, the following expression is maximum and minimum
respectively: 15x4 + 8x3 – 18x2 + 1. Find also the maximum and minimum values
3
of the expression. [Answer: Max. at x = 0, min. at x = –1, and maximum value
5
is 1 and minimum values are – 10, – 1.808]

178
 Differentiation

20. Examine f(x) = x3 – 9x2 + 24x – 12 for maximum or minimum values. [Answer:
Maximum at x = 2, max. value f(2) = 8 and minimum at x = 4, min. value f(4) = 4]
21. Show that f(x) = x5 – 5x4 + 5x3 – 1 is a maximum when x = 1, a minimum
when x = 3 and neither when x = 0.
22. The profit function of a company can be represented by p(x) = x – 0.00001x2,
where x is unit sold. Find the optimal sales volume and the amount of profit to be
expected at that volume. [Answer: 50000 & 25000]
23. Total cost of producing x units is x3 – 10x2 + 17x + 66 and the price at which
each unit can be sold for Tk. 5. What should be the output for a maximum profit?
Calculate the maximum profit. [Answer: 6 units & Tk. 6]
24. If cost: c(x) = 50x + 30000 and profit: p(x) = 100 – 0.01x, find the functions of
marginal cost and marginal revenue. Also find the value of x when marginal cost is
equal to marginal revenue. [Answer: marginal cost eqn = 50, marginal revenue eqn
= 100 – 0.02x & x = 2500]
25. The cost function and the revenue function of a company are c(x) = 100 + 0.015x 2
and r(x) = 3x, where x is the number of units of product, respectively. Find the
number of units of product that will maximize the profit. What is the maximum
profit? [Answer: 100 units & 50]
26. Show that the function f(x, y) = x2 + y2 – 4x + 6y is minimum at (2, –3).
27. The yearly profit of A company depends upon the number of workers (x) and the
number of units of advertising (y), according to the function
p(x, y) = 412x + 806y – x2 – 4y2 – xy – 50000
(i) Determine the number of workers and the number of units of advertising that
results in maximum the profit. [Answer: x = 166, y = 80]
(ii) Determine the maximum profit. [Answer: 16595]
28. The total cost of making x gallons of oil is C(x) dollars, where
C(x) = 50 + 1.5x + 0.02x2.
a) Write the expression for the marginal cost of the xth gallon.
b) Find the marginal cost of the fifth gallon.
c) Find the marginal cost of the 40th gallon.
[Answer: a) 1.5 + 0.04x, b) $1.7, c) $3.1]
29. Use the Lagrangian function to optimize the function
f(x, y) = 3x2 + 5xy – 6y2 + 26x + 12y; subject to 3x + y = 170
145 139 139
[Answer: x = , y = 25, λ =  and Optimum value =  ]
3 3 3
30. Determine the maximum value of the objective function
f(x, y) = 10x + 4y – 2x2 – y2 subject to 2x + y = 5.
11 4 91
[Answer: Critical point ( , ) and Maximum value = ]
6 3 6

179
 S. M. Shahidul Islam

11
C h a p te r
Integration and its applications

Highlights:
11.1 Introduction 11.8 Definite integral
11.2 Definition of integration 11.9 Properties of definite integral
11.3 Indefinite integral 11.10 Application of integration in
11.4 Fundamental theorem on integration business problems
11.5 Some standard integrals 11.11 Some worked out examples
11.6 Integration by substitution 11.12 Exercise
11.7 Integration using partial fractions

11.1 Introduction: Integral calculus is also the most important part of mathematics.
There are two types of integration. One is indefinite integration and the other is definite
integration. Indefinite integration deals with the inverse operation of differentiation, i.e.,
anti-derivative. Definite integration is the limit of a special type of addition process of
infinitesimal parts of a region. So, it may be expressed as the area enclosed by a set of
curves. Integral calculus has a great use in business problems. For example with a given
marginal cost function it would be possible to find cost function. In this chapter we discuss
nature of integration, how to find the integral value of some given functions and the use in
business problems.

11.2 Definition of integration: If F(x) be any differentiable function of x such that

d
F ( x)  f ( x)
dx
then F(x) is called an anti-derivative or an indefinite integral or simply an integral of
f(x). Symbolically, we write this as follows:
F(x) =  f ( x)dx
and is read as ‘F(x) is the integral of f(x) with respect to x’. Here, the function f(x) is
known as integrand.
The process of finding the integral of a given function is called integration and the given
function is the integrand.

180
 Integration

Note: 1. Integration is the sum of a certain infinite series. The symbol  used for
integral is a distorted form of the letter S, the first letter of the word ‘Sum’.
2. The symbol  dx is the integral operator with respect to x.

3. It is clear from the definition that

d
dx
 f ( x)dx  f ( x) .
11.3 Indefinite integral: Let us consider x3, x3 + 2 and x3 + c, c is a constant. Since
integration is anti-derivative, we have
d 3
dx
(x ) = 3x2,  3x 2 dx = x3;
d 3
dx
(x + 2) = 3x2,  3x 2 dx = x3 + 2;
d 3
(x + c) = 3x2,  3x dx = x + c.
2 3
And
dx
So,  3x 2 dx does not give a definite value. For this such type of integration is called

 3x x3 + c,
2
indefinite integral. The general value of dx is where c is called
integral constant. So, every indefinite integral of F(x) can be obtained from f(x) + c, by
taking a suitable value of c. After this we shall use ‘c’ as integral constant in this
chapter.

11.4 Fundamental theorem on integration:

1.  kf ( x)dx  k  f ( x)dx ; where k is constant.
2.   f ( x)  f
1 2 ( x)      f n ( x)dx   f1 ( x)dx   f 2 ( x)dx       f n ( x)dx
d 
3.  (uv)dx  u  vdx    dx (u) vdxdx ; where u and v are functions of x. This is

known as integration by parts formula.

11.5 Some standard integrals:

d
1.
dx
(x) = 1   dx = x + c

d n x n 1
( x )  nx n 1   x dx   c [n ≠ – 1]
n
2.
dx n 1
d d 1 1
 x dx   x dx  ln x  c  log e x  c
1
3. (log ex) = (ln x) = 
dx dx x

181
 S. M. Shahidul Islam

d x
(e ) = ex  e dx = ex + c
x
4.
dx
d mx e mx
5.
dx
(e ) = memx   e mx dx 
m
c

d x ax
(a ) = ax loge a   a dx  c
x
6.
dx log e a
d mx a mx
  a dx  m log e a  c
mx mx
7. (a ) = m a logea
dx
Note: We may use these standard derivatives as formula where necessary.

Example: Evaluate  10 x 5 dx
x 51 10 x 6 5
10 x dx = 10  c = x 6  c (Answer)
5
Solution: +c =
5 1 6 3
Example: Evaluate  ( x  3x  5x)dx
5 2

Solution:  ( x 5  3x 2  5x)dx =  x dx   3x dx   5xdx

5 2

 x dx  3 x dx  5 xdx
5 2
=
x 51 x 21 x11
=  3. 5 c
5 1 2 1 11
1 5
= x6  x3  x2  c (Answer)
6 2

Example: Evaluate ∫(1+ x)(1 – 3x)dx

Solution: ∫(1+ x)(1 – 3x)dx = ∫(1 – 3x + x – 3x2)dx
= ∫(1 – 2x – 3x2)dx
= ∫1dx – 2∫xdx – 3∫x2dx
x2 x3
= x – 2. – 3. +c
2 3
= x – x2 – x3 + c (Answer)

Example: Evaluate ∫(3x-1 + 4x2 – 3x + 8)dx

Solution: ∫(3x-1 + 4x2 – 3x + 8)dx = 3∫x-1dx + 4∫x2 dx – 3∫xdx + ∫8dx
x3 x2
= 3lnx + 4. – 3. + 8x + c
3 2
4 3
= 3lnx + x3 – x2 + 8x + c (Answer)
3 2

182
 Integration

Example: Evaluate ∫(2 x + 5x-1 + 3x2 )dx

Solution: ∫(2 x + 5x-1 + 3x2 )dx = 2∫ x dx + 5∫x-1 dx + 3∫x2 dx
1
= 2∫ x dx + 5∫x-1 dx + 3∫x2 dx
2

1
1
x 2
x3
= 2. + 5lnx + 3. +c
1 3
1
2
3
2
x
= 2. + 5lnx + x3 + c
3
2
3
4 2
= x  5 ln x  x 3  c (Answer)
3

Example: Integrate ∫(3x2 + ex + ax)dx

Solution: ∫(3x2 + ex + ax)dx = 3∫x2 dx + ∫ex dx + ∫ax dx
x3 x ax
= 3. +e + +c
3 log e a
3 axx
= x +e + + c (Answer)
log e a

Example: Integrate ∫(3x5 + e(n+1) x + a5x)dx

Solution: ∫(3x5 + e(n+1) x + a5x)dx = ∫x5 dx + ∫e(n+1) x dx + ∫a5x dx
x6 e ( n 1) x a 5x
= + + + c (Answer)
6 n 1 5 log e a

 xe
x
Example: Integrate by parts: dx [AUB-02]
d 
 xe dx = x  e x dx    ( x)  e x dx  dx ;
x [Let first function, u = x
Solution:
 dx  Second function, v = ex]
= xe   1.e dx
x x

= xe x   e x dx
= xe x  e x + c
= e x ( x  1) + c (Answer)

183
 S. M. Shahidul Islam

Example: Evaluate ∫logex dx [RU-88, 92]

Solution: ∫logex dx = ∫(logex).1 dx
d
= (logex) ∫1dx – ∫[ (logex) ∫1dx]dx
dx
1
= (logex) x – ∫ .x dx
x
= xlogex – ∫1dx
= xlogex – x + c
= x(logex – 1) + c (Answer)

x
2
Example: Integrate by parts: ln x dx [AUB-01, NU-94]

x  (ln x) .x
2 2 [Let first function, u = lnx
Solution: ln x dx = dx ; Second function, v = x2]
d 
= ln x  x 2 dx    (ln x)  x 2 dx dx
 dx 
3 3
x 1 x
= ln x.   . dx
3 x 3
1 1 2
= ln x   x dx
3 3
1 1 x3
= ln x  .  c
3 3 3
1
= (3 ln x  x 3 )  c ] (Answer)
9

11.6 Integration by substitution: In this method we first change the variable of

integration to a new variable by taking a proper relationship between two variables so that
the given problem reduces to a known problem. After completing the integration the new
variable is to be changed by the old one. Experience is the best guide to choose the
suitable substitution.

Example: Evaluate ∫(2x + 3)5dx

du d
Solution: Let u = 2x + 3 =>  (2x + 3)
dx dx
du
=>  2
dx
du
=>  dx
2

184
 Integration

du
So, ∫(2x + 3)5dx = ∫u5
2
1 5
= ∫u du
2
1 u6
= . c
2 6
1
= (2x + 3)6 + c [Putting value of u] (Answer)
12

dx
Example: Evaluate  (5x  6) 3

du du
Solution: Let u = 5x – 6  5 =>
=>  dx
dx 5
dx -3 du 1 u 2 1 1
So,  (5x  6) 3
=∫u
5
= .
5 2
+ c =  (5x – 6)-2 + c =
10 10(5 x  6) 2
+c

dw
Example: Evaluate  (7  2w) 3

du du
Solution: Let u = 7 – 2w =>  – 2 =>  dw
dw 2
dw du 1 u 2 1 1
So,  (7  2w) 3
= ∫ u-3
2
  .
2 2
+ c = (7 – 2w)-2 + c =
4 4(7  2w) 2
+c

Example: Evaluate ∫x2(2x3 + 3)5dx

du du
Solution: Let u = 2x3 + 3 => =>  6x2 =>  x2dx
dx 6
So, ∫x2(2x3 + 3)5dx = 3 5 2
∫(2x + 3) .x dx
du
= ∫ u5
6
1
= . ∫u5 du
6
1 u6
= . +c
6 6
1
= (3x3 + 3)6 + c [Putting the value of u]
36

185
 S. M. Shahidul Islam

Example: Integrate ∫e2x – 1 dx

du du
Solution: Let u = 2x – 1 =>  2 => = dx
dx 2
du 1 1
So, ∫e2x – 1 dx = ∫ eu = eu + c = e2x – 1 + c (Answer)
2 2 2

1
Example: Evaluate  ax  b dx [AUB-03]

du du
Solution: Let u = ax + b  a =>
=> = dx
dx a
1 1 du 1 1
So,  ax  b dx = u a a
. = logeu + c =
a
loge(ax + b) + c (Answer)

1 1
Therefore,  ax  b dx = a
log e (ax  b)  c
1 1 We can use these as formulae
Similarly,
 ax  b dx = a log e (ax  b)  c

Example: Integrate ∫ 2x ln(a + x2)dx [AUB-98]

du
Solution: Let u = a + x2 =>  2x => du = 2x dx
dx
So, ∫ 2x ln(a + x2)dx = ∫ ln(a + x2) 2x dx
= ∫ ln u du
= ∫ (ln u).1 du
d
= (ln u) ∫1du – ∫[ (ln u) ∫1du]du
du
1
= (ln u) u – ∫ [ .u] du
u
= u ln u – ∫1du
= u ln u – u + c
= u(ln u – 1) + c
= (a + x2)[ln(a + x2) – 1] + c (Answer)

11.7 Integration using partial fractions: When a fraction function is given, first we try
to integrate it by substitution method, if it is not possible then we find the partial fractions
and integrate then.

186
 Integration

3x  2
Example: Integrate
(1  x)(1  2 x)
3x  2 A B
Solution: Let = +
(1  x)(1  2 x) (1  x) (1  2 x)
Multiplying both sides by (1 + x)(1 + 2x), we get
3x + 2 = A(1 + 2x) + B(1 + x)
1
Putting x = –1 and  , we have
2
– A = – 1 => A = 1
1 1
And B= => B = 1
2 2
3x  2 1 1
So, = + , these are the partial fractions.
(1  x)(1  2 x) (1  x) (1  2 x)
3x  2 1 1
Hence ∫ dx = ∫ dx + ∫ dx
(1  x)(1  2 x) (1  x) (1  2 x)
1 1
=∫ dx + ∫ dx
( x  1) (2 x  1)
1
= loge(1 + x) + loge(1 + 2x) + c (Answer)
2

x3
Example: Evaluate ∫ dx
( x  a)( x  b)( x  c)
x3 A B C
Solution: Let =1+ + +
( x  a)( x  b)( x  c) ( x  a) ( x  b) ( x  c)
Multiplying both sides by (x – a)(x – b)(x – c), we get
x3 = (x – a)(x – b)(x – c) + A(x – b)(x – c) + B(x – a)(x – c) + C(x – a)(x – b)
Putting x = a, b and c, we have
a3 b3 c3
A= , B= and C =
(a  b)(a  c) (b  a)(b  c) (c  a)(c  b)
x3 a3 b3
So, =1+ +
( x  a)( x  b)( x  c) (a  b)(a  c)( x  a) (b  a)(b  c)( x  b)
3
c
+
(c  a)(c  b)( x  c)

187
 S. M. Shahidul Islam

x3 a3
Hence, ∫ dx = ∫1dx + ∫ dx +
( x  a)( x  b)( x  c) (a  b)(a  c)( x  a)
b3 c3
∫ dx + ∫ dx
(b  a)(b  c)( x  b) (c  a)(c  b)( x  c)
a3 b3
= x + loge(x – a) + loge(x – b) +
(a  b)(a  c) (b  a)(b  c)
c3
loge(x – c) + k ; k is arbitrary constant.
(c  a)(c  b)

11.8 Definite integral: Let f(x) be a continuous function on the interval [a, b] and F(x)
is an anti-derivative for f(x) on [a, b] then
b
a
f ( x)dx  F (b)  F (a)

is called the definite integral of f(x) from a to b. The number a and b are respectively
called the lower limit and the upper limit of the definite integral. Here, the arbitrary
(integral) constant, c disappears.
b
Note:  f ( x)dx
a
represents the area under the curve f(x) from a to b.

11.9 Properties of definite integral:

b b
1.  a
f ( x)dx   f (u )du ; i.e., Definite integral is independent of variable.
a
a
2.  a
f ( x)dx  0 ;
b a
3.  a
f ( x)dx   f ( x)dx ;
b
b c b
4.  a
f ( x)dx   f ( x)dx   f ( x)dx ; where a < c < b.
a c
a a
5.  0
f ( x)dx   f (a  x)dx ;
0
a a
6.  a
f ( x)dx  2 f ( x)dx ; if f(x) is an even function, i.e., f(–x) = f(x).
0
a
7.  a
f ( x)dx  0 ; if f(x) is an odd function, i.e., f(–x) = – f(x).
All the properties can be proved simply by the definition of definite integral.

188
 Integration

3
Example: Evaluate 2
x 3 dx
3 3
3 x 31 x4 34 2 4 81 16 65
  
3
Solution: x dx = = = = = (Answer)
2 3 1 2 4 2 4 4 4 4 4

10 1
Example: Evaluate 6 x2
dx
10 1
 dx = log e ( x  2) 6
10
Solution:
6 x2
= log e (10  2)  log e (6  2)
= log e 12  log e 8
 12 
= log e  
8
3
= log e   (Answer)
2

3
e
3x
Example: Evaluate dx
2
3
3 e 3x e 33 e 32 1 1
Solution:  e dx = 3x
=  = (e 9  e 6 ) = e 6 (e 3  1) (Answer)
2 3 2 3 3 3 3

3 4
Example: Evaluate 1 2x  3
dx
du du
Solution: Let u = 2x + 3 2 => =>
= dx
dx 2
When x = – 1, u = 1 and when x = 3, u = 9
3 4 9 4 du
So, 1 2 x  3 dx = 1 u . 2
91
= 2  du
1 u
9
= 2 ln u 1
= 2( ln 9  ln 1)
= 2(2.1972 – 0)
= 4.3944 (Answer)

189
 S. M. Shahidul Islam

1
Example: Compute the area under f(x) = 3 x 2 – 2 over the interval x = 4 to x = 16.
Solution: The required area is computed as follows:
16 1
Area = 
4
(3x 2
 2)dx
f(x)
16
3
x 2
= 3.  2x
3 f(x) = 3x1/2 – 2
2 4
3 16
= 2x 2
 2x
4

= 2(16) 2  2(16)  2(4) 2  2(4)

3 3

    4 16 x
= (2  64 – 32) – (2  8 – 8) Figure 11.1
= 96 – 8
= 88 square unit (Answer)

2
Example: Evaluate  2
x 3 dx
2 2
2 x 31 x4 2 4 (2) 4 16 16
Solution:  x dx = 3
= =  =  = 0 (Answer)
2 3  1 -2 4 -2 4 4 4 4
Another way: f(x) = x3 is an odd function, because f(–x) = (–x)3 = – x3 = – f(x).
2
So, by property (7),  2
x 3 dx = 0. (Answer)

11.10 Application of integration in business problems:

Integral is the anti-derivative of a function. So, the business functions which are related
with differentiation are also associated with integration. For example: we know that if the
total revenue function, say r(x), is given then the marginal revenue function is the first
derivative of total revenue function, say r/(x). Therefore, it follows that the total revenue
function is the integral of the marginal revenue function.
That is, total revenue function, r(x) = ∫ r/(x) dx.
r ( x)
And the average revenue function = .
x

Example: Let the marginal revenue function and marginal cost function of a firm are
r/(x) = 16 – x2 and c/(x) = 3x2 – 2x + 8 respectively; where x is the quantity of product.
And fixed cost of the firm is Tk. 500, i.e., c(0) = 500, then find
(i) the total revenue function.
(ii) the average revenue function.
(iii) the demand function.

190
 Integration

(iv) the maximum total revenue.

(v) the total revenue from product 1 to 3.
(vi) the total cost function.
(vii) the total profit function.
Solution: Given that, the marginal revenue function, r/(x) = 16 – x2.
(i) The total revenue function: r(x) = ∫ r/(x) dx.
= ∫ (16 – x2)dx
1
= 16x – x3 + c
3
We know that, total revenue = 0 if we produce no product, i.e., x = 0. So, c = 0
1
Therefore, the total revenue function: r(x) = 16x – x3 (Answer)
3
r ( x) 1 1
(ii) The average revenue function = = (16x – x3)/x = 16 – x2. (Answer)
x 3 3

(ii) We know that the average revenue from a product is the demand of that product.
1
So, the demand function, d(x) = 16 – x2. (Answer)
3

(iv) We shall find the maximum or minimum total revenue where the marginal revenue,
r/(x) = 0. i.e., 16 – x2 = 0 => x2 = 16 => x = 4 and – 4
But, in Economics negative production is not possible. So, x = 4 is acceptable only.
Here, second derivative of the total revenue function, r(x) is
d
r//(x) = (16 – x2) = – 2x.
dx
And r//(4) = – 2(4) = – 8; which is negative.
So, the maximum revenue will be get for x = 4 unit products.
4

 (16  x
2
Hence, the maximum total revenue = )dx
0
4
x3
= 16 x 
3 0

 (4) 3   (0) 3 
= 16(4)  
 16( 0)  
 3   3 
64
= 64 
3
128
= (Answer)
3

191
 S. M. Shahidul Islam

3
(v) The total revenue from product 1 to 3 =  (16  x 2 )dx
1
3
x3
= 16 x 
3 1

 (3)   3
(1) 3 
= 16(3)    16(1)  
 3   3 
47
= 39 –
3
70
= (Answer)
3
(vi) Given that, the marginal cost function: c/(x) = 3x2 – 2x + 8 and c(0) = 500.
So, the total cost function, c(x) = ∫ c/(x) dx.
= ∫ (3x2 – 2x + 8) dx
= x3 – x2 + 8x + c
But given, c(0) = 500. So, c = 500
Hence, the total cost function: c(x) = x3 – x2 + 8x + 500. (Answer)

(vii) We know that, total profit = total revenue – total cost

So, the total profit function, p(x) = r(x) – c(x)
1
= (16x – x3)– (x3 – x2 + 8x + 500)
3
4
= – x3 + x2 + 8x – 500 (Answer)
3

Consumer’s surplus & producer’s surplus: We know that demand is decreasing and
supply is increasing with respect to the price. Let (x0, p0) is the equilibrium point at which
the demand, d(x) and the supply, s(x) are equal. Then
p

s(x)
Price Equilibrium point Supply function

p0
(x0, p0)
Demand function
x0 d(x)
0 Quantity x
Figure 11.2

192
 Integration

x0

The consumer’s surplus =  d ( x)dx  x

0
0  p0 . That is, the consumer’s surplus is the

difference between the cost consumers are willing to pay for a commodity and what they
actually pay.
x0

The producer’s surplus = x0  p0   s( x)dx . That is, the producer’s surplus is the
0
difference between the revenue producers actually receive and what they have been
willing to receive.

Example: The demand function for a product is p = 20 – x – x2. Find the consumer’s
surplus when the demand is 3.
Solution: When demand x0 = 3, the price p0 = 20 – 3 – 32 = 8
x0

So, the consumer’s surplus =  d ( x)dx  x

0
0  p0

 (20  x  x )dx  3  8
2
=
0
3
x2 x3
= 20 x    24
2 3 0

3 2 33
= 20(3)    24
2 3
9
= 60   9  24
2
45
= (Answer)
2

Rate of sales: When the rate of sales of a product is a known function of t, say f(t)
where t is a time measure, the total sales of this product over a time period T is
T

 f (t )dt
0

Example: Suppose the rate of sales of a new product is given by f(t) = 200 – 90e-t, where
t is the number of days the product is on the market. Find the total sales during the first 5
days.
5
Solution: The total sales during the first 5 days =  f (t )dt
0

193
 S. M. Shahidul Islam

5
=  (200  90e t )dt
0
5
= 200t  90e t
0
5
= 1000  90e – 90e0
= 1000 + 90(0.0067) – 90
= 910.603 units. (Answer)

Amount of an annuity: If an annuity consists of equal annual payments Tk. P in which an

interest rate of Tk. j for one taka per annum is compounded continuously, the amount
Tk. F of the annuity after n payment is
n

 Pe
jt
F= dt
0
Example: A bank pays interest at the rate of 10% per annum compounded continuously.
If a person places Tk. 1000 in a saving account each year, how much will be in the
account after 6 years?
10
Solution: Here, P = 1000, n = 6 and j = = 0.1
100
6
The amount after 6 years is F =  1000e 0.1t dt
0
6
e 0.1t
= 1000 
0.1 0
= 10000{e0.1(6) – e0.1(0)}
= 10000(1.822 – 1)
= Tk. 8220 (Answer)

11.11 Some worked out examples:

xdx
Example (1): Evaluate  [NU-97, CU-87]
1 x2
xdx
Solution: Let, I = 
1 x2
1 du
=  . Let, u = 1 + x2
u 2 du du
1 
1   2 x =>  xdx
=  u 2 du dx 2
2

194
 Integration

1
2
1u
= c
2 1
2
1
= (1  x )  c 2 2

= (1  x 2 )  c (Answer)

x 1
e
Example (2): Evaluate  x 1
dx [NU-96]
x 1
e
Solution: Let I =  x 1
dx Let, u = x 1
du 1
=  e .2du
u  
dx 2 x  1
= 2 e u du => 2du 
1
dx
= 2eu + c x 1
= 2e x 1  c (Answer)

1
Example (3): Evaluate e x
1
dx [RU-81]
1
Solution: Let, I = e
1
dx
x

ex
=  x x dx [Multiplying numerator and denominator by e-x]
e (e  1)
ex
=  1  e  x dx du
Let, u = 1 + e-x

1   e  x
=   du dx
u =>  du  e  x dx
=  log e u  c
=  log e (1  e  x )  c (Answer)

Example (4): ∫ (x2 + 1)x dx [CU-87]

2
Solution: Let, I = ∫ (x + 1)x dx
du Let, u = x2 + 1
= ∫ u. du du
2   2 x =>  xdx
dx 2

195
 S. M. Shahidul Islam

1
= ∫ u du
2
1 u2
= . +c
2 2
1
= (x2 + 1)2 + c (Answer)
4

Example (5): Evaluate ∫ x2e2x dx [RU- ’93, ‘81]

Solution: Let, I = ∫ x2e2x dx
d 2
= x2 ∫ e2x dx – ∫ [ (x ) ∫ e2x dx]dx
dx
e2x e2x
= x 2. – ∫ [2x. ]dx
2 2
1
= x2e2x – ∫ xe2x dx
2
1 d
= x2e2x – [x ∫ e2x dx – ∫ { (x) ∫ e2x dx}dx
2 dx
2x
1 e e2x
= x2e2x – [x. – ∫ 1. dx]
2 2 2
1 1 1
= x2e2x – [ xe2x – ∫ e2x dx]
2 2 2
1 1 1 e2x
= x2e2x – [ xe2x – . ]+c
2 2 2 2
1 1
= e2x [ x2 – x + ] + c (Answer)
2 2

3x 2  2
Example (6): Evaluate  3 dx [RU-88]
x  2x  5
3x 2  2 Let, u = x3 + 2x + 5
Solution: Let I =  3 dx du
x  2x  5  = 3x2+ 2
1 dx
=  du => du = (3x2+ 2)dx
u
= logeu + c
= loge(x3 + 2x + 5) + c (Answer)

Example (7): Evaluate ∫ x2ex dx [RU-80]

Solution: Let, I = ∫x e
2 x
dx

196
 Integration

d 
= x 2  e x dx    ( x 2 )  e x dx dx ;
 dx 
= x e   2 x.e dx
2 x x

= x 2 e x  2 xe x dx
d
= x 2 e x  2[ x  e x dx   {
( x)  e x dx}dx]
dx
= x e  2[ xe   1.e dx]
2 x x x

= x 2 e x  2[ xe x   e x dx]
= x 2 e x  2[ xe x  e x ]  c
= e x ( x 2  2 x  2]  c (Answer)

1
Example (8): Evaluate ∫ log e (log e x)dx
x Let, u = logex
1
Solution: Let I = ∫ log e (log e x)dx 
du 1

1
=> du = dx
x dx x x
= ∫ logeu du
d
= (logeu) ∫1du – ∫ [ (logeu) ∫1du]du
du
1
= (logeu)u – ∫ .u du
u
= (logeu)u – ∫ du
= (logeu)u – u + c
= u(logeu – 1) + c
= (log e x){log e (log e x)  1}  c (Answer)

1
Example (9): Evaluate ∫ dx [RU-81]
x(1  log e x) 3
1
Solution: Let, I = ∫ dx
x(1  log e x) 3 Let, u = 1 + logex
du 1 1
=
1
∫ 3 du   => du = dx
u dx x x
= ∫ u-3du
u 2
= +c
2

197
 S. M. Shahidul Islam

1
=  (1  log e x) 2 + c (Answer)
2

3 2 xdx
Example (10): Evaluate  1 x
1 2
[RU-80]
32 xdx Let, u = 1 + x2
Solution: Let, I =1 1  x 2 du
10
  2 x => du  2 xdx
du dx
=  u = 2 when x = 1
2
u
10
and u = 10 when x = 3
= ln u 2
= ln 10  ln 2
10
= ln
2
= ln 5 (Answer)

2
x
Example (11): Evaluate x
0
2
4
dx [AUB-01, RU-81]

2
x
Solution: Let, I = x
0
2
4
dx
du
Let, u = x2 + 4
du
8   2 x =>  xdx
1 du dx 2
= u. 2
4
u = 4 when x = 0
and u = 8 when x = 2
181
2 4 u
= du

1 8
= log e u 4
2
1
= (log e 8  log e 4)
2
1 8
= log e  
2 4
1
= log e 2 (Answer)
2

ln 2
ex
Example (12): Evaluate 
0 1 ex
dx [RU-96]

198
 Integration

ln 2
ex
Solution: Let, I = 
0 1 ex
dx

3
1 Let, u = 1 + ex
= 2 u du 
du
 e x => du  e x dx
3 dx
= ln u 2
When x = 0, u = 2 and
= ln 3  ln 2 when x = ln 2, u =1 + eln2 =1 + 2= 3
3
= ln
2
= ln 1.5 (Answer)

2
1
Example (13): Evaluate  x( x  1) dx
1
[RU-82]
2
1
Solution: Let, I =  x( x  1) dx
1

1 1 
2
=   x  x  1dx
1
[By taking partial fractions]
2 2
1 1
=  dx   dx
1
x 1
x 1
= log e x 1  log e ( x  1) 1
2 2

= (log e 2  log e 1)  (log e 3  log e 2)

= log e 2  0  log e 3  log e 2
= 2 log e 2  log e 3
= log e 2 2  log e 3
4
= log e   (Answer)
3

5
1
Example (14): Evaluate  y dy
0
[RU-86]

5
1
Solution: Let, I =  y dy
0
5
= ln y 0
= ln 5  ln 0
= Not found, because the value of ln 0 is not known.

199
 S. M. Shahidul Islam

Example (15): Find the area bounded by the curve f(x) = 15 – 2x – x2 and the straight
line h(x) = 9 – x. [AUB-02]
Solution: Taking f(x) = h(x), we get
15 – 2x – x2 = 9 – x
Or, x2 + x – 6 = 0
Or, x2 + 3x – 2x – 6 = 0
Or, x(x + 3) – 2(x + 3) = 0
Or, (x + 3)(x – 2) = 0
Or, x = – 3 and x = 2
So, – 3 and 2 are the x – coordinates of the points of intersection of f(x) and h(x).
Therefore, the area bounded by the curve f(x) and the straight line h(x) is
2

 [ f ( x)  h( x)]dx
3
2 f(x) = 15 – 2x – x2
 [(15  2 x  x )  (9  x)]dx
2
=
3
2
h(x) = 9 – x
 [(6  x  x
2
= )dx
3
2 X
x2 x3 -3 0 2
= 6x  
2 3 3 Figure 11.3
 (2) (2)  
2 3
(3) (3) 
2 3
= 6(2)     6(3)   
 2 3   2 3 
8 9
= (12  2   18   9)
3 2
125
= square unit. (Answer)
6
Example (16): If marginal cost of a firm is c/(x) = x2 and c(0) = 0.1 find the total cost
function. [RU-82]
Solution: Given that the marginal cost function, c/(x) = x2.
So, the total cost function, c(x) = ∫ c/(x) dx
= ∫ x2 dx
1
= x3  c
3
And given that c(0) = 0.1
1 3
 (0)  c = 0.1 => c = 0.1
3
1
Therefore, the total cost function is c(x) = x 3  0.1 (Answer)
3

200
 Integration

6
Example (17): Let the marginal revenue function of a firm is r/(x) = 5 + ; where
( x  2) 2
x is the quantity of sold products. Find the total revenue function. [NU- 98 Mgt.]
6
Solution: Given that the marginal revenue function: r/(x) = 5 +
( x  2) 2
6
So, total revenue function: r(x) = ∫[5 + ]dx
( x  2) 2
6
= 5x – +c
( x  2)
Since, revenue = 0 when x = 0 unit product is sold, i.e., r(0) = 0.
6
So, 0 = 5(0) – + c => c = 3
(0  2)
6
Hence, the total revenue function is r(x) = 5x – + 3 (Answer)
( x  2)

Example (18): Find the consumer’s surplus and producer’s surplus under pure
competition for the demand function p = 61 – x2 and supply function p = 11 + x2, where
p is the price and x is quantity.
Solution: Under pure competition, market equilibrium conditions can be obtained by
equating the demand and supply.
 61 – x2 = 11 + x2
Or, 2x2 = 50
Or, x2 = 25
Or, x = 5, – 5 [x = – 5 is not acceptable]
 x0 = 5
So, p0 = 61 – 25 = 36
x0

The consumer’s surplus =  d ( x)dx  x

0
0  p0 .

 (61  x )dx  5  36 .
2
=
0
5
x3
= 61x   180
3 0

(5) 3
= 61(5)   180
3
250
= (Answer)
3

201
 S. M. Shahidul Islam

x0

And the producer’s surplus = x0  p0   s( x)dx

0
5
= 5  36   (11  x 2 )dx
0
5
x3
= 180  11x 
3 0

(5) 3
= 180  11(5) 
3
250
= (Answer)
3

11.12 Exercise:
1. Define integral and definite integral. Give examples.
2. What is difference between differentiation and integration?
3. Why does arbitrary constant c disappear in the definite integral? Explain by a
example.
4. Evaluate the following integrals:
x11
(i) ∫ x10dx [Answer: c]
11
2 3
(ii) ∫ x1/2dx [Answer: x 2 + c]
3
1 1
(iii) ∫ 3 dx [Answer:  2 + c]
x 2x
4 5
(iv) ∫5x dx [Answer: x + c]
(v) ∫(4x3 + 3x2 + 2x + 3)dx [Answer: x4 + x3 + x2 + 3x + c]
1 2
(vi) ∫(x2 – 1)2dx [Answer: x5 - x3 + x + c]
5 3
3
 1 4
x 3x 2 1
(vii) ∫  x   dx [Answer:   3 ln x  2  c ]
 x 4 2 2x
x 1
4
x 3
1
(viii) ∫ 2
dx [Answer:  c]
x 3 x
x 2 2 3 1
(ix) ∫( x   )dx [Answer: x 2  x 2  4 x  c ]
2 x 3 4
1 x 4 1 2x 1 3 2
(x) ∫( 2  e   3 )dx [Answer:
x
 e  x  4 log e x  x 3  c ]
2 x x log e 2 2 2

202
 Integration

5. Integrate by parts:
(i) ∫ln x dx [Answer: x(ln x – 1) + c]
1 1
(ii) ∫ x lnx dx [Answer: x 2 ln x  x 2  c ]
2 4
n 1
x x n 1
(iii) ∫ xnlogex dx [Answer: log e x  c]
n 1 (n  1) 2
(iv) ∫ xe-x dx [Answer: – e-x(x + 1) + c ]
x 2 e 3 x 2 xe 3 x 2e 3 x
(v) 2 3x
∫ x e dx [Answer:   c]
3 9 27
6. Evaluate the following integrals by method of substitution:
1 1
(i)  x ln xdx [Answer: (ln x) 2  c ]
2
 3 1
(ii)  (5  2 x) 2 dx [Answer: (5  2 x) 12  c ]
(ax 2  2bx  2c) 6
 (ax  2bx  2c) (ax  b)dx [Answer:  c]
2 5
(iii)
12
dx 1
(iv)  1  4x [Answer:
4
ln(1  4 x)  c ]
xdx 1
(v)  2x 2
3
[Answer: ln( 2 x 2  3)  c ]
4
8x 2 4
(vi)  ( x 3  2) 3 dx [Answer: 
3( x  2) 2
3
c]

4x 3  2x
 x 4  x 2  2 dx [Answer: log e ( x  x  2)  c ]
4 2
(vii)
7. Integrate using partial fractions:
1 1
(i)  ( x  1)( x  3) dx [Answer: 2 log e ( x  1)  log e ( x  3)  c ]
1 1  x2 
(ii) ∫ dx [Answer: log 
e
c]
2 
x  x3 2 1 x 
x2
(iii)  x 2  13x  42 dx [Answer: 19 log e ( x  7)  18 log e ( x  6)  c ]
x 1
(iv)  ( x  1)(2 x  1) dx [Answer: 6 {2 log e ( x  1)  log e (2 x  1)}  c ]
x2 a2 b2
(v)  ( x  a)( x  b) dx [Ans: x  a  b log e ( x  a)  a  b log e ( x  b)  c ]

203
 S. M. Shahidul Islam

8. Evaluate the following definite integrals:

3 26
1 x dx
2
(i) [Answer: ]
3
4
(ii) 2
(7 x 3  x)dx [Answer: 426]
2

 (x
2
(iii)  x)dx [Answer: 2]
1
3 1 4 2
e e (e  1) ]
2x
(iv) dx [Answer:
2 2
3 2x
(v)  1 x
1 2
dx [Answer: ln10 – ln2]
2
1 4
 (e  3x 2 )dx (e  e 2  14) ]
2x
(vi) [Answer:
1
2
6 60dx
(vii) 1 (3x  2) 2
[Answer: 3]
2
1 6
 (e  4 x 2 )dx (e  e 3  28) ]
3x
(viii) [Answer:
1
3
8 1
9. Prove that  3 x3
dx  log e 5 .
185
10. Find the area under f(x) = x1/3 + 5 over the interval x = 1 to x = 8. [Answer: ]
4
256
11. Find the area bounded by the curve f(x) = 16 – x2 and the x-axis. [Answer: ]
3
12. Find the area bounded by the curve f(x) = 3x3 – 3x and the straight line h(x) = x.
2
0 3
8
[Answer:
3
] [Hints: Area =  [ f ( x)  h( x)]dx 
2
 [ f ( x)  h( x)]dx ]
0

3

13. If marginal cost c/(x) = 8x3, and c(1) = 5, find the total cost function. [Answer:
c(x) = 2x4 + 3]
14. If marginal cost c/(x) = 2 + x + x2, and c(0) = 50, find the total cost function.
1 1
[Answer: c(x) = 2x + x2 + x3 + 50]
2 3
15. Let the marginal cost function (in thousand taka) of a firm is c/(x) = 4x2 – 6x + 1;
where x is the quantity of products, and the fixed cost of the firm is Tk. 400000,
i.e., c(0) = 400000. Find the total cost of the firm for making x = 12 unit product.
[Answer: Tk. 2284000]

204
 Integration

16. If the marginal revenue function of a firm r/(x) = 5 + 3x – x2 and r(6) = 112, find
the total revenue function and the average revenue function. [Answer: revenue:
3 1 r ( x) 100 3x x 2
r(x) = 100 + 5x + x 2 – x 3 and avg. revenue:  5  ]
2 3 x x 2 3
17. Let the marginal cost and the marginal revenue functions of a firm are
c/(x) = 0.1x2 – 4x + 110 and r/(x) = 150 – x; where x is the number of unit product.
If total cost is Tk. 4000 for making 30 units product, find
(i) fixed cost.
(ii) condition for r(0) = 0
(iii) profit function.
(iv) number of unit production to maximize the profit.
[Answer: (i) Tk. 1600, (ii) When no product is sold,
3 1 3
(iii) p(x) = 40x + x 2 - x - 1600 and (iv) 40 ]
2 30
18. Demand function is p = 20 – 2x; where x is the quantity. If x = 12, find the
consumer’s surplus. [Answer: 36]
19. Assume that in 1995 the annual world use of natural gas was 50 trillion cubic feet.
The annual consumption of gas is increasing at a rate of 3% compounded
continuously. How long will it take to use all available gas, if it is known that in
1995 there were 2200 trillion cubic feet of proven reserves? Assume that no new
t
discoveries are made. [Answer: 28.1 years] [Hints: Find t of  50e 0.03t dt  2200 ]
0
20. A bank pays interest at the rate of 6% per annum compounded continuously. Find
how much should be deposited in the bank each year in order to accumulate
3
Tk. 6000 in 3 years. [Answer: Tk. 1818.18] [Hints: Find P of 6000   Pe 0.06t dt ]
0

205
 S. M. Shahidul Islam

12
C h a p te r
Coordinate Geometry
Highlights:
12.1 Introduction 12.10 Area of a quadrilateral
12.2 Directed line 12.11 Straight line
12.3 Quadrants 12.12 Slope or gradient of a straight line
12.4 Coordinates 12.13 Different forms of equations of the
12.5 Coordinates of mid point straight line
12.6 Distance between two points 12.14 Circle
12.7 Section formula 12.15 Some worked out examples
12.8 Coordinates of the centroid 12.16 Exercise
12.9 Area of a triangle

12.1 Introduction: Coordinate geometry is the contribution of French mathematician

Renatus Cartesius and so it is some time called Cartesian geometry. It is now the main
branch of geometry in which two real numbers, called coordinates, are used to indicate the
position of a point in a plane. The main contribution of coordinate geometry is that it has
enabled the integration of algebra and geometry. For this algebraic methods are employed
to represent and prove the fundamental geometrical theorems. At present coordinate
geometry is considered as more powerful tool of analysis than the Euclidian geometry. For
this coordinate geometry is termed as analytical geometry. Now it is very useful to solve
complex business problems. In this chapter we shall discuss some of them.

12.2 Directed line: A directed line is a straight line with number units positive, zero and
negative is called directed line or number line. The point of origin is the number 0. The
arrow indicates its direction. On the right side of the arrow are the positive numbers and
on the other side are the negative numbers. It is like a real number scale illustrated below:

–7 –6–5 –4 –3 –2 –1 0 1 2 3 4 5 6 7
Negative Positive
Origin
Directed line
Figure – 12.1

206
 Coordinate Geometry

A directed line can be horizontal normally indicated by X/OX axis and vertically indicated
normally by YOY/ axis. The point where the two lines intersect each other is called the
point of origin and is denoted by (0,0).

12.3 Quadrants: The two directed lines, when they intersect at right angles at the point of
origin, divide their plane into four parts or regions. These four pasts are known as
quadrants.
Y_
7
_
6
_
Second Quadrant 5 First Quadrant
_
4
_
(–, + ) _
3 (+, + )
2
_
1
_
| | | | | | | | | | | | | | |
/ _
X –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 X
_
–2
_
(–, –) –3 (+,–)
_
–4
Third Quadrant _ Fourth Quadrant
–5
_
–6
_
–7

/
Figure –Y12.2

12.4 Coordinates: In a two dimensional figure a point in plane has two coordinates. The
exact position of the point can be located by the unit size of these coordinates. The first
coordinate is known as x-coordinate and second coordinate is known as y-coordinate.
y x-coordinate

 P(x, y)
y-coordinate
x / 0 x

y/
Figure – 12.3

Example: Plot the points (2, 3), (-5, 4), (-3, -2) and (4, -3) in the Cartesian plane.
Solution:

207
 S. M. Shahidul Islam

Y_
7
_
6
_
5
 (-5, 4) 4
_
_
3
_
 (2, 3)
2
_
1
_
| | | | | | | | | | | | | | |
_
X/ –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7
_ X
(-3, -2)  –2
_
–3
_
 (4, -3)
–4
_
–5
_
–6
_
–7

Figure – 12.4
Y/
12.5 Coordinates of mid point: We can find out the coordinates of a mid point from the
coordinates of any two points using the following formula :
Let P(x1, y1) and Q(x2, y2) be two given points and R(xm, ym) be mid point of them, then
x  x2 y  y2
xm = 1 and ym = 1
2 2
Example: The coordinates of the mid point of the points (-2, 5) and (6, 3) are
26 53
( , ) , i.e, (2, 4)
2 2

12.6 Distance between two points:

P(x1, y1)
Let P (x1, y1) and Q (x2, y2) be any Y
two points. Two perpendiculars PN Q(x2,y2)
and QM are drawn on the axis OX R
from the points P and Q respectively.
From Q an another perpendicular QR y2 y1
is drawn on PN. From the right angle
triangle PQR, we have
PQ2 = PR2 + QR2 X O x2 M x1 N X
2 2
= (PN – RN ) + ( ON – OM )
= (PN – QM)2 + (ON – OM)2 Y
Figure – 12.5

208
 Coordinate Geometry

= ( y1 – y2 )2 + ( x1 – x2 )2
= ( x1 – x2)2 + (y1 – y2 )2
PQ = ( x 1 - x 2 ) 2  (y1 - y 2 ) 2
PQ = ( Difference of abscissas) 2  ( Difference of ordinates ) 2

Example: Find the distance between the points A (8, – 3) and B (– 7, 5) .

Solution: Let the distance between the two points be d
Therefore , d = [(Difference of abscissas )2 + (Difference of ordinates)]
= [{ 8– (– 7)}2 + {– 3– (5)}2]
= [(8+7)2 + {–3–5}2]
= [152 + (–8)2]
=  (225+64)
= 289
= 17.
Hence, the required distance is 17 units. (Answer)

12.7 Section formula: The coordinates of a point R(x, y) dividing a line in the ratio of
m1 : m2 connecting the points P(x1, y1) and Q(x2, y2) are
m x  m2 x1 m y  m2 y1
x= 1 2 and y = 1 2
m1  m2 m1  m2
Proof: Three perpendiculars PL, RM and QN are drawn on the axis OX from the points
P, R and Q respectively. From P and R two other perpendiculars PK and RT are drawn on
RM and QN respectively. From the figure we see that  PKR and  RTQ are similar
triangles.
PK RK PR m1 Q (x2, y2)
So,    (x, y)
RT QT RQ m2 y R m2 T
(x1, y1)
PK m1 P m1
When,  K
RT m2
x  x1 m1
Or,  x/ 0 L M N x
x 2  x m2
Or, m2x – m2x1 = m1x2 – m1x
Or, x(m1 + m2) = m1x2 + m2x1 y/
m x  m2 x1 Figure – 12.6
So, x= 1 2
m1  m2
RK m1
Similarly, when 
QT m2

209
 S. M. Shahidul Islam

y  y1 m1
Or, 
y 2  y m2
Or, m2y – m2y1 = m1y2 – m1y
Or, y(m1 + m2) = m1y2 + m2y1
m y  m2 y1
So, y= 1 2
m1  m2
Note: If this division be external then
m x  m2 x1 m y  m2 y1
x= 1 2 and y = 1 2
m1  m2 m1  m2

Example: The coordinates of the point R which divides the connecting line of the points
P(3, 4) and Q(-3, -4) in the ratio of 2 : 3 are
 2.(3)  3.3 2.(4)  3.4  3 4
 ,  i.e,  , 
 23 23  5 5
Example: Find the coordinates of the point which externally divides the line joining the
points (4, -5) and (6, 8) in the ratio 2 : 1.
Solution: Let A(4, -5), B(6, 8) and the point P(x, y) divides AB externally in the ratio
2 : 1.
2  6  1 4 2  8  1  (5)
So, x = = 8 and y = = 21
2 1 2 1
Hence, the coordinates of the required point are (8, 21) (Answer)

12.8 Coordinates of the centroid of a triangle: The coordinates of the centroid of a

triangle whose vertices are A(x1, y1), B(x2, y2) and C(x3, y3) is
 x1  x2  x3 y1  y 2  y3  A(x1, y1)
 , 
 3 3 
Proof:
Let us consider the vertices of the triangle F G (x, y) E
A(x1,y1), B(x2,y2) and C (x3, y3) as shown
in diagram. In the diagram median AD
bisects the base BC at D with coordinates
B(x2, y2) D C(x3, y3)
 x2  x2 y 2  y 2 
D , . Figure – 12.7
 2 2 
We know that centroid is that point in a median which divides the median in the ratio 2 :1.
Let G(x, y) be the centroid of the triangle ABC. So, G divides the median AD in the ratio
2 :1. Hence by the section formula, the coordinates of G are

210
 Coordinate Geometry

x 2  x3
2.  1.x1
2 x  x 2  x3
x=  1
2 1 3
y  y3
2. 2  1. y1
2 y  y 2  y3
and y=  1
2 1 3
 x  x2  x3 y1  y 2  y3 
Therefore, the coordinates of the centroid are  1 , .
 3 3 
Example: The coordinates of the centroid of the triangle whose vertices are (3, 5), (5, 6)
35 4 5 65
and (4, -5) are  ,  i.e., (4, 2)
 3 3 

12.9 Area of a triangle: The area of a triangle whose vertices are A(x1, y1), B(x2, y2)
x1 y1 1
1
and C(x3, y3) is the absolute value of . x 2 y 2 1 .
2 A(x1,y1)
x3 y 3 1 y
Proof:
B(x2,y2) C(x3,y3)
Let, A(x1, y1), B(x2, y2) and C(x3, y3)
be the coordinates of the vertices of
the triangle ABC. We draw x/ 0M L N x
perpendiculars AL, BM and CN
from A, B and C on the x-axis. It is
clear from the figure, y/
Figure – 12.8

The area of  ABC = Area of trapezium ABML + Area of trapezium ALNC

– Area of trapezium BMNC.
1
Since, the area of the trapezium = (Sum of the parallel sides) 
2
(Perpendicular distance between the parallel sides)
So, the area of the triangle ABC can be given as
1 1 1
 ABC = (BM + AL)ML + (AL + CN)LN – (BM + CN)MN
2 2 2
1 1 1
= (y2 + y1)(x1 – x2) + (y1 + y3)(x3 – x1) – (y2 + y3)(x3 –x2)
2 2 2
1
= (x1y2 + x1y1 – x2y2 – x2y1 + x3y1 + x3y3 – x1y1 – x1y3 – x3y2 – x3y3 + x2y2 + x2y3)
2

211
 S. M. Shahidul Islam

1
= ( x1y2 – x2y1 + x2y3 – x3y2 + x3y1 – x1y3)
2
1
= [x1(y2 – y3) – x2(y1 – y3) + x3(y1 – y2)]
2
x1 y1 1
1
= . x 2 y2 1 (Proved)
2
x3 y 3 1

Example: Find the area of the triangle whose vertices are A(2, 3), B(5, 7) and C(-3, 4).
2 3 1
1
Solution: The area of the  ABC = 5 7 1
2
3 4 1
1
= [2(7 – 4) – 5(3 – 4) – 3(3 – 7)]
2
1
= [6 + 5 + 12]
2
1
= (23)
2
= 11.5 square units (Answer)

Note: Three points must be collinear if they form a triangle whose area is zero.

Example: Show that three points A(3, 5), B(-7, 5) and C(1, 5) are collinear.
3 5 1
1
Solution: Area of the  ABC = 7 5 1
2
1 5 1
1
= [3(5 – 5) + 7(5 – 5) + 1(5 – 5)]
2
1
= .0
2
=0
So, the points A, B and C are collinear. (Proved)
Another way: AB = (3  7) 2  (5  5) 2  100  0  100  10
BC = (7  1) 2  (5  5) 2  64  0  64  8
AC = (3  1) 2  (5  5) 2  4  0  4  2

212
 Coordinate Geometry

Here, BC + AC = 8 + 2 = 10 = AB
So, the points A, B and C are collinear. (Proved)

12.10 Area of a quadrilateral: The area of a quadrilateral which vertices are A(x1, y1),
B(x2, y2), C(x3, y3) and D(x4, y4) is
1
the absolute value of [x1y2 – x2y1 + x2y3 – x3y2 + x3y4 – x4y3 + x4y1 – x1y4]
2

Example: Find the area of a quadrilateral whose vertices are A(1, 1), B(3, 5), C(4, -1) and
D(2, 2).
1
Solution: Area of quad. = │ [1.5 – 3.1 + 3.(-1) – 4.5 + 4.2 – 2.4 + 2.1 – 1.2]│
2
1
= │ [5 – 3 – 3 – 20 + 8 – 8 + 2 – 2]│
2
1
= │ (-21) │
2
= │-10.5│
= 10.5 square units (Answer)
y B
12.11 Straight line: The minimum Straight line
distance between two distinct A
points is known as straight line. In
the figure AB is a straight line.
x/ 0 x

y/
Figure – 12.9
12.12 Slope or gradient of a straight line: The slope or gradient of a straight line is the
tangent of the angle formed by the line above the x – axis towards its positive direction
whatever be the position of the line. If  is the angle formed by the line and the positive
direction of the x – axis, the slope of the line is
y  B (x2, y2)
m = tan 
In terms of the coordinates, the slope of the line joining two points  A (x1, y1)
A(x1, y1) and B(x2, y2) is
y 2  y1 Difference of ordinates 
m= =
x 2  x1 Difference of abscissae 0 x
/
x
The slope of a line is generally denoted by m.
y/
Figure – 12.10

213
 S. M. Shahidul Islam

12.13 Different forms of equations of the straight line: y

x=0
1. Equations of the coordinate axes: The
y=b
value of y ordinates of all points on the x-
b x=a
axis is always 0 (zero). And the value of x
ordinates of all points on the y-axis is x/ 0 a y=0 x
always 0 (zero). Therefore, the equation of
the x-axis is y = 0 and
The equation of the y-axis is x = 0. y/
Figure – 12.11
2. Equations of lines parallel to the coordinate axes:
Let P(x, y) be any point on a line parallel to x-axis at a distance b from it. For any
position of the point P lying on this line y = b.
So, the equation of this line is y = b.
Similarly, the equation of the line parallel to the y-axis at a distance a from it is
x = a. y
3. Origin-slope form: The equation of a
y = mx + c y = mx
line passing through the origin and
c m = tan 
having slope m is : y = mx. 
x/ 0 x
4. Slope intercept form: The equation of
the line with the slope m and an intercept
on y-axis is : y = mx + c. This is the y/
general equation of straight line. Figure – 12.12

5. Two intercepts form: The equation of a y

line having intercepts a and b on the x y
 1
x y a b
coordinate axes is :  1 b
a b
0 a x
x/
6. Slope-point form: The equation of a
straight line having a slope m and passing
through the point (x1, y1) is y/
y
y – y1 = m(x – x1) Figure – 12.13
 (x1, y1)
y – y1 = m(x – x1)

x/ 0 x

y/
Figure – 12.14

214
 Coordinate Geometry

7. Two points form: The equation of a straight line passing through two points
(x1, y1) and (x2, y2) is y
y1  y 2  A (x1, y1)
y – y1 = (x – x1)
x1  x 2  B(x2, y2)

x/ 0 x

y/
Figure – 12.15

8. Parallel line form: The equation of the parallel line to the line ax + by + c = 0 is
ax + by + k = 0; k is a constant.
9. Perpendicular line form: The equation of the perpendicular line to the line
ax + by + c = 0 is bx – ay + k = 0; k is a constant.

Note: Let m1 and m2 are slopes of two straight lines respectively. These two lines
will be perpendicular to each other if m1  m2 = -1 and will be parallel to each other if
m1 = m2 .

12.14 Circle: The circle is the locus of a point which moves in such a way that its distance
from a fixed point always remains constant. The fixed point is known the centre of the
circle and the constant distance is called the radius of the circle.
The equation of the circle whose centre is (h, k) and the radius is a is
(x – h)2 + (y – k)2 = a2

Example: Find the equation of the circle whose centre is (2, 3) and the radius is 5.
Solution: The required equation is (x – 2)2 + (y – 3)2 = 52
=> x2 – 4x + 4 + y2 – 6y + 9 = 25
=> x2 + y2 – 4x – 6y – 12 = 0 (Answer)

12.15 Some worked out examples:

Example (1): Show that the points (6, 6), (2, 3) and (4, 7) are the vertices of a right angled
triangle. [AUB-01]
Solution: Let A, B and C be the points (6, 6), (2, 3) and (4, 7) respectively, then
AB2 = (6 – 2)2 + (6 – 3)2 = 16 + 9 = 25
BC2 = (2 – 4)2 + (3 – 7)2 = 4 + 16 = 20
CA2 = (4 – 6)2 + (7 – 6)2 = 4 + 1 = 5
BC2 + CA2 = 20 + 5 = 25 = AB2
 AB2 = BC2 + CA2.
So, ACB  1 right angle

215
 S. M. Shahidul Islam

Hence, the points A(6, 6), B(2, 3) and C(4, 7) are the vertices of a right angled triangle.

Example (2): Find the coordinates of the circum centre of a triangle whose coordinates
are (3, -2), (4, 3) and (-6, 5). Hence find the circum radius and circumference.
Solution: Let A(3, -2), B(4, 3) and C(-6, 5) be the vertices of the triangle and P(x, y) be
the circum centre. So, by definition,
PA = PB = PC
 PA2 = PB2 = PC2.
Now by the distance formula
PA2 = (x – 3)2 + (y + 2)2 = x2 – 6x + 9 + y2 + 4y + 4 = x2 + y2 – 6x + 4y + 13
PB2 = (x – 4)2 + (y – 3)2 = x2 – 8x + 16 + y2 – 6y + 9 = x2 + y2 – 8x – 6y + 25
PC2 = (x + 6)2 + (y – 5)2 = x2 + 12x + 36 + y2 – 10y + 25 = x2 + y2 + 12x – 10y + 61
Now PA2 = PB2.
 x2 + y2 – 6x + 4y + 13 = x2 + y2 – 8x – 6y + 25
 2x + 10y = 12
 x + 5y = 6 [Dividing by 2]
 x = 6 – 5y - - - (i)
And PB = PC2.
2

 x2 + y2 – 8x – 6y + 25 = x2 + y2 + 12x – 10y + 61
 -20x + 4y = 36
 -5x + y = 9 [Dividing by 4]
 -5(6 – 5y) + y = 9 [Using (i)]
 -30 + 25y + y = 9
 26y = 39
39
 y=
26
3
 y=
2
Putting the value of y in (i), we get
3 3
x = 6 – 5. = –
2 2
3 3
Therefore, the coordinates of the circumcentre P is (– , ) (Answer)
2 2
3 3
Now, the circum radius of  ABC, r = (  3) 2  (  2) 2 [ PA]
2 2
81 49
= 
4 4
130
=
4

216
 Coordinate Geometry

130
= (Answer)
2
And the circumference = 2  r
130
= 2
2
=  130

Example (3): Find the area of the triangle whose vertices are A(3, 1), B(2k, 3k), C(k, 2k)
and prove that these three points will be collinear if k = – 2. [AUB-02]
3 1 1
1
Solution: Area of the  ABC = 2k 3k 1
2
k 2k 1
1
= [3(3k – 2k) – 2k(1 – 2k) + k(1 – 3k)]
2
1
= (3k – 2k + 4k2 + k – 3k2)
2
1 2
= (k +2k)
2
1
= k(k + 2) square units
2
Now , the points A, B, C will be collinear if the area of the triangle is zero,
1
i.e., k(k + 2) = 0
2
Or, k(k + 2) = 0
So, k = 0 or, k + 2 = 3 => k = – 2.
But k = 0 makes B and C a point.
So, k=–2 (Proved)

Example (4): The coordinates of the three towns A, B and C are (4, -2), (2, -2) and (6, -
2). Find the distances of the towns from one to another. And prove that these three towns
are situated on a straight line. [RU-80, 82 A/C]
Solution: The distance of town A and B, AB = (4  2) 2  (2  2) 2  4  2 units
The distance of town B and C, BC = (2  6) 2  (2  2) 2  16  4 units
The distance of town A and C, AC = (6  4) 2  (2  2) 2  4  2 units
Here, AB + AC = 2 + 2 = 4 = BC
So, these three town are situated on a straight line. (Proved)

217
 S. M. Shahidul Islam

Example (5): Find the equation and the slope of the straight line joining the points (3, 5)
and (2, 3). [RU-80, 82 A/C]
Solution: The equation of the line which passes through the points (3, 5) and (2, 3) is
53
y–5= (x – 3)
32
Or, y – 5 = 2(x – 3)
Or, y = 2x – 1 (Answer)
We know that, m is the slope of the line y = mx + c.
So, the slope of the line y = 2x – 1 is 2. (Answer)

Example (6): Prove that the first degree equation ax + by + c = 0 always represents a
equation of a straight line, i.e., the general equation of straight line.
Proof: Let A(x1, y1), B(x2, y2) and C(x3, y3) be three points on the locus represented by
the equation ax + by + c = 0.
 ax1 + by1 + c = 0 - - - (i)
ax2 + by2 + c = 0 - - - (ii)
ax3 + by3 + c = 0 - - - (iii)
Doing (i) – (ii), we have a(x1 – x2) + b(y1 – y2) = 0
y1  y 2 a
Or, 
x1  x 2 b
Again doing (ii) – (iii), we have a(x2 – x3) + b(y2 – y3) = 0
y 2  y3 a
Or, 
x 2  x3 b
y1  y 2 y  y3
 = 2
x1  x 2 x 2  x3
Or, (y1 – y2)(x2 – x3) = (y2 – y3)(x1 – x2)
Or, x2y1 – x2y2 – x3y1 + x3y2 – x1y2 + x1y3 + x2y2 – x2y3 = 0
1
Or, [x1(y2 – y3) – x2(y1 – y3) + x3(y1 – y2)] = 0
2
x1 y1 1
1
Or, . x 2 y2 1 = 0
2
x3 y 3 1
That is, the area of the triangle formed by A, B and C is zero. Hence, the points A, B and
C are collinear. But A, B and C are any three points on the locus of ax + by + c = 0.
Therefore, the equation ax + by + c = 0 always represents a straight line. (Proved)

Example (7): Find the equation of the straight line passing through the point (4, 5) and
the sum of its intercepts on the axes is 18.

218
 Coordinate Geometry

Solution: Let us consider the equation of the required line be

x y
  1 - - - (i)
a b
This line passes through the point (4, 5), therefore, we have
4 5
  1 - - - (ii)
a b
And a + b = 18 Or, a = 18 – b - - - (iii)
Using (iii) in (ii), we get
4 5
 1
18  b b
4b  5(18  b)
Or, 1
b(18  b)
Or, 4b + 90 – 5b = 18b – b2
Or, b2 – 19b + 90 = 0
Or, b2 – 10b – 9b + 90 = 0
Or, b(b – 10) – 9(b – 10) = 0
Or, (b – 10)(b – 9) = 0
 b – 10 = 0 Or, b = 10
Or, b–9=0 Or, b=9
When b = 10, a = 18 – 10 = 8
And when b = 9, a = 18 – 9 = 9
So, the required equation are
x y x y
  1 and,  1 (Answer)
8 10 9 9

Example (8): Find the equation of the straight line passing through the point (-3, 1) and
perpendicular to the line 5x – 2y + 7 = 0. [AUB-02]
Solution: The equation of the line perpendicular to the line 5x – 2y + 7 = 0 is
2x + 5y + k = 0 - - - (i)
Since, line (i) passes through the point (-3, 1)
2.(-3) + 5.1 + k = 0
Or, -6 + 5 + k = 0
 k=1
Therefore, the equation of the required line is 2x + 5y + 1 = 0 (Answer)

Example (9): Find the equation of the line passing through (2, 5) and (5, 6). And prove
that this line is perpendicular to the line passing through (-4, 5) and (-3, 2).
Solution: The equation of the line passing through the points (2, 5) and (5, 6) is
y 5 x2

56 25

219
 S. M. Shahidul Islam

y 5 x2
Or, 
1 3
Or, - 3y + 15 = - x + 2
Or, x – 3y + 13 = 0; which is the required equation. (Answer)
1 1
Let the slope of this line, m1 = 
3 3
And, let the slope of the line joining the points (-4, 5) and (-3, 2),
52 3
m2 =   3
 4  (3)  1
1
 m1  m2 =  (3)  1
3
Hence, the two lines are perpendicular to each other. (Proved)

Example (10): Find the equation of the straight line passing through the point of
intersection of the two lines 2x + 3y – 1 = 0 and x – 2y + 3 = 0 and intersects equal
portion from both axes.
Solution: Given that 2x + 3y – 1 = 0 - - - (i)
x – 2y + 3 = 0 - - - (ii)
Doing (i) – 2  (ii), we have 7y – 7 = 0
 y=1
Putting the value of y in (ii), we have
x – 2.1 + 3 = 0
 x = -1
So, the point of intersection of the given two straight line is (-1, 1)
Let the equation of the line which intersects equal parts from both axes be
x y
 1
a a
x y x y
That is,   1 - - - (iii) Or,   1 - - - (iv)
a a a a
Since both the lines pass through (-1, 1), hence putting (-1, 1) in equation (iii), we have
1 1
  1 Or, a = 0
a a
x y
So, the line is   1 , which is meaning less.
0 0
Now, putting (-1, 1) in equation (iv), we have
1 1
 1 Or, a = - 2
a a
x y
So, the equation (iv) becomes  1
2 2

220
 Coordinate Geometry

Or, x – y = -2
i.e., x – y + 2 = 0, which is the required equation. (Answer)

Example (11): Find the coordinates of the points at which straight line 3x – 4y + 12 = 0
intersects the coordinates axes. What is the slope of the line? And find the length of the
intersecting part. [NU-01 A/C]
Solution: The equation of the line is given by 3x – 4y + 12 = 0 - - - (i)
When y = 0 then by (i), we have x = 4
and when x = 0 then by (i) we have y = 3
So, the coordinates of the intersecting points are (4, 0) and (0, 3) (Answer)
We can write equation (i) as follows:
3
y = x + 3 - - - (ii)
4
Comparing equation (ii) with the equation y = mx + c, we get the slope of the given line
3
as follows: m = (Answer)
4
And the distance between the points (4, 0) and (0, 3) is
(4  0) 2  (0  3) 2  25  5
So, the length of the intersecting part = 5 units. (Answer)

Example (12): Find the equation of the circle whose centre is (4, 6) and it passes through
the point (1, 1). [RU-89 A/C]
Solution: We know that the equation of a circle is
(x – h)2 + (y – k)2 = a2 - - - (i)
The centre of the circle is (4, 6), so, h = 4 and k = 6. Putting the value of h and k in the
equation, we get
(x – 4)2 + (y – 6)2 = a2 - - - (ii)
Since, the circle passes through the point (1, 1),
(1 – 4)2 + (1 – 6)2 = a2
 9 + 25 = a2
 a2 = 34
So, the equation of the circle is
(x – 4)2 + (y – 6)2 = 34
 x2 – 8x + 16 + y2 – 12y + 36 = 34
 x2 + y2 – 8x – 12y + 18 = 0 (Answer)

Example (13): A firm invested Tk. 20000 in a new factory that has a net return of Tk.
2000 per year. An investment of Tk. 40000 would yield a net income of Tk. 8000 per year.
What is the linear relationship between investment and annual income? What would be the
return on an investment of Tk. 30000? [AUB-02 MBA]

221
 S. M. Shahidul Islam

Solution: Let x and y coordinates represent the investment and the annual income
respectively. Then the required linear relationship between investment and annual income
is the equation of the straight line joining the points (20000, 2000) and (40000, 8000) and
the equation is
2000  8000
y – 2000 = (x – 20000)
20000  40000
Or, y – 2000 = 0.3(x – 20000)
Or, y – 2000 = 0.3x – 6000
i.e., 0.3x – y – 4000 = 0, this is the required relationship. (Answer)
Again when the investment x = 30000, the annual income y can be found by putting the
value of x in the above equation, i.e.,
0.3(30000) – y – 4000 = 0
Or, 9000 – y – 4000 = 0
Or, y = 5000
So, the required income = Tk. 5000 (Answer)

Example (14): The total expense of a firm y, are partly constant and partly proportional to
the number of the products x. The total expenses are Tk. 5000 when 20 products are made
and Tk. 7500 when 40 products are made. [AUB-03 MBA]
(i) Find the linear relationship between x and y.
(ii) Find the variable cost for a product and the fixed cost.
(iii) What would be the total expenditure if 30 products are made?
Solution: (i) Then the required linear relationship between products and expenses is the
equation of the straight line joining the points (20, 5000) and (40, 7500) and the equation
5000  7500
is y – 5000 = (x – 20)
20  40
Or, y – 5000 = 125(x – 20)
Or, y – 5000 = 125x – 2500
i.e., y = 125x + 2500, this is the required relationship. (Answer)
(ii) If x represents the number of products and y represents the cost to produce x units
product then the equation y = mx + c means that, m is the variable cost per unit product
and c is the fixed cost.
So, the equation y = 125x + 2500 shows that the variable cost = Tk. 125 and the fixed
cost = Tk. 2500. (Answer)
(iii) When x = 30 units products are made, the total expenditure,
y = Tk. [125(30) + 2500] = Tk. 6250 (Answer)

222
 Coordinate Geometry

12.16 Exercise:
1. Plot the points with the following coordinates: A(-5, -5), B(3, 2), C(-3, 4) and
D(5, 0) in the Cartesian plane.
2. Find the coordinates of the mid-point of the join of points (-1, 5) and (7, 3)
[Answer: (3, 4)]
3. Find the distance between two points (4, - 1) and (7, 3). [Answer: 5]
4. Find the distance between two points (1+ 2 ,2) and (1,1– 2 ). [Answer: 5+2 2 ]
5. Find the coordinates of the point which internally divides the line joining the
16 11
points (4, -5) and (6, 8) in the ratio 2 : 1. [Answer: ( , )]
3 3
6. Find the coordinates of the points which divides the connecting line of points (8, 9)
and (-7, 4) internally in the ratio 2 : 3 and externally in the ratio 4 : 3. [Answer:
(2, 7) and (-52, -11)]
7. Find the coordinates of the centroid of the triangle whose vertices are (3, 2),
(-1, -4) and (-5, 6). [Answer: (-1, 4/3)]
8. Show that the points (1, 4), (3, -2) and (-3, 16) are collinear.
9. Show that the points A(-6, 6), B(2, 6) and C(2, 10) are the vertices of a right
angled triangle.
10. If A( -3, 5), B(6, 5) and C(a, -7) are the vertices of the triangle ABC whose
ABC  1 right angle, find the values of a. [Answer: 6]
11. Prove that the points (2a, 4a), (2a, 6a) and (2a + 3 a, 5a) are vertices of an
equilateral triangle.
12. Show that the triangle whose vertices are (1, 10), (2, 1) and (-7, 0) is an isosceles
triangle.
13. Show that the points (2,-2), (8, 4), (5, 7) and (-1, 1) are the vertices of a
parallelogram. [NU-97]
14. If a warehouse P(x, y) is equidistant from the three markets A(4, 2), B(5, 3) and
1 13
C(6, 5), find the coordinates of the warehouse. [Answer: ( , )]
2 2
15. If the points A(x, y), B(3, 4) and C(-2, 3) form an equilateral triangle, find the
1 3 7  5 3 1 3 7  5 3
coordinates of the point A. [Answer: ( , ) or ( , )]
2 2 2 2
16. Find the area of the triangle whose vertices are A(2, -1), B(-3, -4) and C(0, 2).
[Answer: 10.5 square units]
1 1
17. Prove that the points (a, 0), (0, b) and (1, 1) will be collinear if   1.
a b
18. Find the area of the quadrilateral whose vertices are (-2, -1), (1, 3), (5, 6) and
(2, 2). [Answer: 7 square units]
19. Find the equation of the straight line parallel to the x-axis and passing through
(4, 5) [Answer: y = 5]

223
 S. M. Shahidul Islam

20. Find the equation of the straight line parallel to the y-axis and passing through
(3, 7) [Answer: x = 3]
21. Find the equation of the straight line which passes through the two points (4, 5)
and (3, 7) [Answer: 2x + y – 13 = 0]
22. Find the length and the equation of the line which passes through the two points
A(9, -2) and B(3, -3). And also find the coordinates of the points at which the line
intersects the coordinate axes. [Answer: Length = 37 , eqn: x – 6y – 21 = 0, the
line intersects the x-axis at (21, 0) and intersects the y-axis at (0, -7/2)]
23. In a triangle with vertices P(0, 6), Q(-2, -2) and R(4, 2), find the equation of the
perpendicular bisector of the side QR. [Answer: 3x + 2y –3 = 0]
24. Find the equation of the straight line which is parallel to 2x – 3y – 5 = 0 and
passing through (4, 5). [Answer: 2x – 3y + 7 = 0]
25. Find the equation of the straight line which is perpendicular to 2x + 5y – 5 = 0 and
passing through (3, -2). [Answer: 5x – 2y – 19 = 0]
26. Find the equation of the circle whose centre is (0, 0) and it passes through the
point (4, 3). [Answer: x2 + y2 = 25]
27. Find the equation of the circle whose centre is (0, 4) and it passes through the
point (-3, 1). [Answer: x2 + y2 – 8y – 2 = 0]
28. Find the equation of the circle whose centre is (4, 5) and which passes through the
centre of the circle x2 + y2 + 4x + 6y – 12 = 0.
[Answer: x2 + y2 –8x – 10y – 59 = 0]
29. As the number of units manufactured increases from 4000 to 6000, the total cost of
production increases from Tk. 22000 to Tk. 30000. Find the linear relationship
between the cost (y) and the number of units made (x).
[Answer: 4x – y + 6000 = 0] [RU-83, 87 A/C]
30. A firm invested Tk. 10 millions in a new factory that has a net return of Tk. 0.5
million per year. An investment of Tk. 20 millions would yield a net income of Tk.
2 millions per year. What is the linear relationship between investment and annual
income? What would be the return on an investment of Tk. 50 millions?
[Answer: 3x – 20y – 20 = 0, Tk. 6.5 millions] [NU-02 A/C]
31. An investment of Tk. 90000 in a certain business yields an income of Tk. 8000. An
investment of Tk. 50000 yields an income of Tk. 5000. What is the linear
relationship between investment and income? [Answer: 3x – 40y + 50000]
32. The total expense of a firm y, are partly constant and partly proportional to the
number of the products x. The total expenses are Tk. 1040 when 12 products are
made and Tk. 1600 when 20 products are made.
a. Find the linear relationship between x and y.
b. Find the variable cost for a product and the fixed cost.
c. What would be the total expenditure if 15 products are made?
[Answer: (a) 70x – y + 200 = 0, (b) Tk. 70 & Tk. 200 and (c) Tk. 1250]

224
 Linear programming

13
C h a p te r
Linear Programming
Highlights:
13.1 Introduction 13.9 Simplex
13.2 What is optimization 13.10 Development of a minimum
13.3 Summation symbol feasible solution
13.4 Linear programming 13.11 The artificial basis technique
13.5 Formulation 13.12 Duality in linear programming
13.6 Some important definitions problem
13.7 Standard form of LP problem 13.13 Some worked out examples
13.8 Graphical solution 13.14 Exercise

13.1 Introduction: Linear programming is a technique for determining an optimum

schedule (such as maximizing profit or minimizing cost) of interdependent activities in
view of the available resources. Programming is just another word for “Planning” and
refers to the process of determining a particular plan of actions from amongst several
alternatives. A linear programming problem with only two variables presents a simple
case, for which the solution can be derived using a graphical method. We can solve any
linear programming problem by simplex method. A linear programming problem may
have a solution (a definite and unique solution or an infinite number of optimal solutions
or an unbounded solution) or not. In this chapter we shall discuss the formulation and the
solution techniques of general types of linear programming problems.

13.2 What is optimization: The fundamental problem of optimization is to arrive at the

‘best’ possible decision under some given circumstances. Some typical optimization
problems, taken from a variety of practical field of interest. In each of these cases, we
have to optimize (maximize or minimize) certain functions under certain constraints.

13.3 Summation symbol: The Greek capital letter ∑ (sigma) is the mathematical symbol
for summation. If f(i) denotes some quantity whose value depends on the value of i, the
expression
4
i
i 1

225
 S. M. Shahidul Islam

is read as ‘sigma i, i going from 1 to 4’ and means to insert 1 for i, then 2 for i, then 3 for
i, 4 for i, and sum the results. Therefore,
4
 i = 1 + 2 + 3 + 4 = 10.
i 1
In the above example, the i under the sigma symbol is called the index of the summation.
In this way,
3

 (i
i 1
2
 1) = (12 – 1) + (22 – 1) + (32 – 1) = 0 + 3 + 8 = 11
4
 p = p + p + p + p = 4p
i 1
n

 p = np
i 1
4
x
i 1
i = x1 + x2 + x3 + x4.

13.4 Linear programming: The general linear programming problem is to find a vector
(x1, x2, . . ., xj, . . ., xn) which minimizes or maximizes the linear form
c1x1 + c2x2 + . . . + cjxj + . . . + cnxn
subject to the linear constraints
a11x1 + a12x2 + . . . + a1jxj + . . . + a1nxn = b1
a21x1 + a22x2 + . . . + a2jxj + . . . + a2nxn = b2
--- --- ---
ai1x1 + ai2x2 + . . . + aijxj + . . . + ainxn = bi
--- --- ---
am1x1 + am2x2 + . . . + amjxj + . . . + amnxn = bm
and xj ≥ 0; j = 1, 2, 3, . . ., n.
where the aij, bi and cj are given constants and m < n. We shall always assume that the
constraints equation have been multiplied by (–1) where necessary to make all bi ≥ 0,
because of the variety of notations in common use, one will find the general linear
programming problem stated in many forms. The more common forms are the following:
n
(a) Minimize c
j 1
j xj
n
subject to a
j 1
ij x j  bi ; i = 1, 2, 3, . . ., m

and xj ≥ 0 ; j = 1, 2, 3, . . ., n
(b) Minimize f(X) = CX
subject to AX = b
and X ≥0

226
 Linear programming

Where A = (aij), C = (c1, c2, . . ., cn) is a row vector, X = (x1, x2, . . ., xn) is a column
vector, b = (b1, b2, . . ., bm) is a column vector and 0 is an n-dimensional null column
vector.
(c) Minimize CX
subject to x1P1 + x2P2 + . . . + xnPn = Po
and X≥0
Where Pj for j = 1, 2, . . ., n is the jth column of the matrix A and Po = b.

13.5 Formulation: As in any other planning problem, the operations researcher must
analyze the goals and the system in which the solution must operate. The complex of inter
related components in a problem area, referred to by operations researchers as a ‘system’
is the environment of a decision, and it represents planning premises. To solve any
business problem or production problem, we have to transfer it as mathematical problem.
This problem transformation is known as formulation.

Example: (Production planning problem)

A firm manufactures 3 products A, B and C. The profit per unit sold of each product is
Tk. 3, Tk. 2 and Tk. 4 respectively. The time required to manufacture one unit of each of
the three products and the daily capacity of the two machines P and Q is given in the table
below:
Machine Time per unit (minutes) product Machine capacity
(minutes/day)
A B C
P 4 3 5 2000
Q 2 2 4 2500

It is required to determine the daily number of units to be manufactured for each product,
so as to maximize the profit. However, the firm must manufacture at least 100 A’s, 200
B’s and 50 C’s but no more that 150 A’s. It is assumed that all the amounts produced are
consumed in the market. Formulate this problem as linear programming problem.
Solution: Step–1: We study the situation to find the key decision to be made and in this
connection looking for variables helps considerably.
Step-2: Select symbols for variable quantities identified in Step-1. Let the number of units
of the products A, B and C manufactured daily be designated x1, x2 and x3 respectively.
Step-3: Express feasible alternatives mathematically in terms of the variables. These
feasible alternatives are those which are physically, economically and financially possible.
Since it is not possible to manufacture any negative quantities, it is quite obvious that in
the present situation feasible alternatives are sets of variables of x 1, x2 and x3 satisfying
x1 ≥ 0, x2 ≥ 0 and x3 ≥ 0.
Step-4: Identify the objective quantitatively and express it as a linear function of variables.
The objective here is to maximize the profit. In view of the assumption that all the units
produced are consumed in that market, it is given by the linear function
z = 3x1 + 2x2 + 4x3

227
 S. M. Shahidul Islam

Step-5: Express in words the influencing factors or constraints (or restrictions) which
occur generally because of the constraints on availability (resources) or requirements
(demands). Express these restrictions also as linear equalities/inequalities in terms of
variables. Here in order to produce x1 units of product A, x2 units of product B and x3
units of product C, the total times needed on machines P and Q are given by
4x1 + 3x2 + 5x3 and 2x1 + 2x2 + 4x3 respectively.
Since the manufacturer does not have more than 2000 minutes available on machine P and
2500 minutes available on machine Q, we must have
4x1 + 3x2 + 5x3 ≤ 2000 and 2x1 + 2x2 + 4x3 ≤ 2500 .
Also, additional restrictions are 100 ≤ x1 ≤ 150, x2 ≥ 200, x3 ≥ 50.
Hence, the manufacturer’s problem can be put in the following mathematical form:
Maximize z = 3x1 + 2x2 + 4x3
Subject to 4x1 + 3x2 + 5x3 ≤ 2000
2x1 + 2x2 + 4x3 ≤ 2500
100 ≤ x1 ≤ 150, x2 ≥ 200, x3 ≥ 50.

Example: (Blending problem)

A firm produces an alloy having the following specifications
(i) specific gravity ≤ 0.98
(ii) chromium ≥ 8%
(iii) melting point ≥ 4500 C
Raw materials A, B and C having the properties shown in the table can be used to make
the alloy:

Property Raw material

A B C
Specific gravity 0.92 0.97 1.04
Chromium 7% 13% 16%
Melting point 4400 C 4900 C 4800 C

Cost of the various raw materials per unit ton are: Tk. 90 for A, Tk. 280 for B and Tk. 40
for C. Find the proportion in which A, B and C be used to obtain an alloy of desired
properties while the cost of raw materials is minimum.
Solution: Step-1: Key decision to be made is how much (percentage) or raw materials A,
B and C be used for making the alloy.
: Key decision to be made is how much (percentage) or raw materials A, B and C be used
for making the alloy.
Step-2: Let the percentage contents of A, B and C be x1, x2 and x3 respectively.
Step-3: Feasible alternatives are sets of values of x1, x2 and x3.
Step-4: Objective is to minimize the cost, i.e.,
minimize z = 90x1 + 280x2 + 40x3
Step-5: Constraints are imposed by the specifications required for the alloy. They are
0.92x1 + 0.97x2 + 1.04x3 ≤ 0.98

228
 Linear programming

7x1 + 13x2 + 16x3 ≥ 8

440x1 + 490x2 + 480x3 ≥ 450
and x1 + x2 + x3 = 100
Hence, the blending problem can be put in the following mathematical form:
Minimize z = 90x1 + 280x2 + 40x3
Subject to 0.92x1 + 0.97x2 + 1.04x3 ≤ 0.98
7x1 + 13x2 + 16x3 ≥ 8
440x1 + 490x2 + 480x3 ≥ 450
x1 + x2 + x3 = 100
x1, x2, x3 ≥ 0

13.6 Some important definitions:

1. Convex set: A subset S  Rn is said to be convex set if for each pair of point x, y
in S, the line segment [x : y] = {ax + (1 – a)y, 0 ≤ a ≤1} joining the points x and y
is contained in S.

x
x y y
y x
Convex set Convex set Non convex set

Figure – 13.1
2. Feasible solution: A feasible solution to a linear programming problem (LPP) is a
solution which satisfies the constraints (equality or inequality constraints and the
non-negativity constraints)
3. Basic solution: A basic solution to an LPP with m constraints in n variables is a
solution obtained by setting n – m variables equal to zero and solving for the
remaining m variables, provided that the determinant of the coefficients of these m
variables is non zero. The m variables are called basic variables. Basic solution
may or may not be feasible solution and conversely feasible solution may or may
not be a basic solution.
4. Basic feasible solution: A basic feasible solution is a basic solution which also
satisfies all basic variables are non-negative.
5. Non degenerate basic feasible solution: A non-degenerate basic feasible solution
is a basic feasible solution with exactly m positive xi; that is, all basic variables are
positive.
6. Optimum solution: A optimum solution to an LPP is a feasible solution which
optimizes (minimizes or maximizes) the value of the objective function.
7. Slack variables: Let the constraints of a general LPP be
n

a
j 1
ij x j  bi ; i = 1, 2, . . ., m

229
 S. M. Shahidul Islam

Then the non-negative variables xn+i which satisfy

n

a
j 1
ij x j  xn i  bi ; i = 1, 2, . . ., m

are called slack variables.

Note: The variables xn+i are called slack variables, because
Slack = Requirement – Production.
8. Surplus variables: Let the constraints of a general LPP be
n

a
j 1
ij x j  bi ; i = 1, 2, . . ., m

Then the non-negative variables xn+i which satisfy

n

a
j 1
ij x j  xn i  bi ; i = 1, 2, . . ., m

are called surplus variables.

Note: The variables xn+i are called surplus variables, because
Surplus = Production – Requirement.

13.7 Standard form of LP problem: The characteristics of the standard form of linear
programming problem are
1. All the constraints are expressed in the form of equations, except the non-
negativity constraints which remain inequalities (≥ 0)
2. The right hand side of each constraints equation is non-negative.
3. All the decision variables are non-negative.
4. The objective function is of the maximization or minimization type.
Example: Express the following LPP into standard form:
Maximize z = 3x1 + 2x2
Subject to 2x1 + x2 ≤ 2
3x1 + 4x2 ≥ 12
x1 , x 2 ≥ 0
Solution: Introducing slack variable s1 and surplus variable s2, the problem in the standard
form can be expressed as
Maximize z = 3x1 + 2x2
Subject to 2x1 + x2 + s1 = 2
3x1 + 4x2 – s2 = 12
x1, x2, s1, s2 ≥ 0

13.8 Graphical solution: The solution of any linear programming problem with only two
variables can be derived using a graphical method. This method consists of the following
steps:
(1)Represent the given problem in mathematical from, i.e., formulate an LP model for the
given problem.

230
 Linear programming

(2) Represent the given constraints as equalities on x 1, x2 co-ordinates plane and find
the convex region formed by them.
(3) Plot the objective function.
(4) Find the vertices of the convex region and also the value of the objective function at
each vertex . The vertex that gives the optimum value of the objective function gives the
optimal solution to the problem.
Note: In general, a linear programming problem may have
(i ) a definite and unique optional solution,
(ii) an infinite number of optimal solutions,
(iii) an unbounded solution, and
(iv) no solution.

Example: (product Allocation Problem): A factory uses three different resources for the
manufacture of two different products 20 unites of the resource A, 12 units of B and 16
units of C being available. 1 unit of the first product requires 2, 2 and 4 units of the
respective resources and 1 unit of the second product requires 4, 2 and 0 units of the
respective resources. It is known that the first product gives a profit of two monetary units
per unit and the second 3. Formulate the linear programming problem. How many units of
each product should be manufactured for maximizing the profit? Solve it graphically.
Solution: Mathematical formulation of the problem:
Step-1: The key decision is to determine the number of units of the two products.
Step-2: Let x1 units of the first product and x2 units of the second product be
manufactured for maximizing the profit.
Step-3: Feasible alternatives are the sets of the values of x1 and x2 satisfying x1  0
and x2  0, as negative number of production runs are meaningless ( and thus not feasible).
Step-4: The objective is to maximize the profit realized from both the products, i.e., to
maximize z = 2x1 + 3x2
Step-5: Since 1 unit of the first product requires 2, 2 and 4 units, 1 unit of the second
product requires 4, 2 and 0 units of the respective resources and the units available of the
three resources are 20, 12 and 16 respectively, the constraints (or restrictions) are
2x1 + 4x2 < 20 Or, x1 + 2x2 ≤ 10
2x1 + 2x2 < 12 Or, x1 + x2 ≤ 6
4x1 + 0x2 < 16 Or, 4x1 < 16
Hence the manufacturer’s problem can be put in the following mathematical form:
Maximize z = 2x1 + 3x2
Subject to the constraints:
x1 + 2x2 ≤ 10
x1 + x2 ≤ 6
4x1 < 16
x1, x2 ≥ 0

231
 S. M. Shahidul Islam

Graphical solution of the problem:

Step-1: Construct the graph. Consider a set of rectangular Cartesian axes OX1X2 in the
plane. Each point has coordinate of the type (x1, x2) and conversely every ordered pair
(x1, x2) of real numbers determines a point in the plane.
It is clear that any point which satisfies the conditions x1 ≥ 0 and x2 ≥ 0 lies in the first
quadrant and conversely for any point (x1, x2) in the first quadrant, x1 ≥ 0 and x2 ≥ 0. Thus
our search for the number pair (x1, x2) is restricted to the points of the first quadrant only.
Step-2: To graph each constraint in the first quadrant satisfying the constraints, we first
treat each in equation as though it were an equation and then find the set of points in the
first quadrant satisfying the constraints.
Considering, now, the first constraint as an equation, we have x1 + 2x2 = 10. Clearly, this
is the equation for a straight line and any point on the straight line satisfies the inequality
also. Taking into consideration the point (0, 0), we observe that 0 + 2 (0) = 0 < 10, i.e.,
origin also satisfies the inequality. This indicates that any point satisfying the inequality
x1 + 2x2 ≤ 10 lies in the first quadrant on that side of the line, x1 + 2x2 = 10 which contains
the origin.
In a similar way, we see that all points satisfying the constraint x 1 + x2 ≤ 6 are the points in
the first quadrant lying on or below the line x1 + x2 = 6.
The set of points satisfying the inequality x1 ≤ 4 lies on or towards the left of the line
x1 = 4.
Step-3: All points in the area shown shaded in figure satisfy the three constraints
x1 + 2x2 ≤ 10 and x1 + x2 ≤ 6 and x1 ≤ 4 and also the non-negative restrictions x1 ≥ 0,
x2 ≥ 0 X2
_
10
_
9
_
8
_
7
_
6
_
5 C
_D
4
_
3
_
2
_ B
1
| | | A| | | | | | | | X1
0 1 2 3 4 5 6 7 8 9 10 11
Figure – 13.2

This area is called the convex region or the solution space or the region of feasible
solutions. Any point in this region is a feasible solution the given problem. The convex
region OABCD is bounded by the lines x1 = 0, x2 = 0, x1 + 2x2 = 10, x1 + x2 = 6 and
x1 = 4. The five vertices of the convex region are O = (0, 0), A = (4, 0), B = (4, 2),
C = (2, 4) and D = (0, 5)
Putting O = (0, 0), A = (4, 0), B = (4, 2), C = (2, 4) and D = (0, 5) in the objective
function 2x1 + 3x2, we find 0, 8, 14, 16 and 15 respectively. Here, 16 is the maximum

232
 Linear programming

value. Hence, the solution of the problem is x1 = 2, x2 = 4 and the maximum value of the
objective function is z = 16.

Example: A furniture company makes tables and chairs. Each table takes 4 hours of
carpentry and 2 hours in painting and varnishing shop. Each chair requires 3 hours in
carpentry and 1 hours in painting and varnishing. During the current production period
240 hours of carpentry and 100 hours. of painting and varnishing time are available. Each
table sold yields a profit of Tk. 420 and each chair yields a profit of Tk. 300. Determine
the number of tables and chairs to be made to maximize the profit. [AUB-03]
Solution: Let x1 = the number of tables and x2 = the number of chairs.
 the profit from the tables 420x1 and the profit from the chairs 300x2
So, the total profit 420x1 + 300x2 which is the objective function. We have to maximize
the objective function z = 420x1 + 300x2
Required carpentry hours for tables 4x1 and required carpentry hours for tables 3x2
 the required carpentry hours 4x1 + 3x2
Since 240 carpentry hours are available
4x1 + 3x2 ≤ 240.
Similarly, for painting and varnishing, we have
2x1 + x2 ≤ 100.
The non negative conditions x1, x2 ≥ 0
So, the linear programming problem (LPP) of the given problem is

Maximize z = 420x1 + 300x2

Subject to the constraints:
4x1 + 3x2 ≤ 240
2x1 + x2 ≤ 100
x1, x2 ≥ 0

Graph of the problem:

X1
_
_
_ 2x1 + x2 ≤ 100
A(0, 80)_
_
_
_ B(30, 40)
_
_
_ 4x1 + 3x2 ≤ 240
| | | | | | | | | | | X1
0 C(50, 0)

Figure – 13.3

233
 S. M. Shahidul Islam

The solution space satisfying the given constraints is shown shaded on the graph paper.
Any point in the shaded region is a feasible solution to the given problem. The coordinates
of the four vertices of the bounded convex region are
A (0,80), B(30,40),C(50,0) and 0(0,0)
The values of the objective function z = 420x1 + 300x2 at four vertices are 24000 at A,
24600 at B, 21000 at C and 0 at O.
Since the maximum value of the objective function is 24600 which occurs at the vertices
B (30,40). Hence, the solution to the given problem x1 = 30, x2 = 40 and the maximum
value = 24600.
Therefore, to maximize the profit, the furniture company should make 30 tables and 40
chairs.

Example: Maximize 3x1 + 2x2

Subjcet to 2x1 – x2 ≥ -2
x1 + 2x2 ≤ 8
x1, x2 ≥ 0
Solution: The graph of the problem is as follows:

y
From the graph we get the vertices A(0,2),
B(4/5,18/5), C(8,0) and O(0,0). The values of
the objective function at these points are
B(4/5,18/5) 4 at A, 48/5 at B, 24 at C and 0 at O. Here, 24 is
A(0,2) the maximum value of the objective function
C(8,0) which occurs at the vertices C(8,0). Hence, the
x/ x solution of the given problem is x1 = 8, x2 = 0
and max. value of z = 24 (Answer)

y/

Figure – 13.4

13.9 Simplex: A simplex is an n dimensional convex polyhedron with exactly (n +1)

extreme points,
(i) If n = 0, then the convex polyhedron is a point.
(ii) If n = 1, then the convex polyhedron is a straight line.
(iii) If n = 2, then the convex polyhedron is a triangle.
(iv) If n = 3, then the convex polyhedron is tetrahedron.

234
 Linear programming

Example: Minimize z = 2x1 - x2 + x3 - 5x4 + 22x5

Subject to: x1 - 2x4 + x5 = 6 (1.1)
x2 + x4 - 4x5 = 3 (1.2)
x3 + 3x4 + 2x5 = 10 (1.3)
xj  0
Solution: The initial basic feasible solution is x1 = 6, x2 = 3, x3 = 10, x4 = 0, x5 = 0, with
the value of the objective function for this solution by the unrestricted variable z = 2x1- x2
+ x3 = 19. We would like to determine if a different basic feasible solution will yield a
smaller value of the objective function or whether the current solution is the optimum.
Note that this is equivalent to asking whether one of the non-basic variable, here x4 and x5,
which are now set equal to zero, should be allowed to take on a positive value, if possible.
From the above equation we solve for the current basic variable in terms of the non-basic
variable to obtain

x1 = 6 + 2x4 - x5 - - - (2.1)
x2 = 3 - x4 + 4x5 - - - (2.2)
x3 = 10 - 3x4 - 2x5 - - - (2.3)

We next rewrite the objective function in terms of only the non basic variables by
substituting for x1, x2 and x3 the corresponding right-hand side expressions above to obtain

z = 2 (6 + 2x4 - x5) - (3 - x4 + 4x5) + (10 - 3x4 - 2x5) - 5x4 + 22x5

Or, z = 19 - 3x4 + 14x5
Or, z = 19 - (3x4) - (- 14x5)
With x4 = 0 and x5 = 0, z = 19, which is the value for the current basic feasible solution.
We see from the last transformed expression of z that if x4 can be made positive, the
objective will decrease 3 units for each unit increase of x 4: while any positive unit
increase to x5 will increase the value of the objective function by 14 units. Since we are
minimizing, it would appear to be appropriate to determine a new basic feasible solution,
i.e., an extreme point solution, involving x4 a positive level, if positive .
We next generate a new extreme point solution by replacing x 2 by x4 to obtain the
solution

x1 + 2x2 - 7x5 = 12 [(1.1) + 2(1.2)]

x2 + x4 - 4x5 = 3
-3x2 + x3 +14x5 = 1 [(1.3)-3(1.2)]

Or, x1 = 12 - 2x2 + 7x5

x4 = 3 - x2 + 4x5
x3 = 1 + 3x2 - 14x5

235
 S. M. Shahidul Islam

The new basic feasible solution is x1 = 12, x2 = 0, x3 = 1, x4 = 3, x5 = 0 . Substituting

the expressions of the basic variables x1, x3, x4 in terms of the non-basic variables x2 and
x5 into the objective function, we now have

z = 2 (12 - 2x2 + 7x5) - x2 + (1 + 3x2 - 14x5) - 5 (3 - x2 + 4x5) + 22x5

Or, z = 10 - (-3x2) - ( -2x5)
From this last expression we see that any increase in the values of the non-basic variable
x2 and x5 would increase the value of the objective function. We thus conclude that the
new basic feasible solution is an optimum, with the value of the objective function of
z = 10. Since z = 10 is less than z = 19. Hence the minimum value of the objective
function is 10 .
The above process is just the direct application of the elimination procedure on the linear
programming system of equations, which now includes an explicit expression of the
objective function.

13.10 Development of a minimum feasible solution:

We assume that the linear programming problem is feasible that every basic feasible
solution is non degenerate, and that we are given a basic feasible solution. These
assumptions, as will be discussed later, are made without any loss in generality. Let the
given solution be X0 = ( x10, x20, . . ., xmo) and associated set of linearly independent
vectors be p1 , p2, . . ., pm, we then have
x10 p1 + x20 p2 + . . . + xmopm = p0 - - - (1)
x10c1 + x20c2 + . . . + xmocm = z0 - - - (2)

Where all xi0  0, the cj are cost the coefficients of the objective function, and z0 is the
corresponding value of the objective function for the given solution. Since the set
p1, p2, . . ., pm is linearly independent and thus forms a basis, we can express any vector
from the set p1, p2, . . ., pn in terms of p1, p2, . . ., pm. Let pj be given by
x1j p1 + x2j p2 + . . . + xmj pm = pj ; j = 1, 2, . . ., n - - - (3)
and we define
x1j c1 + x2j c2 + . . . + xmj cm = zj ; j = 1, 2, . . ., n - - - (4)

Where the cj are the cost coefficients corresponding to the pj .

Theorem 1: If for any fixed j, the condition zj - cj  0 holds, them a set of feasible
solutions can be constructed such that z  z0 for any member of the set, where the lower
bound of z is either finite or infinite (z is the value of the objective function for a particular
member of the set of feasible solutions).
Case-1: If the lower bound is finite, a new feasible solution consisting of exactly m
positive variables can be constructed whose value of the objective function is less than the
value for the preceding solution.

236
 Linear programming

Case-2: If the lower bound is finite, a new feasible solution consisting of exactly m +1
positive variables can be constructed whose value of the objective function can be made
arbitrarily small.
Proof: The following analysis applies to the proof of both cases:
Multiplying (3) by some number  and subtracting from (1), and similarly multiplying (4)
by the some number  and subtracting from (2), for j = 1, 2, . . ., n, we get

(x10 – x1j)p1 + (x20 – x2j)p2 + . . . + (xm0 – xmj)pm + pj = p0 - - - (5)

(x10 – x1j)c1 + (x20 – x2j)c2 + . . . + (xm0 – xmj)cm + cj = z0 – (zj – cj) - - - (6)

Where cj has been added to both sides of (6). If all the coefficients of the vectors
p1, p2, . . ., pm, pj in (5) are non negative, then we have determined a new feasible solution
whose value of the objective function is by (6) z = z0 – (zj – cj). Since the variables
x10, x20, . . ., xm0 in (5) are all positive, it is clear that there is a value of   0 (either finite
or infinite) for which the coefficients of the vectors in (5) remain positive. From the
assumption that, for a fixed j, zj – cj  0, we have
z = z0 –  ( zj – cj )  z 0 , for   0.
Proof of case-1: If, for the fixed j, at least one xij  0 in (3) for i = 1, 2, . . ., m, the largest
value of  for which all coefficients of (5) remain non negative is given by
x
0 = min i 0  0 for xij  0 - - - (7)
i xij
Since we assumed that the problem is non degenerate, i.e., that all basic feasible solutions
have m positive elements, the minimum in (7) will be obtained for a unique i. If 0 is
substituted for  in (5) and (6), the coefficient corresponding to this unique i will vanish.
We have then constructed a new basic feasible solution consisting of pj and m -1 vectors
of the original basis. The new basis can be used as the previous one. If a new z j – cj  0
and a corresponding xij 0, another solution can be obtained which has a smaller value of
the objective function. This process will continue either until all zj – cj  0, or until, for
some zj – cj  0, all xij  0. If all zj – cj  0, the process terminates.
Proof of case-2: If at any stage we have, for some j, zj – cj  0 and all xij  0, then there is
no upper bound to  and the objective function has a lower bound of – . We see for
this case that, for any   0, all the coefficients of (5) are positive. We then have a
feasible solution consisting of m + 1 positive elements. Hence, by taking  large
enough, the corresponding value of the objective function given by the right hand side of
(6) can be made arbitrarily small.
Example: Minimize x2 – 3x3 + 2x5
Subject to x1 + 3x2 – x3 + 2x5 =7
– 2x2 + 4x3 + x4 = 12
– 4x2 + 3x3 + 8x5 + x6 = 10
xj  0, j= 1, 2, . . ., 6

237
 S. M. Shahidul Islam

Solution: Our initial basis consists of p1, p4, p6 and the corresponding solution is
X0 = (x1, x4, x6 ) = (7, 12, 10)
Since c1 = c4 = c6 = 0, the corresponding value of the objective function, z0 equals zero.
p3 is selected to go into the basis, because
max (zi – ci ) = z3 – c3 = 3  0
j

 is the minimum of xi0 /xi3 for xi3  0, that is, min. (12/4, 10/3) = 12/4 = 0 = 3 and
hence p4 is eliminated. We transform the tableau and obtain a new solution.
X 0/ = (x1, x2, x6 ) = (10, 3, 1 )
and the value of the objective function is –9. In the second step, since
1 10
max ( z /j  c j )  z 2/  c2   0 and  0  , p2 is introduced into the basis and p1 is
j 2 5
12
eliminated. We transform the second-step values of tableau and obtain the solution
X 0// = (x2, x3, x6 ) = ( 4, 5, 11 )
Serial Basis c p 0 1 -3 0 2 0 
0

p1 p2 p3 p4 p5 p6
1 p1 0 7 1 3 -1 0 2 0
p4
2 0 12 0 -2 4 1 0 0 12/4=3=  0
p6 Pivot
3 0 10 0 -4 3 0 8 1 10/3
zi – ci 0 0 -1 3
Greatest 0 -2 0
1 p1 0 10 1 5/2 0 ¼ 2 0 10
Pivot
=4=  0
p3 5
2 -3 3 0 -1/2 1 ¼ 0 0 12

3 p6 0 1 0 -5/2 0 -3/4 8 1

z /j  c j -9 0 ½
Greatest
0 -3/4 -2 0
1 p2 1 4 2/5 1 0 1/10 4/5 0
p3
2 -3 5 1/5 0 1 3/10 2/5 0

3 p6 0 11 1 0 0 -1/2 10 1
z //j  c j -11 -1/5 0 0 -4/5 -12/5 0
 
with a value of the objective function equal to –11. Since max z //j  c j  0 , this solution
is the minimum feasible solution.

238
 Linear programming

13.11 The artificial basis technique: Up to this point, we have always assumed that the
given linear-programming problem was feasible and contained a unit matrix that could be
used for the initial basis. Although a correct formulation of a problem will usually
guarantee that the problem will be feasible, many problems do not contain a unit matrix.
For such problems, the method of the artificial basis is a satisfactory way to start the
simplex process. This procedure also determines whether or not the problem has any
feasible solutions.
The general linear programming problem is to minimize
c1x1 + c2x2 + . . . + cnxn
Subject to
a11x1 + a12x2 + . . . + a1nxn = b1
a21x1 + a22x2 + . . . + a2nxn = b2
... ... ...
am1x1 + am2x2 + . . . + amnxn = bm
and xj  0 ; j = 1, 2, . . ., n.
For the method of the artificial basis we augment the above system as follows:
Minimize :
c1x1 + c2x2 + . . . + cnxn + wxn+1 + wxn+2 + . . . + wxn+m
Subject to :
a11x1 + a12x2 + . . . + a1nxn + xn+1 = b1
a21x1 + a22x2 + . . . + a2nxn + xn+2 = b2
... ... ... ... ...
am1x1 + am2x2 + . . . + amnxn + xn+m = bm
and xj  0 ; j = 1, 2, . . ., n, n+1, . . ., n+m

The quantity w is taken to be an unspecified large positive number. The vectors pn+1,
pn+2, . . ., pn+m form a basis (an artificial basis) for the augmented system. Therefore, for
the augmented problem, the first feasible solution is
x0 = (xn+1, 0, xn+2, 0, . . ., xn+m, 0) = (b1, b2, . . ., bm )  0
 xn+1, 0 pn+1 + xn+2, 0 pn+2 + . . . + xn+m, 0 pn+m = p0 - - - (1)
w xn+1, 0 + w xn+2, 0 + . . . + w xn+m, 0 = z0 - - - (2)
And also,
x1j pn+1 + x2j pn+2 + . . . + xmj pn+m = pj - - - (3)
w x1j + w x2j + . . . + w xmj = zj - - - (4)

Multiplying (3) by  and then subtracting from (1) we have,

(xn+1,0 -  x1j ) pn+1 + (xn+2,0 -  x2j )pn+2 + . . . + . . . + (xn+m,0 -  xmj )pn+m +  pj = p0 --- (5)
Multiplying (4) by  and then subtracting from (2) we have
(xn+1,0 -  x1j )w + (xn+2,0 -  x2j )w + . . . + (xn+m, 0 -  xmj)w + cj = z0 -  (zj - cj) - - - (6)
m
Where, zj = w  xij
i 1

239
 S. M. Shahidul Islam

m
 (zj - cj ) = w  xij - cj
i 1
For the first solution each zj – cj will then have a w coefficient which are independent of
each other. We next set up the associated computational procedure as the given table. For
each j, the row free from w component and the row with w component of z j – cj have
been placed in the (m+1)st and (m+2)nd rows, respectively of that column.
We treat this table exactly like the original simplex table except that the vector introduced
into the basis is associated with the largest positive element in the (m+2)nd row. For the
m
first iteration, the vector corresponding to max
j
x
i 1
ij is introduced into the basis.

We continue to select a vector to be introduced into the basis, using the element in the
(m+2)nd row as criterion, until either (a) all artificial vectors are eliminated from the basis
or (b) no positive (m+2)nd element exists. The first alternative implies that all the
elements in the (m+2)nd row equal to zero and that the corresponding basis is a feasible
basis for the original problem.

Serial Basis c p0 c1 c2 . ck . cn w . w . w
p1 p2 . pk . pn pn+1 . pn+l . pn+m
1 pn+1 w pn+1,0 x11 x12 . x1k . x1n 1 . 0 . 0
2 pn+2 w pn+2,0 x21 x22 . x2k . x2n 0 . 0 . 0
.. .. .. .. .. .. . .. . .. .. . .. . ..
.. .. .. .. .. .. . .. . .. .. . .. . ..
l pn+l w pn+l,0 xl1 xl2 . xlk . xln 0 . 1 . 0
.. .. .. .. .. .. . .. . .. .. . .. . ..
.. .. .. .. .. .. . .. . .. .. . .. . ..
m pn+m w pn+m,0 xm1 xm2 . xmk . xmn 0 . 0 . 1
m+1 0 -c1 -c2 . -c3 . -cn 0 . 0 . 0
m+2  x n i , 0 x x i1 i2
. x ik
. x in
0 . 0 . 0

We than apply the regular simplex algorithm to determine the minimum feasible solution.
If the second alternative, if the (m+2, 0 ) element, i.e., the artificial part of the
corresponding value of the objective, is greater than zero, then the original problem is not
feasible.

Example: Minimize 2x1 + x2 [AUB-02]

Subject to 3x1 + x2 - x3 = 3
4x1 + 3x2 - x4 = 6
x1 + 2x2 - x5 = 2
and xj  0 : j = 1, 2, . . ., 5

240
 Linear programming

Solution: For finding a artificial basis we may rewrite the problem as following:
Minimize 2x1 + x2 + 0x3 + 0x4 + 0x5 + wx6 + wx7 + wx8
Subject to 3x1 + x2 - x3 + x6 =3
4x1 + 3x2 - x4 +x7 =6
x1 + 2x2 - x5 + x8 = 2
and xj  0 : j = 1, 2, . . ., 8
Using the above problem we find the following tableau
Sl. Ba c Po 2 1 0 0 0 w w w 
sis P1 P2 P3 P4 P5 P6 P7 P8
1 P6 w 3 3 1 -1 0 0 1 0 0 3/3=1=  o

2 P7 w 6 4 3 0 -1 0 0 1 0 6/4

3 P8 w 2 1 2 0 0 -1 0 0 1 2/1=2
4+1 zi – ci 0 -2 -1 0 0 0 0 0 0
4+2 11 Greatest
8 6 -1 -1 -1 0 0 0
1 P1 2 1 1 1/3 -1/3 0 0 0 0 1/(1/3)=3

2 P7 w 2 0 5/3 4/3 -1 0 1 0 2/(5/3)=6/5

3 P8 w 1 0 5/3 1/3 0 -1 0 1 1/(5/3)=3/5

4+1 zi – ci 2 0 -1/3 -2/3 0 0 0 0
4+2 3 0 10/3 5/3
Greatest
-1 -1 0 0
1 P1 2 4/5 1 0 -2/5 0 1/5 0

2 P7 w 1 0 0 1 -1 1 1 1/1=1

3 P2 1 3/5 0 1 1/5 0 -3/5 0 (3/5)/(1/5)=3

4+1 zi – ci 11/5 0 0 -3/5 0 -1/5 0

4+2 1 0 0 1
Greatest -1 1 0
1 P1 2 6/6 1 0 0 -2/5 3/5

2 P3 0 1 0 0 1 -1 1

3 P2 1 2/5 0 1 0 1/5 -4/5

4+1 zi – ci 14/5 0 0 0 -3/5 Greatest
2/5
4+2 0 0 0 0 0 0
1 P1 2 3/5 1 0 -3/5 1/5 0

2 P5 0 1 0 0 1 -1 1

3 P2 1 6/5 0 1 4/5 -3/5 0

4+1 z– ci 12/5 0 0 -2/5 -1/5 0

The above tableau gives us the extreme point (3/5, 5/6, 0, 0, 1).

241
 S. M. Shahidul Islam

So, the solution of the problem is

x1 = 3/5, x2 = 5/2, x3 = 0, x4 = 0, x5 = 1
and the minimum value is 12/5 (Answer)

13.12 Duality in linear programming problem: The term duality implies that every
linear programming problem whether of maximization or minimization has associated
with it another linear programming problem based on the same data. The original problem
is called the primal problem while the other is called its dual problem. It is important to
note that in general, either problem can be considered as primal and the other as its dual.
Thus, the two problems constitute the pair of dual problem.
Primal problem Dual problem
Minimize Maximize
12x1 + 20x2 100w1 + 120w2
Subject to Subject to

6w1 + 7w2 ≤ 12
6x1 + 8x2  100
8w1 + 12w2 ≤ 20
7x1 + 12x2  120

w1, w2  0
x1, x2  0

Example: The primal problem:

Maximize 5x1 – 3x2
Subject to 4x1 + 5x2 ≤ 45
3x1 – 7x2 ≤ 15
x1 , x 2 ≥ 0
The dual problem of the above primal problem is
Minimize 45w1 + 15w2
Subject to 4w1 + 3w2 ≥ 5
5w1 – 7w2 ≥ – 3
w1, w2 ≥ 0

Note: If either the primal or the dual problem has a finite optimum solution, then the other
problem has a finite solution and the value of the objective functions are same, i.e., min.
f(x) = max. g(w).
If either primal or dual has an unbounded solution then the other has no solution.

242
 Linear programming

Example: Solve the following LPP and find also the solution of its dual problem.
Maximize x 1 + x2 + x 3 [AUB-03]
Subject to 2x1 + x2 + 2x3 ≤ 3
4x1 + 2x2 + x3 ≤ 2
x1, x2, x3 ≥ 0
Solution: Adding slack variables x4, x5 we can rewrite the problem as follows:
Maximize x 1 + x2 + x 3
Subject to 2x1 + x2 + 2x3 + x4 =3
4x1 + 2x2 + x3 + x5 = 2
xj ≥ 0; j = 1, 2, . . ., 5
Converting the problem into simplex table, we get the following tableau

Sl. Basis c Po 1 1 1 0 0 
P1 P2 P3 P4 P5
1 P4 0 3 2 1 2 1 0 3/2 =  o
2 P5 0 2 4 2 1 0 1 2/1 = 2
zj – cj 0 -1 -1 -1 0 0
1 P3 1 3/2 1 1/2 1 1/2 0 (3/2)/(1/2) = 3

2 P5 0 1/2 3 3/2 0 -1/2 1 (1/2)/(3/2)=1/3= o
zj – cj 3/2 0 -1/2 0 1/2 0
1 P3 1 4/3 0 0 1 2/3 -1/3
2 P2 1 1/3 2 1 0 -1/3 2/3
zj – cj 5/3 1 0 0 1/3 1/3

Therefore, the extreme point solution of the primal is (1/3, 4/3) and the maximum value
of the primal is 5/3.
The dual problem of the given problem is
Minimum 3w1 + 2w2
Subject to 2w1 + 4w2 ≥ 1
w1 + 2w2 ≥ 1
2w1 + w2 ≥ 1
w1, w2 ≥ 0
In the first table, P4 and P5 were in basis. In the last table the zj – cj values of P4 and P5
are 1/3 and 1/3 respectively. So, the solution of the dual problem is w1 = 1/3, w2 = 1/3
and the minimum value of the dual is 5/3.
Note: If we solve the dual problem by simplex method, we shall get the same solution.

13.13 Some worked out examples:

Example (1): Minimize 5x1 – 10x2
Subjcet to 2x1 – x2 ≥ -2
x1 + 2x2 ≤ 8
x1, x2 ≥ 0

243
 S. M. Shahidul Islam

Solution: The graph of the problem is as follows:

y From the graph we get the vertices A(0,2),

B(4/5,18/5), C(8,0) and O(0,0). The values of
the objective function at these points are
– 20 at A, – 32 at B, 40 at C and 0 at O. Here, –
B(4/5,18/5) 32 is the minimum value of the objective
A(0,2) function which occurs at the vertices B(4/5,
C(8,0) 18/5).
x/ x Hence, the solution of the given problem is x 1
= 4/5, x2 = 18/5 and min. value of z = –32
(Answer)

y/

Figure – 13.5
Example (2): A furniture company makes tables and chairs. Each table takes 5 hours of
carpentry and 10 hours in painting and varnishing shop. Each chair requires 20 hours in
carpentry and 15 hours in painting and varnishing. During the current production period
400 hours of carpentry and 450 hours. of painting and varnishing time are available. Each
table sold yields a profit of $45 and each chair yields a profit of $80. Using simplex
method determine the number of tables and chairs to be made to maximize the profit.
Solution: Let x1 = the number of tables and x2 = the number of chairs.
So, the total profit 45x1 + 80x2 which is the objective function. We have to maximize the
objective function z = 45x1 + 80x2
The required carpentry hours 5x1 + 20x2
Since 400 carpentry hours are available
4x1 + 3x2 ≤ 240.
Similarly, for painting and varnishing, we have
10x1 + 15x2 ≤ 450.
The non negative conditions x1, x2 ≥ 0
So, the linear programming problem (LPP) of the given problem is
Maximize 45x1 + 80 x2
Subject to 5x1 + 20x2  400
10x1 + 15x2  450
x1, x2  0
We can rewrite the problem as follows:
Minimize – 45x1 – 80x2
Subject to x1 + 4x2 + x3 = 80
2x1 + 3x2 + x4 = 90
xj,  0 ; j = 1, 2, 3, 4

244
 Linear programming

Sl. Basis c Po -45 -80 0 0 

P1 P2 P3 P4
1 P3 0 80 1 4 1 0 80/4 = 20 =  o
2 P4 0 90 2 3 0 1 90/3 = 30
3 zj – cj 0 45 80 0 0
1 P2 -80 20 1/4 1 1/4 0 20/(1/4) = 80

2 P4 0 30 5/4 0 -3/4 1 30/(5/4)=24= o
3 zj – cj -1600 25 0 -20 0
1 P2 -80 14 0 1 2/5 -1/5
2 P1 -45 24 1 0 -3/5 4/5
3 zj – cj -2200 0 0 -5 -20

The above tableau gives the extreme point (24, 14), i.e., x1 = 24, x2 = 14. So, the company
will earn maximum profit $2200 if 24 tables and 14 chairs are made.

Example (3): Solve the following LPP:

Minimize x1 – x2 + x3
Subject to x1 – x4 – 2x6 = 5
x2 + 2x4 – 3x5 + x6 = 3
x3 + 2x4 – 5x5 + 6x6 = 5
xi  0 ; i = 1, 2, . . .,6
Solution: Form the given problem, we get the following tableau:
Sl Basis c Po 1 -1 1 0 0 0 
P1 P2 P3 P4 P5 P6
1 P1 1 5 1 0 0 -1 0 -2
2 P2 -1 3 0 1 0 2 -3 1 3/1 = 3
3 P3
1
5 0 0 1 2
-5 6 5/6 =  o < 3
4 zj – cj 7 0 0 0 -1 -2 3
1 P1 1 20/3 1 0 1/3 -1/3 -5/3 0
2 P2 -1 13/6 0 1 -1/6 5/3 -13/6 0
3 P6 0 5/6 0 0 1/6 1/3 -5/6 1
4 zj – cj 9/2 0 0 -1/2 -2 1/2 0

Since in the second step we find a positive value ½ for z j – cj but there is no positive
number above ½ . So, we can say that the feasible region is unbounded.
Therefore, the optimum solution is at infinity and also the minimum value is infinity.
Example (4): Maximize x1 + 2x2 + 3x3 – x4
Subject to x1 + 2x2 + 3x3 = 15
2x1 + x2 + 5x3 = 20
x1 + 2x2 + x3 + x4 = 10
and xj  0; j = 1, 2, 3, 4.

245
 S. M. Shahidul Islam

Solution: Since we are always minimizing, the corresponding objective is to minimize

– x1 – 2x2 – 3x3 + x4
Subject to x1 + 2x2 + 3x3 = 15
2x1 + x2 + 5x3 = 20
x1 + 2x2 + x3 + x4 = 10
and xij  0; j = 1, 2, 3, 4.
Since in our constraint we get only one basis vector so, we have to take two arbitrary basis
vectors. For this we rewrite our problem as follows:
Minimize –x1 – 2x2 – 3x3 + x4 + wx5 + wx6
Subject to x1 + 2x2 + 3x3 + x5 = 15
2x1 + x2 + 5x3 + x6 = 20
x1 + 2x2 + x3 + x4 = 10
and xj  0; j = 1, 2, . . ., 6

Serial Basis c p0 –1 –2 –3 1 w w 
p1 p2 p3 p4 p5 p6
1 P5 w 15 1 2 3 0 1 0 15/3=5
2 P6 w 20 2 1 5 0 0 1 20/5=4=  0

3 P4 1 10 1 2 1 1 0 0 10/1=10
4+1 zi – ci 10 2 4 4 0 0 0
4+2 35 3 3 8
Greatest 0 0 0
1 P5 w 3 –1/5 7/5 0 0 1 3/(7/5)=15/7=  0
2 P3 –3 4 2/5 1/5 1 0 0 4/(1/5)=20
3 P4 1 6 3/5 9/5 0 1 0 6/(9/5)=30/9
4+1 zi – ci –6 2/5 16/5 0 0 0
4+2 3 –1/5 7/5
Greatest 0 0 0
1 P2 –2 15/7 –1/7 1 0 0
2 P3 –3 25/7 3/7 0 1 0 (25/7)/(3/7)=25/3
3 P4 1 15/7 6/7 0 0 1 (15/7)/(6/7)=5/2=  0

4+1 zi – ci –90/7 6/7 0 0 0

4+2 0 0 0 0 0
1 P2 –2 5/2 0 1 0 1/6
2 P3 –3 5/2 0 0 1 –3/6
3 P1 –1 5/2 1 0 0 7/6
4+1 zi – ci –15 0 0 0 –1

 (5/2, 5/2, 5/2, 0 ) is the required extreme point. The minimum value of the new
objective function is –15. Hence the maximum value of our given objective function is
–(–15) = 15. So, the solution of the problem is x1 = 5/2, x2 = 5/2, x3 = 5/2, x4 = 0 and the
max. value is 15. (Answer)

246
 Linear programming

Example (5): Solve the following LPP :

Maximize 2x1 – 6x2
Subject to 3x1 + 2x2 ≤ 6
x1 – x2 ≥ -1
-x1 – 2x2 ≥ 1
x1 , x 2 ≥ 0
Solution: We can rewrite the problem as follows:
Maximize 2x1 – 6x2 + 0x3 + 0x4 + 0x5 + wx6
Subject to 3x1 + 2x2 + x3 =6
x1 – x2 + x4 = -1
-x1 – 2x2 – x5 + x6 = 1
xi ≥ 0; i = 1, 2, . . ., 6
Sl Basis c Po -2 6 0 0 0 w 
P1 P2 P3 P4 P5 P6
1 P3 0 6 -3 2 1 0 0 0
2 P4 0 1 -1 1 0 1 0 0
3 P5 1 -1 -2 0 0
w -1 1
4 zj – cj 0 2 -6 0 0 0 0 Term free of w
5 1 -1 -2 0 0 -1 0 Term with w

Since all elements of fifth row are non-positive but w appears in the basis at a non-zero
value 1. Hence, the problem has no feasible solution.
Example (6): Solve the following LPP:
Minimize x1 + 2x2
Subject to x1 – 3x2 ≤ 6
2x1 + 4x2 ≥ 8
x1 – 3x2 ≥ – 6
x1 , x 2 ≥ 0
Solution: We can rewrite the problem as follows:
Minimize x1 + 2x2
Subject to x1 – 3x2 ≤ 6
2x1 + 4x2 ≥ 8
– x1 + 3x2 ≤ 6
x1 , x 2 ≥ 0
Introducing slack variables x3, x6, surplus variable x4 and artificial variable x5, we get the
problem as follows:
Minimize x1 + 2x2 + 0x3 + 0x4 + wx5 + 0x6
Subject to x1 – 3x2 + x3 =6
2x1 + 4x2 – x4 + x5 =8
x1 – 3x2 + x6 = – 6
xj ≥ 0; j = 1, 2, . . ., 6

247
 S. M. Shahidul Islam

Now, we convert the problem into the simplex table as follows:

Sl Basis c Po 1 2 0 0 w 0 
P1 P2 P3 P4 P5 P6
1 P3 0 6 1 -3 1 0 0 0
2 P5 w 8 2 4 0 -1 1 0 8/4 = 2 =  o
3 P6 6 -1 3 0 0 6/3 = 2
0 0 1
4 zj – cj 0 -1 -2 0 0 0 0
5 8 2 4 0 0 0 0
1 P3 0 12 5/2 0 1 -3/4 0 12/(5/2) = 24/5
2 P2 2 2 ½ 1 0 -1/4 0 2/(1/2) = 4 =  o
3 P6 0 0 -5/2 0 0 ¾ 1
4 zj – cj 4 0 0 0 -1/2 0
5 0 0 0 0 0 0

From the above table, we find the extreme point (0, 2, 12, 0, 0, 0). Therefore, the
optimum solution is x1 = 0, x2 = 2 with minimum value of the objective function 4.
We can bring P1 vector in the basis for same objective value because in 4th row of P1
column of 2nd step we get 0 and above this 0 we get two positive elements. Taking ½ as
pivot we get the following table.
Sl Basis c Po 1 2 0 0 w 0 
P1 P2 P3 P4 P5 P6
1 P3 0 2 0 -5 1 ½ 0
2 P1 1 4 1 2 0 0 0
3 P6 10 0 5 0 -1/2
0 1
4 zj – cj 4 0 0 0 -1/2 0

So, the second optimal solution is x1 = 4, x2 = 0 and zmax = 4.

There are two different optimal solutions. Hence there will exist an infinite number of
optimal solutions.

13.14 Exercise:
1. Find the numerical values of
5 n
(i) a
a 1
[Answer: 14] (ii)  a if
a 1
n is 6 [Answer: 20]
5 10
(iii)  3ii 1
[Answer: 45] (iv) p
p 5
2
[Answer: 355]

2. Write the expanded form of

5
(i) x
a 1
a [Answer: x1 + x2 + x3 + x4 + x5]

248
 Linear programming

n
(ii) a x
i 1
i i  b [Answer: a1x1 + a2x2 + . . . +anxn = b]

3. Express in compact summation form:

6
(i) y1 – y2 + y3 – y4 + y5 – y6 [Answer:  ( )
i 1
i 1
yi ]
n
(ii) a1x1 + a2x2 + . . . + anxn = c [Answer: a x
i 1
i i c]

4. A dietitian is planning the menu for the evening meal at a university dining hall.
Three main items will be served, all having different nutritional content. The
dietitian is interested to providing at least the minimum daily requirement of each
of three vitamins in this one meal. The following table summarizes the vitamin
content per ounce of each type of food, the cost per ounce of each food, and
minimum daily requirements (MDR) for the three vitamins. Any combination of
the three foods may be selected as long as the total serving size is at least 9 ounces.
Vitamins
Food 1 2 3 Cost per ounce, $
1 50 mg 20 mg 10 mg 0.10
2 30 mg 10 mg 50 mg 0.15
3 20 mg 30 mg 20 mg 0.12
MDR 290 mg 200 mg 210 mg

The problem is to determine the number of ounces of each food to be included in the meal.
The objective is to minimize the cost of each meal subject to satisfying minimum daily
requirements of the three vitamins as well as the restriction on minimum serving size.
Give the formulation of the problem.
[Answer: Minimize z = 0.10x1 + 0.15x2 + 0.12x3
Subject to 50x1 + 30x2 + 20x3 ≥ 290
20x1 + 10x2 + 30x3 ≥ 200
10x1 + 50x2 + 20x3 ≥ 210
x1 + x2 + x3 ≥ 9
x1, x2, x3 ≥ 0]
5. Solve the following linear programming problems using graphical method:
(i) Maximize z = 2x1 + 3x2 (ii) Maximize z = 2x1 – 6x2
Subject to x1 + x2 ≤ 30 Subject to 3x1 + 2x2 ≤ 6
x2 ≥ 3 x1 – x2 ≥ -1
x2 ≤ 12 - x1 – 2x2 ≥ 1
x1 – x2 ≥ 0 x1, x2 ≥ 0
0 ≤ x1 ≤ 20 [Answer: No solution]
[Answer: x1 = 18, x2 = 12, z = 72]

249
 S. M. Shahidul Islam

(iii) Maximize z = 2x1 + 3x2 (iv) Minimize z = x1 + 6x2

Subject to -x1 + 2x2 ≤ 4 Subject to x1 – 3x2 ≤ 6
x1 + x2 ≤ 6 2x1 + 4x2 ≥ 8
x1 + 2x2 ≤ 9 x1 – 3x2 ≥ -6
x1, x2 ≥ 0 x1, x2 ≥ 0
[Answer: x1 = 9/2, x2 = 3/2, and z = [Answer: Many solutions. In particular
27/2] x1 = 0 or 4, x2 = 2 or 0 and min. value =
4]

(v) Minimize z = 2x1 + x2 (vi) Maximize z = x1 + x2

Subject to 3x1 + x2 ≥ 3 Subject to x1 + x2 ≥ 1
4x1 + 3x2 ≥ 6 x1 – x2 ≤ 1
x1 + 3x2 ≤ 3 – x1 + x2 ≤ 1
x1, x2 ≥ 0 x1, x2 ≥ 0
[Answer: x1 = 1, x2 = 2/3, and z= [Answer: Unbounded solution]
8/3]

6. Solve the following LPPs by simplex method:

(i) Minimize z = 2x1 + x2 (ii) Maximize z = 2x1 + 3x2
Subject to 3x1 + x2 ≤ 3 Subject to -x1 + 2x2 ≤ 4
4x1 + 3x2 ≤ 6 x1 + x2 ≤ 6
x1 + 2x2 ≤ 2 x1 + 2x2 ≤ 9
x1, x2 ≥ 0 x1, x2 ≥ 0
[Answer: x1 = 0, x2 = 0, and zmin. = [Answer: x1 = 9/2, x2 = 3/2, and z =
27/2] 27/2]

(iii) Maximize z = 3x1 + 6x2 + 2x3 (iv) Minimize z = 3x1 + 2x2

Subject to 3x1 + 4x2 + x3 ≤ 2 Subject to 2x1 – x2 ≥ -2
x1 + 3x2 + 2x3 ≤ 1 x1 + 2x2 ≤ 8
x1, x2, x3 ≥ 0 x1, x2 ≥ 0
[Answer: (2/5, 1/5, 0) and z max. = [Answer: (8, 0) and zmin. = 24]
12/5]

7. Solve the following LPP by simplex method and hence solve it dual problem.
(iii) Maximize z = x1 + x2 + x3
Subject to 2x1 + x2 + 2x3 ≤ 3
4x1 + 2x2 + x3 ≤ 2
x1, x2, x3 ≥ 0
[Answer: Primal: (0, 1/3, 4/3) and Max. value = 5/3,
Dual: (-1/3, -1/3) and Min. value = 5/3]

250
 Transportation problem

14
Chapter
Transportation Problem
Highlights:

14.1 Introduction 14.6 Degeneracy case

14.2 Transportation problem 14.7 Multiple solutions
14.3 Theorem 14.8 When total supply exceeds
14.4 Northwest corner rule total demand
14.5 Loop 14.9 Maximization problem
14.10 Exercise

14.1 Introduction: It gets its name from its application to problems involving transporting
products from several origins (factories) to several destinations (markets). The two
common objectives of such problems either minimize the cost of shipping n units to m
destinations or maximize the profit of shipping n units to m destinations. Transportation
problem is an especial type of linear programming problem. Though every linear
programming problem can be solved by simplex method, there are more than one solution
methods (northwest-corner method, least cost method and Vogel’s approximation method)
of transportation problem, which are computationally more efficient than the simplex
method. Here, we will discuss only one method named northwest-corner rule in the
context of some examples.

14.2 Transportation problem: The following features characterize the transportation

problem:
Manufacturing plants (Origins) Markets (Destinations)
m in number n in number

O2. D2.
O3. D1. D3.
cij
O1. Oi . xij
. D.
Dj
n
si Om. . dj
. . .

251
 S. M. Shahidul Islam

1. m origins (or, sources) Oi with a total available quantity (supply) si (i =1,2,3, ...,m)
2. n destinations Dj with demand dj (j = 1, 2, 3, . . ., n) of goods
3. a cost cij ≥ 0 for shipping one unit from origin Oi to destination Dj.
The objective is to allocate xij ≥ 0 units from origin Oi to destination Dj such that the
restriction on the availability on the supply at each origin as well as the constraint on the
demand at each destination are met, and at the same time, the total shipping cost is
minimized. Then the transportation problem reduces to the following LP problem.
m n
Minimize Z =  c
i 1 j 1
ij xij
n
Subject to x
j 1
ij  si (i  1, 2, 3, . . ., m)
m

x
i 1
ij  d j ( j  1, 2, 3, . . ., n)

xij  0 (i  1, 2, . . ., m; j  1, 2, . . ., n)

14.3 Theorem: The solution of the transportation problem is achieved only for xij’s
(1≤ i≤ m, 1 ≤ j ≤ n) satisfying the constraints
n

x j 1
ij  si (i  1, 2, 3, . . ., m)
m

x
i 1
ij  d j ( j  1, 2, 3, . . ., n)

xij  0 (i  1, 2, . . ., m; j  1, 2, . . ., n)
 
Proof: Let x *  xij*  R mn be a solution to the transportation problem with
x *
ij  b j0  d j0 for some j0  {1, 2, 3, . . ., m}
Then there exists an i0 such that xi*0 j0 > 0.

 xij if i  i0 , j  j0
*

Let yij =  *
 xij   if i  i0 , j  j0

Choose ε > 0 sufficiently small such that
m

y
i 1
ij  dj
n

y
j 1
ij  si

yij ≥ 0 for all i, j

So, we get the value of the objective function, such that
252
 Transportation problem
m n m n

 c
i 1 j 1
ij yij   cij xij*
i 1 j 1

it contradict the fact that x * minimizes the total cost. Hence, complete the proof of the
lemma.
Note: This lemma state that the transportation problem has solution only if the demand dj
at each destination is met exactly.
Note: Then the transportation problem reduces to the following LP problem:
m n
Minimize Z =  c
i 1 j 1
ij xij
n
Subject to x j 1
ij  si (i  1, 2, 3, . . ., m)
m

x i 1
ij  d j ( j  1, 2, 3, . . ., n)

xij  0 (i  1, 2, . . ., m; j  1, 2, . . ., n)
From the constraint conditions, we get
m m n n

si 1
i ≥  x
i 1 j 1
ij = d
j 1
j

n m
Thus, if d j >
j 1
s
i 1
i , the problem has no feasible solution, otherwise the problem has a

feasible solution.
m n
If, however,  si >
i 1
d
j 1
j , we may create a fictitious (dummy) (n +1)-st destination Dn+1

with demand dn+1 and the cost ci ,n 1 given by

m n
dn+1 = s – d
i 1
i
j 1
j , ci ,n 1 = 0 for i = 1, 2, 3, . . ., m
m n
That is, the demand dn+1 at the dummy destination Dn+1 is  si –
i 1
d
j 1
j , and shipping cost

from each of the origins to Dn+1 is 0 (zero).

So, we can consider the following problem as transportation problem:
m n
Minimize Z =  c
i 1 j 1
ij xij
n
(Primal) Subject to x
j 1
ij  si (i  1, 2, 3, . . ., m) - - - (i)
m

x
i 1
ij  d j ( j  1, 2, 3, . . ., n)

253
 S. M. Shahidul Islam
xij  0 (i  1, 2, . . ., m; j  1, 2, . . ., n)
We can write the problem as follows:
m n
Minimize Z =  c
i 1 j 1
ij xij

Subject to A x = w
x ≥0
where,
 x1 1 
   s1 
1 1 ... ... 1 0 0 ... ... 0 ... 0 0 ... ... 0   x1 2   
  
:   s2 
 0 0 ... ... 0 1 1 ... ... 1 ... 0 0 ... ... 0 
m      : 
   x1n   

A=  0

0 ... ... 0 0 0 ... ... 0 ... 1 1 ... ... 1 

  w  sm 
  x = 
x21   
 1 0 ... ... 0 1 0 ... ... 0 ... 1 0 ... ... 0  :   d1 
     d2 
n  0 1 ... ... 0 0 1 ... ... 0 ... 0 1 ... ... 0   
 x2n 
  
x  : 
  0 
... 1   m1  d 
 0 ... ... 1 0 0 ... ... 1 ... 0 0 ...  n
: 
 (mn) matrix x 
 (m+n  1) matrix
(m + n)
 mn
(mn)  1 matrix
Proposition: The transportation problem (i)
1. has a feasible solution, namely
m n
xij = sidj/p (i = 1, 2, . . ., m; j = 1, 2, . . ., n), p = s = d
i 1
i
j 1
j

2. has a solution of at most m + n –1 positive xij’s giving a bfs (basic feasible

solution); in the non-degenerate case, every bfs has exactly m + n –1 positive xij’s,
otherwise, we have the degenerate case which occurs when some partial sum of si’s
is equal to some partial sum of dj’s.
3. has every bfs integer if all the si’s and dj’s are non-negative integers (not all of
them being zero), and in particular, the problem has integer optimal solution if all
si’s and dj’s are integers.
4. has a finite minimum feasible solution.
Now, the dual problem of problem (i) is
m n
Maximize Z =  si u i   d j v j
i 1 j 1
- - - (ii)

(Dual) Subject to u i + v j  cij (i = 1, 2, . . ., m; j = 1, 2, . . ., n)

u i , v j are unrestricted in sign for all i, j.
We can now display the transportation problem tableau as follows:
254
 Transportation problem

Dj 1 2 ... n Supply ui
Oi
1 c11 c12 : c1n s1 u1
x11 x12 . x1n

2 c21 c22 : c2n s2 u2

x21 x22 . x2n

: : : : : : :
. . . . . . .

m cm1 cm2 : cmn sm um

xm1 xm2 . xmn

Demand d1 d2 ... dn
vj v1 v2 . .. vn

We now recall that, if x *  ( x11

*
, ..., x1*n , x21
*
, ..., x2*n , xm* 1 , ..., xmn
*
) is an optimal solution of
primal problem and if w*  (u1* , u 2* , . . ., u m* , v1* , v2* , . . ., vn* ) , then
m n m n

 cij xij* =
i 1 j 1
 si ui*   d j v *j
i 1 j 1

Also, from the complementary slackness property,

u i* + v *j < cij for every optimal non-basic variable xij* , - - - (1)
u i* + v *j = cij for every optimal basic variable xij* , - - - (2)
The equation (2) represent a system of at most m + n –1 equations in m + n variables.
Therefore, some variables can be arbitrarily specified and unique values are then obtained
from the remaining ones. Since, there is always at least one ‘free’ variable, we fix, by
convention u1 = 0 and then solve the resulting system of equation (2). In the non-
degenerate case of the transportation problem, there would always be exactly one free
variable and the choice u1 = 0 would always lead to a unique solution of the system of
equation (2).

14.4 Northwest corner rule: This rule is developed to solve the transportation problem
by Charnes and Cooper. We discuss the rule follows:
Starting with the original m  n cost matrix together with the (m +1)-st row giving the
demand dj at each destination Dj (j = 1, 2, . . ., n) and the (n +1)-st column giving the
supply s i at each origin Oi (i = 1, 2, . . ., m), we allocate to the cell (1, 1) as many as

255
 S. M. Shahidul Islam
possible without violating the constraints on supply s1 at the origin O1 and demand d 1 at
the destination D1 . Thus
x11 = min ( s1 , d1 ) = s1 ٨ d 1
Three cases may now arise, namely,
(i) s1 > d 1 , (ii) s1 < d 1 and (iii) s1 = d 1 .
In the first case, the demand (that is, the column) constraint is satisfied and we move to the
right to the cell (1, 2) with the remaining s1 – d 1 that must be allocated to the remaining
cells of the first row following the same procedure, and we allocate ( s1 – d 1 ) ٨ d 2 to the
cell (1, 2), that is x12 = ( s1 – d 1 ) ٨ d 2
In the second case, the supply (that is, the row) constraint is met, and we move to the cell
(2,1) and allocate x 21 = ( d 1 – s1 ) ٨ s 2 to this cell.
In the third case, we have degeneracy, which would be treated later.
We continue in this way, allocating as much as possible to the cell (the corresponding
variable) under consideration without violating the constraints: the sum of the i-th row
allocation cannot exceed s i , the sum of the j-th column allocation cannot exceed dj, and no
allocation can be negative. Having allocated to the cell (i, j), we move to the right to the
cell (i, j+1) or to the cell (i+1, j) below according as some supply or some demand
remains; if the i-th row and j-th column constraints are satisfied simultaneously, we have
the degenerate case. Because of the constraint that the total supply equals the total
demand, it is clear that, when we finally reach the last cell (m, n), the allocation to this cell
would simultaneously satisfy the m-th row and n-th column constraints.
Thus, beginning with the northwest-corner cell (1, 1), the northwest corner rule allocates
to this cell (the variable x11 ) s1 ٨ d 1 , satisfying either the row constraint or the column
constraint, and then proceeds so as to satisfy either the row constraint or the column
constraint at each step.
Alternatively, the northwest-corner rule may be viewed as initially allocating to the (1, 1)
cell the number s1 ٨ d 1 , and then following the same rule to either the resulting m  (n –1)
cost matrix or the resulting (m –1)  n cost matrix obtained from the original m  n cost
matrix by crossing off the first column if s1 < d 1 or the first row if s1 > d 1 , and then
proceeding in this way successively.
Thus, in the non-degenerate case, we see that the m + n –1 variables evaluated by the
northwest-corner rule constitute an initial bfs for the transportation problem. These m+n–1
variables are basic, the remaining ones are non-basic. It is to note here that, when
degeneracy is present, the number of basic variables determined by the northwest-corner
rule is less than m + n –1.
After finding the initial bfs by the northwest-corner rule, we now proceed to test whether
this is already optimal for the problem or not. This is based on u i* + v *j < cij and

256
 Transportation problem
u + v = cij . Following the convention, we set u1 = 0, and then solve for u i ’s and v j ’s
*
i
*
j

the system of the system of m + n –1 equations:

u i + v j = cij for each basic variable xij .
Having found the u i ’s and v j ’s from the above system, we then calculate
cij – u i – v j for each non-basic variable xij .
If cij – u i – v j ≥ 0 for each non-basic variable xij , then the initial bfs is optimal.
Otherwise, the initial bfs is not optimal and we have to improve the solution by entering
some non-basic variable in the basis, and thereby making one of the current basic
variables non-basic.
Now, to improve upon the initial bfs, we choose the non-basic cell (variable) with the
most negative of the quantities cij – u i – v j , that is, we choose
ci0 j0 – u i0 – v j0 = min { cij – u i – v j : cij – u i – v j < 0},
and in the next iteration, the variable xi0 j0 is made the basic variable in place of a current
basic variable. To do so, we construct a loop consisting exclusively of the ( i0 , j 0 ) cell and
other (current) basic cells. We then allocate to the ( i0 , j 0 ) cell as much as possible such
that, after making appropriate adjustments to the other cells of the loop, the supply and
demand constraints are not violated, all allocations remain non-negative, and one of the
current basic variables has been reduced to 0, which, henceforth, ceases to be basic.
To see how much is to be allocated to the cell ( i0 , j 0 ), let us first allocate an amount θ > 0
to it. Let the other cells of the loop be ( i0 , t), (u, t), (u, v), . . ., (z, j 0 ). Then, in order to
preserve the supply constraint, we have to reduce the allocation to the cell ( i0 , t) by the
same amount. But then, we have to increase the allocation to the cell (u, t) by θ in order to
maintain the corresponding demand constraint, which in turn would require that the
allocation to the cell (u, v) be reduced by θ accordingly, and so on, and finally, the
allocation to the cell (z, j 0 ) is to be reduced by θ so as not to violate the j 0 -th column
(demand) constraint. Therefore, if we denote by + and – respectively the situations when
the allocation is to be increased and reduced by θ, then we have
cells: ( i0 , j 0 ) ( i0 , t) (u, t) (u, v) . . . (z, j 0 )
signs: + – + – –
Therefore, θ is to be so chosen such that the non-negativity constraints are satisfied for all
the basic variables and at least one basic variable becomes 0 which then ceases to be basic;
if two or more basic variables vanish simultaneously, we make one of them non-basic and
keep the rest at zero level(s). The latter are so chosen that the basic cells do not form
loops.
In this way, we get a new bfs. If this solution is not optimal, we continue the process of
generating new bfs(s) till we get an optimal one.
257
 S. M. Shahidul Islam
Thus, the working rule of the transportation algorithm proceeds as follows:
Step 1 (Initialization): Find an initial bfs by the northwest-corner rule. In the non-
degenerate case, the northwest-corner rule evaluates exactly m + n –1 positive (basic)
variables.
Step 2 (Optimality): For each non-basic cell (variable), calculate the quantities
cij – u i – v j . If all these quantities are non-negative, the current bfs is optimal; if some are
zero, then the given problem possesses multiple solutions, and if all are strictly positive,
the problem under consideration has a unique optimal solution. Otherwise (if at least one
of the quantities cij – u i – v j is negative) go to step 3.
Step 3 (Improved bfs): Choose the non-basic variable (cell) with the most negative value
of cij – u i – v j . This variable (cell) would be basic in the next iteration. This is done, by
constructing a loop consisting of the above non-basic cell and other basic cells. Then
allocate to the non-basic cell as much as possible, making one of the current basic variable
(cell) non-basic by reducing it to zero and making subsequent adjustments so as to
preserve the supply and demand constraints corresponding to the rows and column
involved in the loop.
Return to step 2.
After a finite number of steps, we would reach to an optimal solution to our problem.

14.4.1 Lemma: For the m  n transportation tableau, the northwest-corner rule evaluates
exactly m+n–1 positive variables in the non-degenerate case.
Proof: We prove this lemma by induction method. The result is clearly true for
(m, n) = (1,1), (2,1), (1,2). For (m, n) = (2,2), only one of the following two cases can
arise, depending on whether
(i) s1 < d 1 , (ii) s1 > d 1
In case (i), the allocation to the (1, 1) cell would satisfy the row constraint, and in case (ii),
the allocation to the cell (1, 1) would satisfy the column constraint.
s1 s1 d1 s1 – d 1 s1
d 1 – s1 s 2 –( d 1 – s1 ) s 2 d 2 –( s1 – d 1 ) s2
d1 d2 d1 d2
Now, since, s1 + s 2 = d 1 + d 2 => d 2 = s 2 –( d 1 – s1 ), s 2 = d 2 –( s1 – d 1 ), we see that, in either
case (in the absence of degeneracy), the northwest-corner rule determines three strictly
positive variables, so that, the lemma is also true for (m, n) = (2, 2).
Now, after allocating to the (1,1) cell, the northwest-corner rule applies to either the
m  (n–1) cost matrix or to the (m –1)  n cost matrix. Making the induction hypothesis
that the lemma holds for each of these cost matrices, in either of the above two cases, the
northwest-corner rule evaluates (m + n –2) + 1 = m + n –1 strictly positive variables for
the original m  n cost matrix. This completes the proof by induction.

258
 Transportation problem
14.5 Loop: A loop is a sequence of cells in the original m  n transportation tableau for the
transportation problem such that
1. each pair of consecutive cells lie in either the same row or the same column,
2. no three (or more) consecutive cells lie in the same row or the same column,
3. the first and last cells of the sequence lie in the same row or the same column,
4. no cell appears more than once in the sequence.

Example of loops: Two valid loops are shown in the figure below, each starting form the
cell (1, 3). The loop in the first figure is obtained by the sequence of cells.
(1, 3), (1, 1), (5, 1), (5, 6), (4, 6), (4, 5), (2, 5), (2, 3)
and that in the second figure is formed by the sequence of cells
(1, 3), (1, 6), (4, 6), (4, 1), (3, 1), (3, 2), (5, 2), (5, 4), (3, 4), (3, 3)
1 2 3 4 5 6 1 2 3 4 5 6
1 1
2 2
3 3
4 4
5 5
We note that, in forming a loop starting form any cell, we need an old number of cells
(excluding the initial cell).
From the construction of the initial bfs by the northwest-corner rule, it is clear that no loop
can exist connecting the basic cells (for which the allocations are positive).

Example: A production company has four factories in different places to produce its
products and three market places, where these products are sold. Transport cost of per unit
product from each factory to every market are shown below:

Dj 1 2 3 Supply
Oi si
1 2 7 4 5
2 3 3 1 8
3 5 4 7 7
4 1 6 2 14
Demand 7 9 18
dj

Find the optimal transport system that fulfills the supplies of factories and demands of
markets with minimum transport cost. [AUB-2003 MBA]

259
 S. M. Shahidul Islam
Solution: To solve the problem by northwest corner rule, we form table (1) as follows:
Firstly, in the northwest corner cell (1, 1), we allot minimum of factory-1’s supply and
market-1’s demand, that is, min. { s1 = 5, d 1 = 7} = 5. This allocation fulfills the first
supply constraint but first demand constraint. So, secondly, we allot cell (2, 1) minimum
of market-1’s remaining demand and factory-2’s supply; that is min. {7 – 5, 8} = 2. It
fulfills the first demand constraint but second supply constraint. So, we go to cell (2, 2).
Last allocation 14 in cell (4, 3) satisfies all constraints.
Dj 1 2 3 Supply ui
Oi si
1 2 5 7 (5) 4 (-1) 5 0
(1)
2 3 2 3 6 1 (-5) 8 1

3 5 (1) 4 3 7 4 7 2

4 1 (2) 6 (7) 2 14 14 -3

Demand 7 9 18
dj
vj 2 2 5
We always take u1 = 0 and then calculate other ui’s and vj’s using ui + vj = cij for only
basic cells in the above table. As for example v1 = c11 – u1 = 2 – 0 = 2 and u 2 = c 21 – v1 = 3
– 2 = 1. Then we calculate xij for non-basic variable using cij – ui – vj and put in
parentheses. As for example ( x12 ) = ( c12 – u1 – v 2 ) = (7 – 0 – 2) = (5) and ( x13 ) = ( c13 –
u1 – v3 ) = (4 – 0 – 5) = (–1). There are negative numbers in parentheses in the non-basic
cells. So, the above table is not optimal. We search the smallest negative number in non-
basic cells to build up a loop for getting the next tableau. Here, – 5 is the smallest number
in the non-basic cell (2, 3). So, we form a loop starting from the non-basic cell (2, 3) and
goes through the basic cells (2, 2), (3, 2), (3, 3). In the next table, we make cell (2, 3) basic
allotting the respective max. {6, 3, 4} = 4 which satisfies the supply and demand
constraints. Then we deduce 4 from 6 in cell (2, 2), reduce 3 in cell (3, 2) by 4 and deduce
4 from 4 in cell (3, 3). Now allot of cell (3, 3) becomes 0, that is, this cell is a non-basic
cell in second table. And all other basic cells are unchanged.

260
 Transportation problem
Dj 1 2 3 Supply ui
Oi si
1 2 5 7 (5) 4 (4) 5 0
(2)
2 3 2 3 2 1 4 8 1

3 5 (1) 4 7 7 (5) 7 2

4 1 (-3) 6 (2) 2 14 14 2

Demand 7 9 18
dj
vj 2 2 0
In the same way as in table (1) we calculate ui’s, vj’s and xij’s for non-basic cells in table
(2). Forming a loop through cells (4, 1), (4, 3), (2, 3) and (2, 1) we build up the following
table.

Dj 1 2 3 Supply ui
Oi si
1 2 5 7 (2) 4 (1) 5 0
(3) 2 3 (3) 3 2 1 6 8 -2

3 5 (4) 4 7 7 (5) 7 -1

4 1 2 6 (2) 2 12 14 -1

Demand 7 9 18
dj
vj 2 5 3

Since, in the non-basic cells all are non-negative in the parentheses, hence table (3) is
optimal table. From table (3) we find the following optimal solution:
 x11
* *
x12 *
x13  5 0 0  Origins Destinations
 *   
x x 22 x 23   0 2 6 
* * (5) 1 5 1 (7)
* 2
x =  21 * * * 
= 0 7 0  Or, (14) 4 12 3 (18)
 x31 x32 x33    (8) 2 6 2 2 (9)
 x * x * x *   2 0 12 
 41 42 43    (7) 3 7
with the required minimum cost

261
 S. M. Shahidul Islam
m n
Z* =  c
i 1 j 1
ij xij* = 2  5 + 3 2 + 1 6 + 4  7 + 1  2 + 2  12 = 76

From the optimal table, we also get the solution of the dual problem as follows:
u  (u1* , u 2* , u3* , u 4* ) = (0, -2, -1, -1)
*

v  (v1* , v2* , v3* ) = (2, 5, 3)

*

so that,
m n m n
Z* = s u  d v
i 1
i
*
i
j 1
j
*
j = 0 - 16 - 7 -14 + 14 + 45 + 54 = 76 =  c
i 1 j 1
ij xij*

14.6 Degeneracy case: If from the beginning some sequential partial some of the supplies
si's equals some sequential partial sum of the demand dj's, then we have degeneracy in the
initial bfs. In this case, the northwest-corner rule the basic variables whose number is less
than m+n-1. Even if we start with the non-degenerate initial bfs, we may have degeneracy
in subsequent iterations. This last situation can arise when in the process of making a non-
basic variable, at least two basic variables in the loop reduce to 0 (zero) simultaneously (so
that, at least two current basic variables have values exactly equal to that by which the
non-basic variable in the loop increases).
These degenerate cases are dealt with by the ε-perturbation method, due to Dantzig. In the
ε-perturbation method, the values of si’s and dj’s are perturbed so that the modified
problem reads as
m n
Minimize Z =  ci 1 j 1
ij xij
n
Subject to x
j 1
ij  si   (i  1, 2, 3, . . ., m) - - - (i)
m

x
i 1
ij  d j ( j  1, 2, 3, . . ., n)
m


i 1
xin = dn + m 

xij  0 (i  1, 2, . . ., m; j  1, 2, . . ., n)
Where  > 0 is a small quantity and after the final solution is obtained, we set  = 0. Thus,
the modified problem is obtained from the original one by increasing each supply si by the
small quantity  > 0, keeping the first n-1 demands d1, d2, . . ., dn-1 unchanged and
changing the n-th demand dn to dn+m in order to keep the total supply equal to the total
demand. We then proceed in the same way as that for the non-degenerate case, and after
reaching the optimal solution to the modified problem, we finally set  = 0 to get the
optimal solution to the original problem.

262
 Transportation problem
Another way of dealing with the degenerate case is to allocate values 0 to the additional
variables and treating these variables as basic so that the total number of basic variables
equals m+n-1, some with positive valves and the rest with values 0 assigned to them. The
choice is arbitrary, to a point: basic variables cannot form loops, and in case of tie,
preference is given to variables with the lowest associated costs, but it is not necessary.
We then proceed in the same way as that for the non-degenerate case.

Example: Solve the transportation problem with the following 33 cost matrix, where the
supplies s1 , s 2 and s 3 and the demands d 1 , d 2 and d 3 are as indicated.
Dj 1 2 3 Supply
Oi si [AUB-2002 MBA]
1 8 7 3 60
2 3 8 9 70
3 11 3 5 80
Demand 50 80 80
dj

Solution: Here, since s1 + s 2 = d 1 + d 2 , we have the degenerate case, and the northwest-
corner rule evaluated the following initial bfs:
Dj 1 2 3 Supply ui
Oi si
1 8 50 7 10 3 60

(1) 3 8 70 9 70
2
11 3 5 80 80
3
Demand 50 80 80
dj
vj
To resolve degeneracy, we apply the -perturbation method, and the modified tableau is
given below, where each supply is increased by  and only the demand d 3 is increased by
3,  > 0 being a small quantity.

263
 S. M. Shahidul Islam
Dj 1 2 3 Supply ui
Oi si
1 8 50 7 10+ 3 (-5) 60+ 0

(2) 3 (-6) 8 70- 9 2 70+ 1

2
11 (6) 3 (-1) 5 80+ 80+ -3
3
Demand 50 80 80 +3
dj
vj 8 7 8
Dj 1 2 3 Supply ui
Oi si
1 8 (6) 7 60+ 3 (-5) 60+ 0

(3) 3 50 8 20- 9 2 70+ 1

2
11 (12) 3 (-1) 5 80+ 80+ -3
3
Demand 50 80 80 +3
dj
vj 2 7 8
Dj 1 2 3 Supply ui
Oi si
1 8 (6) 7 60- 3 2 60+ 0

(4) 3 50 8 20+ 9 (5) 70+ 1

2
11 (7) 3 (-6) 5 80+ 80+ 2
3
Demand 50 80 80 +3
dj
vj 2 7 3

264
 Transportation problem

Dj 1 2 3 Supply ui
Oi si
1 8 (12) 7 (6) 3 60+ 60+ 0

(5) 3 50 8 20+ 9 (-1) 70+ 7

2
11 (13) 3 60- 5 20+2 80+ 2
3
Demand 50 80 80 +3
dj
vj -4 1 3
Dj 1 2 3 Supply ui
Oi si
1 8 (11) 7 (6) 3 60+ 60+ 0

(6) 3 50 8 (1) 9 20+ 70+ 6

2
11 (12) 3 80 5  80+ 2
3
Demand 50 80 80 +3
dj
vj -3 1 3

From the last tableau (6), we see that we have reached the optimal solution of the modified
problem. Now, setting  = 0, we get the optimal solution of the original problem, which is
given in the 33 matrix form as well as schematically below:
 0 0 60 
 
x   50 0 20  , with minimum transport cost
*

 0 80 0 
 
m n
Z*=  c
i 1 j 1
ij xij* = 360 + 350 + 920 + 380 = 750

If we want to avoid the -perturbation method of resolving degeneracy we have to make

one more variable basic in initial tableau by assigning the value 0 to a non-basic cell such
that this cell does not constitute a loop with other basic cells of the tableau. The method is
as follows:

265
 S. M. Shahidul Islam

Dj 1 2 3 Supply ui
Oi si
1 8 50 7 10 3 (-6) 60 0

(1) 3 (-6) 8 70 9 (-1) 70 1

2
11 (7) 3 0 5 80 80 -4
3
Demand 50 80 80
dj
vj 8 7 9
Dj 1 2 3 Supply ui
Oi si
1 8 50 7 (6) 3 10 60 0

(2) 3 (-12) 8 70 9 (-1) 70 7

2
11 (1) 3 10 5 70 80 2
3
Demand 50 80 80
dj
vj 8 1 3
Dj 1 2 3 Supply ui
Oi si
1 8 (12) 7 (6) 3 60 60 0

(3) 3 50 8 20 9 (-1) 70 7
2
11 (13) 3 60 5 20 80 2
3
Demand 50 80 80
dj
vj -4 1 3

266
 Transportation problem

Dj 1 2 3 Supply ui
Oi si
1 8 (11) 7 (5) 3 60 60 0

(4) 3 50 8 0 9 20 70 6
2
11 (13) 3 80 5 (1) 80 1
3
Demand 50 80 80
dj
vj -3 2 3
Therefore, we get the optimal solution from the final tableau (4) as before.
Example: There are three television production factories and four market places at
Dinajpur. Supply of each firm, demand of each market per day and transport cost from
each factory to every market are given below:

Dj 1 2 3 4 Supply
Oi si
1 10 7 3 6 3
2 1 6 8 3 5
3 7 4 5 3 7
Demand 3 2 6 4
dj
Find the optimal transport system with minimum transport cost. [AUB-2003 MBA
Production Mgt.]
Solution: Here, since s1 = 3 = d 1 , the problem is of degeneracy type. Calculating the
initial bfs by the northwest-corner rule putting 0 in basic cell (1, 2), we can construct the
the following tableau:
Dj 1 2 3 4 Supply ui
Oi si
1 10 3 7 0 3 (-6) 6 (-1) 3 0

(1) -1
2 1 (-8) 6 2 8 3 3 (-3) 5

3 7 (1) 4 (1) 5 3 3 4 7 4

Demand 3 2 6 4
dj
vj 10 7 9 7
267
 S. M. Shahidul Islam

Dj 1 2 3 4 Supply ui
Oi si
1 10 1 7 2 3 (-14) 6 (-9) 3 0

(2) 2 1 2 6 (8) 8 3 3 (-3) 5 -9

3 (9) (9) 3 4 7 -
7 4 5 3 12

Demand 3 2 6 4
dj
vj 10 7 17 15
Dj 1 2 3 4 Supply ui
Oi si
1 10 (14) 7 2 3 1 6 (5) 3 0
(3)
2 1 3 6 (-6) 8 2 3 (-3) 5 5

3 7 (9) 4 (-5) 5 3 3 4 7 2

Demand 3 2 6 4
dj
vj -4 7 3 1

Dj 1 2 3 4 Supply ui
Oi si
1 10 (15) 7 (6) 3 3 6 (5) 3 0
(4)
2 1 3 6 2 8 0 3 (-3) 5 5

3 7 (10) 4 (1) 5 3 3 4 7 2

Demand 3 2 6 4
dj
vj -5 1 3 1

268
 Transportation problem

Dj 1 2 3 4 Supply ui
Oi si
1 10 (11) 7 (3) 3 3 6 (5) 3 0

(5)
2 1 3 6 2 8 (3) 3 0 5 2

3 7 (6) 4 (-2) 5 3 3 4 7 2

Demand 3 2 6 4
dj
vj -1 4 3 1

Dj 1 2 3 4 Supply ui
Oi si
1 10 (11) 7 (5) 3 3 6 (5) 3 0

(6) 2
2 1 3 6 (2) 8 (3) 3 2 5

3 7 (6) 4 2 5 3 3 2 7 2

Demand 3 2 6 4
dj
vj -1 2 3 1
The tableau (6) is optimal, since all  ij  cij  ui  v j  0 for all non-basic cells.
Therefore, the optimal solution is: Markets: 1 2 3 4
1 0 0 3 0
*  
x = Factories: 2  3 0 0 2  ,
3  0 2 3 2 
Z * = 3  3 + 1  3 + 3  2 + 4  2 + 5  3 + 3  2 = 47 (Answer)

14.7 Multiple solutions: We have already pointed out that, in the final tableau
corresponding to the optimal solution of the transportation problem
non  negative corresponding to a non  basic cell
cij  ui  v j  
0 corresponding to a basic cell

269
 S. M. Shahidul Islam
If ci0 j0 – u i0 – v j0 = 0 for some non-basic cell ( i0 , j0 ), then we have multiple solutions to
the given problem, and another optimal solution is obtained by making the ( i0 , j0 ) cell
basic in the next iteration by considering a loop consisting exclusively of the current non-
basic ( i0 , j0 ) cell and other basic cells in the tableau.
Example: A production firm of computer monitor has four factories (origins) and six
showrooms (destinations) in the city of New York. Supply of each factory, demand of
each showroom and transport cost of a monitor from each factory to every showroom are
as follows:
Dj 1 2 3 4 5 6 Supply
Oi si
1 9 12 9 6 9 10 5
2 7 3 7 7 5 5 6
3 6 5 9 11 3 11 2
4 6 8 11 2 2 10 9
Demand 4 4 6 2 4 2
dj

Find the optimal transportation systems that minimize total transport cost. [AUB-2003]
Solution: We fist find the (non-degenerate) initial bfs by the northwest corner rule and
then form successively the following tableau:
Dj 1 2 3 4 5 6 si ui
Oi
1 9 12
9 6 9 10 5 0
4 1 (-7) (-1) (2) (-5)
2 7 3 7 7 5 5 6 -9
(7) 3 3 (9) (7) (-1)
3 6 5 9 11
3 11 2 -7
(4) (0) 2 (11) (3) (3)
4 6 8 11
2 2 10 9 -5
(2) (1) 1 2 4 2
dj 4 4 6 2 4 2
vj 9 12 16 7 7 15
Dj 1 2 3 4 5 6 si ui
Oi
1 9 4 12 (7) 9 1 6 (6) 9 (9) 10 (10) 5 0

2 7 (0) 3 4 7 2 7 (9) 5 (7) 5 (-1) 6 -2

3 6 (-3) 5 (0) 9 2 11 (11) 3 (3) 11 (3) 2 0

4 6 (-5) 8 (1) 11 1 2 2 2 4 10 2 9 2

dj 4 4 6 2 4 2
vj 9 5 9 0 270
0 8
 Transportation problem

Dj 1 2 3 4 5 6 si ui
Oi
1 9 3 12 (7) 9 2 6 (1) 9 (4) 10 (-3) 5 0

2 7 (0) 3 4 7 2 7 (4) 5 (2) 5 (-6) 6 -2

3 6 (-3) 5 (0) 9 2 11 (6) 3 (-2) 11 (-2) 2 0

4 6 1 8 (6) 11 (5) 2 2 2 4 10 2 9 -3

dj 4 4 6 2 4 2
vj 9 5 9 5 5 13

Dj 1 2 3 4 5 6 si ui
Oi
1 9 1 12 (7) 9 4 6 (1) 9 (4) 10 (3) 5 0

2 7 (0) 3 4 7 0 7 (4) 5 (2) 5 2 6 -2

3 6 (-3) 5 (0) 9 2 11 (6) 3 (-2) 11 (4) 2 0

4 6 3 8 (6) 11 (5) 2 2 2 4 10 (6) 9 -3

dj 4 4 6 2 4 2
vj 9 5 9 5 5 7
Dj 1 2 3 4 5 6 si ui
Oi
1 9 (8) 12 (7) 9 5 6 (4) 9 (7) 10 (3) 5 0

2 7 (3) 3 4 7 0 7 (7) 5 (5) 5 2 6 -2

3 6 1 5 (0) 9 1 11 (9) 3 (1) 11 (4) 2 0

4 6 3 8 (3) 11 (2) 2 2 2 4 10 (3) 9 0

dj 4 4 6 2 4 2
vj 6 5 9 2 2 7

271
 S. M. Shahidul Islam
Since in tableau (5)  ij  cij  ui  v j  0 in non-basic cell, the table is optimal.
*
Therefore, the optimal solution vector x and the associated minimum cost Z * are given
by

Showrooms: 1 2 3 4 5 6
1 0 0 5 0 0 0
 
* 2 0 4 0 0 0 2
x 1 = Factories: 
3 1 0 1 0 0 0
 
4  3 0 0 2 4 0 
Z * = 9  5 + 3  4 + 5  2 + 6  1 + 9  1 + 6  3 + 2  2 + 2  4 = 112
From tableau (5) above, we observe that the cell (3, 2) is non-basic but c32  u3  v2  0 .
This shows that the given transportation problem has multiple optimal solution. One more
integral optimal solution is obtained by making the cell (3, 2) basic in the next iteration.
Constructing a loop through the non-basic cell (3, 2) shown in tableau (5), we then get the
following tableau:
Dj 1 2 3 4 5 6 si ui
Oi
1 9 (3) 12 (7) 9 5 6 (4) 9 (7) 10 (3) 5 0

2 7 (3) 3 3 7 1 7 (7) 5 (5) 5 2 6 -2

3 6 1 5 1 9 (0) 11 (9) 3 (1) 11 (4) 2 0

4 6 3 8 (3) 11 (2) 2 2 2 4 10 (3) 9 0

dj 4 4 6 2 4 2
vj 6 5 9 2 2 7

Hence, the other integral optimal solution to the given problem is

Showrooms: 1 2 3 4 5 6
1 0 0 5 0 0 0
 
* 2 0 3 1 0 0 2
x 1 = Factories: 
3 1 1 0 0 0 0
 
4  3 0 0 2 4 0 
Z * = 9  5 + 3  3 + 7  1 + 5  2 + 6  1 + 5  1 + 6  3 + 2  2 + 2  4 = 112
Remark: In this example, if the variables are not restricted to non-negative integers only,
then the given problem has infinite number of optimal solutions, namely,

272
 Transportation problem
0 0 5 0 0 0
 
0 3   1   0 0 2
x   x1  (1   ) x 2   ,   [0,1]
* * *

1 1   0 0 0
 
3 2 4 0 
 0 0

14.8 When total supply exceeds total demand: In the transportation problem in which
total supply exceeds total demand, we have to create a dummy destination to which the
cost of shipping is zero from each origin. The initial bfs is found by the northwest-corner
rule and the same procedure is applied to the modified problem. In problems with excess
supply, it is clear that not all the items can be shipped and some would remain at some
origin(s). In the optimal solution of the modified problem, the number(s) of items in the
dummy is to be interpreted as those remaining excess at the respective origin(s).

Example: Consider the transportation problem with three origins and four destinations,
where the transportation cost from each origin to each of the destinations, the supply at
each origin, and the demand at each destination are given in the following tableau.

Dj 1 2 3 4 Supply
Oi si
1 15 24 11 12 500
2 25 20 14 16 600
3 12 16 22 13 500
Demand 300 350 350 400
dj
Find the optimal transport system with minimum cost.

Solution: Here, total supply s1 + s2 + s3 = 500 + 600 + 500 = 1600

And total demand = d1 + d2 + d3 + d4 = 300+350 + 350 + 400 = 1400
Total supply = 1600 > 1400 = total demand.
Therefore, we have to introduce a dummy destination to which the cost of transportation
from each of the origins is zero. Then finding the initial bfs by the northwest-corner rule,
we proceed to form the following tableau successively.

Dj 1 2 3 4 Dummy si ui
Oi
1 15 300 24 200 11 (-7) 12 (-8) 0 (-7) 500 0

2 25 (14) 20 150 14 350 16 100 0 (-3) 600 -4

3 12 (4) 16 (-1) 22 (11) 13 300 0 200 500 -7

dj 300 350 350 400 273200

vj 15 24 18 20 7
 S. M. Shahidul Islam

Dj 1 2 3 4 Dummy si ui
Oi
1 15 300 24 100 11 (-7) 12 100 0 (1) 500 0

2 25 (14) 20 250 14 350 16 (8) 0 (5) 600 -4

3 12 (-4) 16 (-9) 22 (3) 13 300 0 200 500 1

dj 300 350 350 400 200

vj 15 24 18 12 -1
Dj 1 2 3 4 Dummy si ui
Oi
1 15 300 24 (9) 11 (2) 12 200 0 (1) 500 0

2 25 (5) 20 250 14 350 16 (-1) 0 (-4) 600 5

3 12 (-4) 16 100 22 (12) 13 200 0 200 500 1

dj 300 350 350 400 200

vj
Dj 1 2 3 4 Dummy si ui
Oi
1 15 100 24 (5) 11 (-2) 12 400 0 (-3) 500 0

2 25 (9) 20 250 14 350 16 (3) 0 (-4) 600 1

3 12 200 16 100 22 (12) 13 (4) 0 200 500 -3

dj 300 350 350 400 200

vj 15 19 13 12 3
Dj 1 2 3 4 Dummy si ui
Oi
1 15 100 24 (5) 11 (-2) 12 400 0 (1) 500 0

2 25 (9) 20 150 14 350 16 (3) 0 200 600 1

3 12 200 16 300 22 (12) 13 (4) 0 (4) 500 -3

dj 300 350 350 400 274200

vj 15 19 13 12 -1
 Transportation problem

Dj 1 2 3 4 Dummy si ui
Oi
1 15 (2) 24 (7) 11 100 12 400 0 (3) 500 0

2 25 (9) 20 150 14 250 16 (1) 0 200 600 3

3 12 300 16 200 22 (12) 13 (2) 0 (4) 500 -1

dj 300 350 350 400 200

vj 13 17 11 12 -3
From the final tableau (6), the optimal solution vector and the associated minimum cost
for the given transportation problem are
 0 0 100 400 
 
x   0 150 250 0 
*

 300 200 0 0 

Z * = 11  100 + 12  400 + 20  150 + 14  250 + 12  300 + 16  200 = 19200.
We also note that out of total available supply of 600 at the origin O2, only 400 units are
shipped under the optimal policy of transportation to minimize the total shipping cost, and
200 units remain unshipped at O2.

14.9 Maximization problem: If the problem is to maximize the objective function, so that
the problem is
m n m n
Maximize Z =  cij xij =   (cij ) xij  minimize
i 1 j 1 i 1 j 1
n
Subject to x
j 1
ij  si (i  1, 2, 3, . . ., m) - - - (i)
m

x i 1
ij  d j ( j  1, 2, 3, . . ., n)

xij  0 (i  1, 2, . . ., m; j  1, 2, . . ., n)
then the optimal solution vector may be obtained by following one of the methods given
below:
275
 S. M. Shahidul Islam
1. The method same as that for minimization problem with the single exception in the
optimality criterion: The current bfs is optimal if  ij  cij  u j  v j ≤ 0 for all non-
basic cells (i, j); otherwise, the current bfs is improved in next iterations.
2. The transportation tableau starts with the costs -cjj, and the initial bfs is determined
with the optimality criterion as minimization problem. The current bfs is optimal if
 ij  cij  ui  v j  0 for all non-basic cell (i, j); otherwise the non-basic cell
( i0 , j0 ) with the most negative value of  ij is chosen to become basic in the next
iteration.
Example: A computer production firm has two factories to produce their products and
three markets to sell the products. By selling a computer, the firm earns different amount
of profit from different markets. The following tableau shows the daily supply of each
factory, daily demand of each market and profit (in thousand taka) per computer at each
market.
Dj 1 2 3 Supply
Oi si
1 4 4 9 25
2 3 5 8 20
Demand(dj) 18 16 11
Find the optimal transport system to maximize the total profit satisfying the supply and
demand constraints. [AUB-2001 MBA]
Solution: First Method:
Dj 1 2 3 Supply ui Here,  ij = cij- ui- vj,
Oi si
1 18 7 (2) 25 0
since, 13 > 0 and
4 4 9
maximum of all
2 3 (-2) 5 9 8 11 20 1  ij >0 is 2 in the cell
(1, 3).
Demand 18 16 11 Hence, we have to
dj make the cell (1,3) as
vj 4 4 7 basic cell
Dj 1 2 3 Supply ui
Oi si
1 4 18 4 (-2) 9 7 25 0

2 3 (0) 5 16 8 4 20 -1

Demand 18 16 11
dj
vj 4 6 9

Since  ij = cij – ui – vj  0 for all non-basic cell (i, j), hence, the table is optimal.

276
 Transportation problem
18 0 7 
The optimal solution is x =   ,
*

 0 16 4 
and the total profit is Z * = (4  18 + 9  7 + 5  16 + 8  4) thousand taka
= 247 thousand taka
= 247000 taka
The problem has multiple solution, since  21 = 0 for non-basic cell.

Second Method:
Dj 1 2 3 Supply ui
Oi si Here we use
- cij = ui + vj
1 -4 18 -4 7 -9 (-2) 25 0
∆ij = - cij - ui - vj
2 -3 (2) -5 9 -8 11 20 -1

Demand 18 16 11
dj
vj -4 -4 -7

Dj 1 2 3 Supply ui
Oi si
1 -4 18 4 (2) -9 7 25 0

2 -3 (0) -5 16 -8 4 20 1

Demand 18 16 11
dj
vj -4 -6 -9
Since  ij = cij – ui – vj ≥ 0 for all non-basic cell (i, j), hence the last tableau is optimal. The
optimal solution and the maximum profit are
18 0 7 
x =   ,
*

 0 16 4 

And Z * = [–{(– 4)  18 + (– 4)  7 + (– 5)  16 + (– 8)  4}] thousand taka

= 247 thousand taka
= 247000 taka.

14.10 Exercises:
1. Define transportation problem.
2. When does a degeneracy case arise in a transportation problem?
3. Discuss northwest corner rule to solve transportation problem.
4. How can we understand to have multiple solutions of a transportation problem?
277
 S. M. Shahidul Islam
5. Solve the following transportation problem:
Dj 1 2 3 4 si  0 0 0 11
Oi * 
[Answer: x =  6 3 0 4  , Z * = 479]
1 14 9 18 6 11  0 7 12 0 
2 10 11 7 16 13  
3 25 20 11 34 19
dj 6 10 12 15

6. Solve the following transportation problem:

Dj 1 2 3 4 si
Oi
1 10 5 6 7 25
2 8 2 7 6 25
3 9 3 4 8 50
dj 15 20 30 35
 0 0 0 25   0 0 0 25 
*    
[Answer: x = 15 0 0 10  Or,  0 15 0 0  , Z * = 535]
 0 20 30 0  15 5 30 0 
   
7. Solve the following transportation problem:
Dj 1 2 3 4 si  0 0 0 11
Oi *  
[Answer: x =  6 3 0 4  , Z * = 479]
1 14 9 18 6 11  0 7 12 0 
2 10 11 7 16 13  
3 25 20 11 34 19
dj 6 10 12 15
8. There are three rice mills and four wholesales market places where rice is sold at
Gazipur. Supply of each rice mill, demand of each market and transport cost of per
ton rice from each mill to every market in U.S $ are given in the following table.

Dj 1 2 3 4 Supply
Oi (in ton)
1 $3 $1 $2 $2 6
2 $5 $2 $5 $6 4
3 $6 $4 $8 $8 8
Demand 4 6 4 4
(in ton)

Find the optimal transport system with minimum total transport cost.

278
 Transportation problem
 0 0 2 4
*  
[Answer: x =  0 2 2 0  (in ton), Z * = $66]
 4 4 0 0
 
9. Solve the following three origins and four destinations transportation problem:
Dj 1 2 3 4 si 5 0 0 2 
Oi *  
1 19 30 50 10 7 [Answer: x =  0 2 7 0  , Z * = 743]
2 70 30 40 60 9  0 6 0 12 
 
3 40 8 70 20 18
dj 5 8 7 14
10. Solve the following transportation problem as follows:
Dj 1 2 3 Supply
Oi si  6 0 4
[Answer: x =   , Z * = 25]
*

1 2 3 2 10 0 2 1
2 4 1 3 7
Demand 6 2 5
dj

11. A furniture firm has two factories to produce cots and three markets to sell the
cots. By selling a cot, the firm earns different amount of profit from different
markets. The following tableau shows the daily supply of each factory, daily
demand of each market and profit (in hundred taka) per cot at each market.
Dj 1 2 3 Supply
Oi si
1 2 3 2 5
2 4 1 3 7
Demand 2 6 4
dj
Find the optimal transport system to maximize the total daily profit satisfying the
supply and demand constraints.
0 5 0
[Answer: x =   , Z * = 3600 taka]
*

 2 1 4

279
 S. M. Shahidul Islam

15
Chapter
Assignment Problem
Highlights:

15.1 Introduction 15.5 The dual of the assignment

15.2 Assignment problem problem
15.3 Algorithm of the Hungarian 15.6 Some worked out example
method 15.7 Exercise
15.4 Justification of Hungarian
method,

15.1 Introduction: A special case of the transportation model is the assignment problem.
This problem is appropriate in a situation, which involves the assignment of resources to
tasks with minimum cost or maximum profit (e.g., assign n persons to n different tasks or
jobs). Also in production and human resource management, it is very necessary to study
assignment problem.

15.2 Assignment problem: There are n jobs, which must be performed, and n persons
available with cij being the cost (for example, the training cost) if the i-th person (i = 1, 2,
3, . . ., n) is assigned to the j-th job (j = 1, 2, 3, . . ., n). The problem is to assign the n
persons to n jobs on a one-one basis (so that, for example, if the i-th person is assigned to
the j-th job then it is unavailable for each of the remaining persons still left unassigned)
such that the total cost is minimized. Let
1, if person i is assigned to job j
xij = 
0, otherwise
Then the linear programming (LP) formulation of the assignment problem is
n n
Minimize: Z =  c
i 1 j 1
ij xij
n
Subject to x
j 1
ij  1 (i  1, 2, 3, . . ., n)
n

x
i 1
ij  1 ( j  1, 2, 3, . . ., n)

280
 Assignment Problem
xij  0 for i, j = 1, 2, 3, . . ., n
From this formulation, we see that the assignment problem is a particular case of the
transportation problem.

15.3 Algorithm of the Hungarian method: The Hungarian method works solely on the
cost matrix C = (cij). The method would be justified later.
Start with the cost matrix
 c11 c12 c13 . . . c1n 
 
 c 21 c 22 c 23 . . . c 2 n 
C = (cij) =  c31 c32 c33 . . . c3n 
 
 ... ... ... 
 
 c n1 c n 2 c n 3 . . . c nn 
Step-1: (Initialization): Make the transformations cij/  cij  min cij , i = 1, 2, 3, . . ., n;
j
//
ij
/
ij
i
 
c  c  min c , j  1, 2, 3, . . ., n . That is, for each row i, find the smallest element and
/
ij

subtract it from each element in that row, thereby getting the new cost matrix C/ = cij/ .  
/
 
/
Next, for each column j of the matrix C = c , find the smallest element and then
ij

subtract it from each element of that column to get the modified cost matrix C // = cij// .  
The cost matrix C// will have at least one zero in each row and each column.
GO TO step-2.
Step-2: (Optimality Criterion): In any modified cost matrix, determine whether there exists
a feasible assignment involving only zero costs, that is, determine whether the modified
cost matrix has n zero entries no two of which lie in the same column. If such an
assignment exists, it is optimal for the given problem. These zeros are boxed in the final
table.

Step-3: (Iterative Step): Cover all zeros in the modified cost matrix with as few horizontal
and vertical lines as possible, each horizontal line covering the zeros of a row and each
vertical line covering the zeros of a column. The total number of lines in this minimal
covering would, of course, be less than n (this number smallest number in the cost matrix
not covered by a line. Subtract this number from every element (of the cost matrix) not
covered by a line and add it to every element covered by two lines (both horizontal and
vertical lines), or equivalently, the minimum cost element is to be subtract from each
element of an uncovered row (including the element of that row covered by a vertical line,
if any) and then added to every covered column.
RETURN TO step-2.

281
 S. M. Shahidul Islam
After a finite number of iterations, the number of minimal covering lines would be exactly
equal to n, the order of the cost matrix C, which would give the optimal assignment
corresponding to the zero costs of the modified cost matrix such that no two (or more)
zeros would be in the same row or column.

Remark-1: If the objective is to maximize the total profit given by some profit matrix
C = (cij) (where cij is now interpreted as the profit obtained when the i-th person is
assigned to the j-th job; i, j = 1, 2, 3, . . ., n) then the above procedure may be applied to
the matrix – C = (– cij) to get the optimal assignment. Alternatively, we may step Step-1 of
the above algorithm as follows:
We make the transformations
cij/  max{cij }  cij , cij//  cij/  min{cij/ }, cij///  cij//  min{cij// }; i, j = 1, 2, 3, . . ., n;
i, j j i

that is, we subtract each element of the profit matrix C = (cij) from the largest element of
C. From the resulting matrix C / we get the matrix C /// by first subtracting from each row
of the matrix C / the smallest element of that row, thereby getting the matrix C // , and then
subtracting from each column of C // the smallest element of that column. Starting with
the modified matrix C /// , we now follow the procedure of Steps-2 and Step-3 of the
original algorithm.

Remark-2: In the case of multiple solutions, there will be more than n number zeros in the
optimal cost matrix. This is the necessary condition but not the sufficient condition of
multiple solutions.

15.4 The dual of the assignment problem: The LP form of the assignment problem is
n n
Minimize: Z =  c
i 1 j 1
ij xij
n
Subject to x
j 1
ij  1 (i  1, 2, 3, . . ., n) (Primal)
n

x
i 1
ij  1 ( j  1, 2, 3, . . ., n)

xij  0 for i, j = 1, 2, 3, . . ., n
The dual of the assignment problem is
n n
Maximize: z =  ui   v j
i 1 j 1
(Dual)

Subject to ui  v j  cij ; i, j  1, 2, 3, ..., n

u i , v j unrestricted in sign for i, j = 1, 2, 3, . . ., n.

282
 Assignment Problem
15.5 Justification of Hungarian method: If x is any feasible solution of the primal
problem, then using the equality constraints of the primal problem, we get
n n n n
Z–z=  c
i 1 j 1
ij xij – u  v
i 1
i
j 1
j

n n
 n  n   n  n 
= 
i 1 j 1
c x
ij ij –   u i   ij 
 x  –   j   xij 
 v 
 i 1  j 1   j 1  i 1 
n n
=  (c
i 1 j 1
ij  u i  v j ) xij

Now, if ( u , v ) is also feasible, from the inequality constraint of the dual, we get
Cij  cij  ui  v j  0 for all i, j = 1, 2, 3, . . ., n.
n n
==>  C
i 1 j 1
ij xij ≥ 0

for all feasible x (satisfying the non-negativity constraint of the primal problem).
Now, for any feasible x and any ( u , v ),
n n n n
Z=  cij xij =
i 1 j 1
 (C
i 1 j 1
ij  u i  v j ) xij ,

so that, x * is optimal for the primal problem if and only if it is optimal for the problem
n n n n
Minimize:  Cij xij =
i 1 j 1
 (c
i 1 j 1
ij  u i  v j ) xij
n
Subject to x
j 1
ij  1 (i  1, 2, 3, . . ., n) (A)
n

x
i 1
ij  1 ( j  1, 2, 3, . . ., n)

xij  0 for i, j = 1, 2, 3, . . ., n
because both problems have same constraints and both objective functions are linear
functions with positive signs and positive coefficients.
Now, let x and ( u , v ) be such that
Cij xij  (cij  ui  v j ) xij  0 for i, j = 1, 2, 3, . . ., n.
If x and ( u , v ) are both feasible, then (since Z – z = 0) they are also optimal for the
primal problem and the dual problem respectively. Therefore, the optimal (minimum)
value of the objective function in (A) is 0 (zero) for the optimal solution x * and ( u * , v * )
of the primal problem and the dual problem respectively with
n n n n

 c
i 1 j 1
*
ij xij = u  v
i 1
*
i
j 1
*
j (B)

283
 S. M. Shahidul Islam
*
Now, the optimal solution x of the primal problem has exactly one 1 in each row and
each column with zeros elsewhere in the remaining components. There, if we subtract u i*
from every element of the i-th row (i = 1, 2, 3, . . ., n) and then subtract v *j from every
element of the j-th column (j = 1, 2, 3, . . ., n), we get and equivalent (with the same
optimal solution as the original primal problem) problem for which the optimal value is 0.
The Hungarian method, starting with the given primal problem with the cost matrix
C = (cij) proceeds to find an equivalent problem with the cost matrix (Cij) which yields the
minimum optimal value 0.
Step-1 of the Hungarian algorithm gives an equivalent problem with the modified cost
matrix with at least one zero in each row and each column. In the modified cost matrix, a
zero-cost assignment; if feasible, gives the minimum value 0, and hence, it must be
optimal. That is, the optimality criterion in Step-2 of the algorithm. Otherwise, the optimal
assignment has not yet been found and we go to Step-3, which is a procedure of
redistributing the zeros and introducing more zeros. Here, we subtract the minimum
uncovered element, say c, from every element of an uncovered row and add c to element
of a covered column. Thus, we get the modified matrix with
(1) at least one more zero entry,
(2) old zeros covered by a single (horizontal/vertical) line being retained,
(3) the rest of old zeros being replaced by c.
c
But these operations are equivalent to subtracting from each uncovered row and each
2
c
uncovered column and then adding to each covered row and each covered column,
2
thereby giving the modified cost matrix of an equivalent problem.
Here, we mention about a simple test, which would indicate whether the zeros in any cost
matrix are well distributed to give the optimal assignment (as in Step-2). We draw the
minimum number of horizontal/vertical lines through row(s) /column(s) so as to cover all
the zeros of the corresponding cost matrix. If this number is equal to the order of the cost
matrix, the optimal solution has been found and is given as follows:
In the final modified cost matrix, we identify the positions of the zeros lying in different
rows and columns (so that no such zeros can lie in the same row or in the same column);
the optimal x * is then found by making the corresponding components 1 and other
components all zeros, or equivalently, we may express the optimal assignment in the
permutation symbol introduced in the above examples.

Remark: We have only considered the n-person n-job problem. The more general
m-person n-job may be treated as follows:
(1) if m > n and we require that all jobs be performed, we introduce m – n dummy jobs
each of which may be done by each person at a cost 0,
284
 Assignment Problem
(2) if m < n and we require all persons to be assigned some jobs, we introduce n – m
dummy persons, each capable of performing every job at zero cost.
We may then apply the Hungarian method to the modified problem.

15.6 Some worked out examples:

Example: Solve the following assignment problem of training cost:
Jobs: 1 2 3 4 5 6
1  9 22 58 11 19 27 
 
2  43 78 72 50 63 48 
3  41 28 91 37 45 33 
Persons:   [AUB-2002 BBA]
4  74 42 27 49 39 32 
 
5  36 11 57 22 25 18 
6  3 56 53 31 17 28 
Solution: We first identify the smallest element in each row and then subtract from each
element of a row the corresponding smallest element. Next, we identify the smallest
element in each column of the modified cost matrix and subtract this number from each
element of that column. We then have
Row min.
 9 22 58 11 19 27  9  0 13 49 2 10 18 
   
 43 78 72 50 63 48  43  0 35 29 7 20 5 
 41 28 91 37 45 33  28 13 0 63 9 17 5 
  ==>  
 74 42 27 49 39 32  27  47 15 0 22 12 5 
 36 11 57 22 25 18   25 0 46 11 14 7 
  11  
 3 56 53 31 17 28  3  0 53 50 28 14 25 
   
Column min. 0 0 0 2 10 5

Subtract:
 0 13 0 0 13 
49
 
 0 35 5 10 0 
29 4
13 0 7 7 0
63 4
=>  
 47 15 20 2 0 
0
 25 0 46 9 4 2 
 4
 0 53 50 26 4 20 
 4
Add: 4 4 4

285
 S. M. Shahidul Islam
We thus get the above modified matrix. The next problem is to find a minimal cover, that
is, the minimum number of horizontal and vertical lines covering all the zeros of the
modified matrix given above. One such minimal cover is shown above. Since the
minimum number of horizontal/vertical lines is 5, which is less than the order of the
original cost matrix (that is, 6), we have to find the minimum of the uncovered elements of
the modified cost matrix. This number is found to be 4. Next, we subtract 4 from all
uncovered elements and add 4 to every twice-covered element. These procedures are done
below.

 4 17 49 0 0 17 
 
 0 35 25 1 6 0 
 13 0 59 3 3 0 
==>  
 51 19 0 20 2 4 
 25 0 42 5 0 2 
 
 0 53 46 22 0 20 
 
Since the number of minimum zero covered lines of the above matrix is 6, equal to the
order of the original matrix, hence this matrix is optimal. From the last matrix above, we
get the following two optimal assignments:

Persons  1 2 3 4 5 6  Persons  1 2 3 4 5 6 
  ;  
Jobs  4 6 2 3 5 1 Jobs  4 1 6 3 2 5
with the minimum cost
Z * = 11 + 48 + 28 + 27 + 25 + 3 = 142 = 11 + 43 + 33 + 27 + 11 + 17.

Example: A production company produced four types of products in four firms. It has
recruited four semi skilled managers as need. They have a little knowledge in different
products. To assign them to their actual posts, company pre-determined their training cost
in taka as follows:
Firms: 1 2 3 4
1  2000 5000 4000 4000 
 
2  0 9000 3000 7000 
Managers:
3  1000 8000 8000 9000 
 
4  9000 4000 1000 6000 
Find their optimal assignment under minimum training cost.[AUB-03MBA(Production
mgt.)]
Solution: The given training cost matrix is to be as follows dividing every element of the
given matrix by 1000:

286
 Assignment Problem
Row min.
 2 5 4 4 2
 
0 9 3 7 0
1 8 8 9 1
 
9 4 1 6 1
 
Subtracting row minimum from every element of that row, we get
0 3 2 2
 
0 9 3 7
==> 
0 7 7 8
 
8 3 0 5
 
Column min. 0 3 0 2
Subtracting column minimum from every element of that column, we get
0 0 2 0
 
0 6 3 5
0 4 4 6
 
 8 0 0 3
 
Since, we can cover all zeros by 3 horizontal/vertical lines, which is less than 4, the order
of the cost matrix; so, the above matrix is not optimal. Here, minimum uncovered element
is 3. Subtracting minimum uncovered element 3 from every uncovered elements and
adding to all twice-covered elements, we get
 3 0 2 0
 
 0 3 0 2
 0 1 4 3
 
11 0 0 3 
 
This is the optimal matrix because there is no way to cover all zeros by less than 4
horizontal/vertical lines. This optimal matrix gives the following optimal assignment:
Managers :  1 2 3 4 
 
Firms :  4 3 1 2
And the minimum training cost = Tk. (4 + 3 + 1 + 4)1000 = Tk. 12000

Example: Five wagons are available at five stations, which are required at five other
stations. The distances from each of the first set of stations to each of the second set are
given below in miles:

287
 S. M. Shahidul Islam
Stations: 1 2 3 4 5
1 10 5 9 18 11 
 
2 13 19 6 12 14 
Stations: 3 3 2 4 4 5
 
4 18 9 12 17 15 
5  11 6 14 19 10 
How should the wagons be transported so as to minimize the total mileage covered?
Solution: The above problem may be viewed as an assignment problem where each
wagon at the first set of five stations is to be assigned to one and only one station of the
second set with the objective of minimizing total miles covered by them. Now, subtracting
from every element of any row the corresponding row minimum number and then
subtracting from every element of any column of the modified matrix the corresponding
column minimum number, and finally, finding the minimal cover (by covering all the
zeros of the final modified matrix by the minimum number of horizontal/vertical lines
each passing through a row/column), we get
Row min.
10 5 9 18 11  5  5 0 4 13 6 
   
13 19 6 12 14  6  7 13 0 6 8 
3 2 4 4 5 2 ==>  1 0 2 2 3 
   
18 9 12 17 15  9 9 0 3 8 6
 11 6 14 19 10  6  5 0 8 13 4 
   
Column min. 1 0 0 2 3

 4 0 4 11 3 
 
 6 13 0 4 5 
==>  0 0 2 0 0 
 
 8 0 3 6 3
 4 0 8 11 1 
 
From the last tableau, we see that the minimum number of horizontal and vertical lines
covering all the zero entries is 3, that is, less than 5, the order of the matrix. The minimum
uncovered element is 1. In the next iteration, we subtract 1 from every uncovered element
and add 1 to every twice-covered element. We then get

288
 Assignment Problem
 3 0 4 10 2 
 
 5 13 0 3 4 
0 1 3 0 0
 
7 0 3 5 2
 3 0 8 10 0 
 
The minimum uncovered number is now 3, and we subtract 3 from every uncovered
element and add 3 to every twice-covered element, getting

 0 0 4 7 2
 
 2 13 0 0 4 
 0 4 7 0 3
 
 4 0 3 2 2
0 0 8 7 0
 
Since the number of minimum zero covered lines of the above matrix is 5, equal to the
order of the original matrix, hence this matrix is optimal. From the last matrix above, we
get the following optimal solution to the problem:
Stations in first set : 1 2 3 4 5 
 
Stations in sec ond set : 1 3 4 2 5 
That is, the wagons at the 1st, 2nd, 3rd, 4th and 5th stations are to be transported respectively
to the 1st, 3rd, 4th, 2nd and 5th stations of the second set so as to minimize the total mileage
covered which is
10 + 6 + 4 + 9 + 10 = 39 miles.

Example: Four salesmen are to be assigned to four districts, one person exactly in one
district. Estimates of sales revenue in U.S $ for each salesman are as follows:
Districts: 1 2 3 4
1  320 350 400 280 
 
2  400 250 300 220 
Salesmen:
3  420 270 340 300 
 
4  25 390 410 350 
Find the optimal assignment of each of the four salesmen to one of the four districts that
maximizes total sales revenue. [AUB-2002 MBA, 2003 BBA]

Solution: First Method:

We start with the following matrix:

289
 S. M. Shahidul Islam
  320  350  400  280 
 
  400  250  300  220 
  420  270  340  300 
 
  250  390  410  350 
 
Subtracting from every element of any row the corresponding row minimum number, and
then subtracting from every element of any column the corresponding column minimum
number, we get successively
 80 50 0 120   80 30 0 60 
   
 0 150 100 180   0 130 100 120 
 0 150 80 120  ==>  0 130 80 60 
   
160 20  160 0 
 0 60   0 0 
In the last matrix, the minimum uncovered element is 60. We now subtract 60 from every
uncovered element and add 60 to every twice-covered element, getting
 140 30 0 60 
 
 0 70 40 60 
 0 70 20 0 
 
 220 0 0 0 
 
The last tableau gives the optimal solution:
Salesmen :  1 2 3 4 
 
Districts :  3 1 4 2 
with the maximum total revenue: $(400+400+300+390) = $1490.

Second Method:
Since the largest element in the original 4  4 matrix is 420, subtracting each element of
the original matrix from 420, we get the first matrix of the following two, and the second
is obtained by subtracting from each row the corresponding row minimum number.
100 70 20 140   80 50 0 120 
   
 20 170 120 200   0 150 10 180 
 0 150 80 120  ==>  0 150 80 120 
   
170 30 10 70  160 20 0 60 
   
Subtracting each column minimum from corresponding column, we get

290
 Assignment Problem
 80 30 0 60 
 
 0 130 100 120 
 0 130 80 60 
 
160 0 
 0 0 
In the last matrix, the minimum uncovered element is 60. We now subtract 60 from every
uncovered element and add 60 to every twice-covered element, getting

 140 30 0 60 
 
 0 70 40 60 
 0 70 20 0 
 
 220 0 0 0 
 
The last tableau gives the optimal solution:
Salesmen :  1 2 3 4 
 
Districts :  3 1 4 2 
with the maximum total revenue: $(400+400+300+390) = $1490.

Remark: If the objective function of the above example is multiplied by some constant α
>0, then the modified problem has the same optimal solution as that example. This
observation allows us to start with the cost matrix
 32 35 40 28 
 
 40 25 30 22 
 42 27 34 30  with smaller entries in above example.
 
 25 39 41 35 
 

Example: A batch of four jobs can be assigned to five different machines. The set-up time
of each job on each machine is given in the following table
Machines: 1 2 3 4 5
1 10 11 4 2 8 
 
2  7 11 10 14 12 
Jobs:
3  5 6 9 12 14 
 
4 13 15 11 10 7 
Find an optimal assignment of jobs to machines, which will minimize the total set-up time.
Solution: Here, the number of jobs is 4 while the number of (and the machine to which is
dummy is assigned under the optimal policy would remain idle for the problem under
consideration). Then the matrix corresponding to the modified problem is

291
 S. M. Shahidul Islam
10 11 4 2 8 
 
 7 11 10 14 12 
 5 6 9 12 14 
 
13 12 11 10 7 
0 0 0 0 0
 
We now subtract from each row the corresponding row minimum and then we subtract
from each element of any column the corresponding column-minimum (which is 0 for
each column for the above matrix). Next, we cover all the zeros of the resulting matrix by
the minimum number of horizontal/vertical lines, as shown in the second matrix below.
Since we can cover all the zeros by 3 such lines, we now subtract the minimum of the
uncovered numbers, that is, 1, from every element in every uncovered row (including
those elements of that row covered by a vertical line only) and add 1 to every covered
column (or, equivalently, we subtract 1 from every uncovered element adding 1 to every
twice-covered element).
Subtract
 10 11 4 2 8  2 10 11 4 2 8 
   
 7 11 10 14 12  7  7 11 10 14 12 
 5 6 9 12 14  5 ==>  5 6 9 12 14 
   
13 15 11 10 7  7 13 15 11 10 7 
0 0 0 0 0 0 0 0 0 0 0
   
21 Subtract: 0 0 0 0 0 0

Subtract
8 9 2 0 6 1
 
0 4 3 7 5 1
0 1 4 7 9 1
 
6 8 4 3 0 1
0 0 
 0 0 0
Add: 1 1 1 1

Finally, we get

292
 Assignment Problem
8 8 1 0 6
 
0 3 2 7 5
0 0 3 7 9
 
6 7 3 3 0
1 0 0 1 1
 
From the above matrix, we see that the machine 3 (to which is assigned the dummy job)
would remain idle, and the optimal solution is
 0 0 0 1 0
 
1 0 0 0 0
1 2 3 4  : Jobs
x * =  0 1 0 0 0  or, symbolically  
  4 1 2 5  : Machines
0 0 0 0 1
 0 0 1 0 0
 
with the minimum set-up time: 2 + 7 + 6 + 7 = 22.
We note that
5 5

 ui*   v *j = 21 + 1 = 22 = total minimum set-up cost,

i 1 j 1

(More precisely, u1*  2  1  3 , u 2*  7  1  8 , u3*  5  1  6 , u 4*  7  1  8 , u5*  0 ,

v1*  0  1  1, v2*  0 , v3*  0 , v4*  0  1  1 , v5*  0  1  1
5 5
  u i*   v *j = 3 + 8 + 6 + 8 + 0 –1 + 0 + 0 – 1 – 1 = 22
i 1 j 1

which satisfies the result of (B) in the previous theorem).

15.7 Exercise:
1. Define assignment problem.
2. Discuss Hungarian algorithm to solve assignment problem.
3. Is Hungarian algorithm accurate to solve assignment problem? And discuss your
answer.
Or, discuss the justification of Hungarian algorithm.
4. Solve the following assignment problems with cost matrices by the Hungarian
method:
 2 5 4 4  3 2 4 2  14 13 17 14 
     
0 9 3 7 5 2 4 8  16 15 16 15 
(i)  (ii)  (iii) 
1 8 8 9 6 1 3 5 18 14 20 17 
     
9 4 1 6 1 2 8 9  20 13 15 18 
     

293
 S. M. Shahidul Islam
2 2 5 7 4 2
1 5 5 2 3  
  1 2 5 7 3 2
1 5 3 1 4 6 4 4 2 1 3
(iv)  4 1 0 2 0 (v)  
  3 5 8 7 1 2
1 4 5 3 3 4 5 3 2 6
2   4 
 5 4 6 6 3 2 2 7 4
 3 
1 2 3 4 *  1 2 3 4 * 1 2 3 4  *
[Answer: (i)   , Z =12 (ii)   , Z = 8 (iii)   , Z = 58
 4 3 1 2  4 2 3 1   1 4 2 3 
1 2 3 4 5   1 2 3 4 5  *  1 2 3 4 5 6  *
(iv)   or   Z = 10 (v)   , Z = 10]
1 4 5 2 3   5 4 3 2 1   2 1 5 6 4 3 
5. The director of data processing for a consulting firm wants to assign four
programming tasks to four of her programmers. She has estimated the total number
of days each programmer would take if assigned each of the programs as follows:
Tasks: 1 2 3 4
1  80 200 150 170 
 
2  150 160 120 100 
Programmers:
3  220 190 160 300 
 
4  250 150 120 90 
Determine the optimal assignment of programmers to programming tasks if the
objective is to minimize the total number of days to complete all four tasks.
Pr ogrammers : 1 2 3 4 
[Answer:   , with minimum number of days = 480]
Tasks : 1 3 2 4 
6. A production company has six production firms in six different countries. It has
recruited six high level officers from many countries for these firms. To assign
them to their actual firms, company pre-determine the total of their adaptability
training cost and production system training cost in taka as follows:
Firms
1 2 3 4 5 6
1  10000 25000 12000 20000 18000 15000 
 
2  15000 20000 13000 18000 25000 20000 
3  25000 15000 10000 19000 30000 21000 
Officers:  
4  7000 35000 14000 17000 10000 18000 
 
5  17000 30000 9000 25000 18000 15000 
6  9000 10000 20000 14000 20000 10000 

294
 Assignment Problem
Determine the optimal assignment of officers to different firms if the objective is
to minimize the total training cost.
Officers : 1 2 3 4 5 6 
[Answer:   and minimum training cost = Tk. 72,000]
Firms : 1 4 2 5 3 6 
7. A production company has six production firms in six different countries. It has
recruited six high level officers from many countries for these firms. To assign
them to their actual firms, company pre-determine the daily profit in U.S $ earned
by them as follows:
Firms
1 2 3 4 5 6
1  10000 25000 12000 20000 18000 15000 
 
2  15000 20000 13000 18000 25000 20000 
3  25000 15000 10000 19000 30000 21000 
Officers:  
4  7000 35000 14000 17000 10000 18000 
 
5  17000 30000 9000 25000 18000 15000 
6  9000 10000 20000 14000 20000 10000 
Determine the optimal assignment of officers to different firms if the objective is
to maximize the total profit.
Officers :  1 2 3 4 5 6 
[Answer:   and maximum profit = $1,45,000]
Firms :  6 5 1 2 4 3 
8. Introducing the appropriate dummies, solve the following assignment problems of
cost matrices.
 65 73 63 57 
 
 67 70 65 58 
18 24 28 32   68 72 69 55 
 
(i) 10 15 19 22  (ii)  
 8 13 17 19   67 75 70 59 
   71 69 75 57 
 
 69 71 66 59 
 
1 2 3  1 2 3  1 2 3 5  *
[Answer: (i)   Or,   , Z * = 50 (ii)   , Z = 254]
1 2 3  1 3 2  1 3 4 2 

295
 Annexure – 1

Trigonometry
Highlights:
A1.1 Introduction A1.5 Inverse trigonometric ratios
A1.2 Trigonometric ratios A1.6 Limit
A1.3 Fundamental relations A1.7 Differentiation
A1.4 Trigonometric ratios of some standard A1.8 Integration
angles A1.9 Exercise

A1.1 Introduction: Trigonometry is the branch of Mathematics which deals with the
measurement of angles. It is the most powerful tool of mathematics. But till now it has no
mentionable application in business section. We attached this section in this book for more
interested students.

A1.2 Trigonometric ratios: The basic measurement of the trigonometric ratios from a
right angled triangle is as follows: A
perpendicu lar hypotenuse Hypotenuse
sin θ = , cosec θ =
hypotenuse perpendicu lar Perpendicular
base hypotenuse
cos θ = , sec θ = θ B
hypotenuse base C Base
perpendicu lar base
tan θ = , cot θ = Figure A1.1
base perpendicu lar
From the above measurement of the trigonometric ratios, we get
1 1 1 sin  cos 
cosec θ = , sec θ = , cot θ = , tan θ = and cot θ = .
sin  cos  tan  cos  sin 
Note: Since hypotenuse always greater than or equal to perpendicular and base,
sin θ ≤ 1, cos θ ≤ 1, cosec θ ≥ 1 and sec θ ≥ 1.

A1.3 Fundamental relations:

(i) sin2 θ + cos2 θ = 1, (ii) sec2 θ + tan2 θ = 1, (iii) cosec2 θ + cot2 θ = 1
(iv) sin(-θ) = - sin θ, (v) cos(-θ) = cos θ, (vi) tan(-θ) = - tan θ
For multiple angles:
(a) sin(A + B) = sinA cosB + cosA sinB, (b) sin(A – B) = sinA cosB – cosA sinB
(c) cos(A + B) = cosA cosB – sinA sinB, (d) cos(A – B) = cosA cosB + sinA sinB
(e) sin 2θ = 2 sin θ. cos θ, (f) cos 2θ = cos2 θ – sin2 θ = 1 – 2 sin2 θ = 2cos2 θ – 1,
CD CD CD CD
(g) sin C + sin D = 2 sin .cos , (h) sinC – sinD = 2 cos .sin
2 2 2 2

i
 S. M. Shahidul Islam

CD CD CD DC

(i) cosC + cosD = 2 cos .cos , (j) cosC – cosD = 2 sin .sin
2 2 2 2
A1.4 Trigonometric ratios of some standard angles:
Angle 00 300 450 600 900
sin θ 0 1 1 3 1
2 2 2
cos θ 1 3 1 1 0
2 2 2
tan θ 0 1 1 3 ∞
3
cosecθ ∞ 2 2 2 1
3
sec θ 1 2
2 2 ∞
3
cot θ ∞ 3 1 1 0
3

Example: If tan θ + sin θ = m and tan θ – sin θ = n, prove that m2 – n2 = 4 mn .

Solution: L.H.S = m2 – n2
= (tan θ + sin θ)2 – (tan θ – sin θ)2
= 4 tan θ. sin θ
= 4 (tan sin  ) 2
= 4 tan 2  (1  cos 2  )
= 4 tan 2   sin 2 
= 4 (tan  sin  )(tan  sin  )
= 4 mn = R.H.S (Proved)
Example: Find the value of A: 2 sin2A – 5 cosA + 1 = 0; A < 900.
Solution: Given that 2 sin2A – 5 cosA + 1 = 0
Or, 2 (1 – cos2A) – 5 cosA + 1 = 0
Or, 2 – 2cos2A – 5cosA + 1 = 0
Or, 2cos2A + 5cosA – 3 = 0
Or, 2cos2A– cosA + 6cosA – 3 = 0
Or, 2cosA(cosA – 1) + 3(cosA – 1) = 0
Or, (cosA – 1)(2cosA + 3) = 0
Or, cosA – 1 = 0 Or, cosA = 1 = cos 00, So, A = 00.
3
And 2cosA + 3 = 0 Or, cosA = – [Not acceptable]
2
Therefore, A = 00. (Answer)
A1.5 Inverse trigonometric ratios: If sin θ = x, θ = sin-1 x = sin-1(sin θ). Similarly,
θ = cos-1(cos θ) = tan-1(tan θ). Again if θ = sin-1 x, x = sin θ = sin( sin-1 x). sin-1 x,
cos-1 x, tan-1 x, cosec-1 x, sec-1 x and cot-1 x are known as inverse trigonometric ratios.

ii
 Annexure – 1

A1.6 Limit: In the chapter of limit and continuity, we have seen the fundamental formulae
of limit. To find the limit of trigonometric functions the following formula is very
important.
sin x tan x
Lim  Lim  Lim cos x =1
x 0 x x 0 x x0

The proof of this formula is beyond the scope of this book.

tan x
Example: Evaluate the limit lim .
x 0 2 x

tan x 1  tan x 
Solution: lim = lim  
x 0 2 x x 0 2
 x 
1  tan x 
= lim   [ . . . lim kf ( x)  k lim f ( x) ]
2 x 0  x  x 0 x 0

1 tan x
= .1 [. . . lim  1]
2 x 0 x
1
= (Answer)
2
1  cos x
Example: Evaluate the limit lim .
x  (  x ) 2

Solution: Let x = π + h. So h  0 as x  π.
1  cos x 1  cos(  h)
lim = lim
x  (  x ) 2 h 0 {  (  h)}2

1  cosh
= lim [We know that, cos(π + x) = – cosx]
h 0 h2
2 sin 2 h 2
= lim 2
[. . . 1 – cos2x = 2sin2x]
h0 h
2 sin 2 h 2
= lim
h0 4( h ) 2
2

1  sin h 2 sin h 2 
= lim  . 
2 h 0  h 2 h
2 

1  sin h 2   sin h 2 
= lim    lim   [. . .  0 as h  0]
2 h 2 0 h 2  h 2 0 h 2 
h
2

1
=  1 1
2
1
= (Answer)
2

iii
 S. M. Shahidul Islam

A1.7 Differentiation: We have learnt fundamental theorems on differentiation and

various differentiation techniques in chapter 10. Here we shall learn how to differentiate a
trigonometric function.
Example: Find the differential coefficient of sin x by first principle.
Solution: Here, f(x) = sin x, f(x + h) = sin(x + h)
d f ( x  h)  f ( x )
So, (sin x)  lim
dx h 0 h
sin( x  h)  sin x
= lim
h 0 h
2 sin 2 . cos x  h2 x . .
xh x
= lim [ . sinC – sinD = 2sin D2 C .cos C 2 D ]
h 0 h
2 sin h2 . cos( x  h2 )
= lim
h 0 h
h
sin
= lim h 2  lim cos( x  h2 )
h 0 h 0
2
sin h2
= 1.cos x [. . . lim  1 and lim cos( x  h2 )  cos x ]
h 0 h h 0
2
= cos x (Answer)
In the similar way we find the following results:
d d
(i) (sin x) = cos x (ii) (cos x) = - sin x
dx dx
d d
(iii) (tan x) = sec2 x (iv) (cot x) = - cosec2 x
dx dx
d d
(v) (sec x) = sec x. tan x (vi) (cosec x) = - cosec x. cot x
dx dx
d d
(vii) (sin mx) = m cos mx (viii) (cos mx) = - m sin mx
dx dx
d 1 d 1
(ix) (sin-1 x) = (x) (cos-1 x) = –
dx 1 x2 dx 1 x2
d 1 d 1
(xi) (tan-1 x) = (xii) (cot-1 x) = –
dx 1 x 2
dx 1 x2
d 1 d 1
(xiii) (sec-1 x) = (xiv) (cosec-1 x) = –
dx x x 12 dx x x2 1
The above results are used as formula.

Example: Find the differential coefficient of sin(ax + c).

Solution: Let u = ax + c

iv
 Annexure – 1

d d d
 [sin(ax + c)] = (sin u). (ax + c)
dx du dx
= cos u. a
= a cos u
= a cos(ax + c) (Answer)

dy
Example: Find the of y = log(sin x2) [NU – 01 A/C]
dx
dy d
Solution: = [ log(sin x2)]
dx dx
1 d
= 2
. ( sin x2)
sin x dx
1 d 2
= 2
.cos x2. (x )
sin x dx
cos x 2
= .2x
sin x 2
= 2x cot x 2 (Answer)
x  sin x
Example: Differentiate with respect to x: . [NU-00 A/C]
1  cos x
x  sin x
Solution: Let y =
1  cos x
dy d x  sin x
 = ( )
dx dx 1  cos x
d d
(1  cos x) ( x  sin x)  ( x  sin x) (1  cos x)
= dx dx
(1  cos x) 2
(1  cos x)(1  cos x)  ( x  sin x)( sin x)
=
(1  cos x) 2
1  2 cos x  cos 2 x  x sin x  sin 2 x
=
(1  cos x) 2
1  2 cos x  x sin x  1
= [. . . sin2 x + cos2 x = 1]
(1  cos x) 2

2  2 cos x  x sin x
= (Answer)
(1  cos x) 2

v
 S. M. Shahidul Islam

m sin 1 x 2 d2y dy
Example: If y = e , then show that (1 – x ) 2
–x – m2y = 0 [RU-90]
dx dx
m sin 1 x
Solution: Given that y = e
dy d 1
 = ( e m sin x )
dx dx
dy 1 d
Or, = e m sin x (m sin-1 x)
dx dx
dy 1
Or, = y. m
dx 1 x2
dy
Or, 1 x2 = my
dx
2
2  dy 
Or, (1 – x )   = m2y2 [Doing square]
 dx 
Again differentiating with respect to x, we get
2
d  dy  d
[(1 – x2)   ] = ( m2y2)
dx  dx  dx

2 2
d  dy   dy  d d
Or, (1 – x ) [   ] +   . (1 – x2) = m2 (y2)
2
dx  dx   dx  dx dx

2
2 dy d dy  dy  dy
  (– 2x) = m .2y.
2
Or, (1 – x ).2 . ( ) +
dx dx dx  dx  dx

2
dy d 2 y  dy  2 dy
Or, 2(1 – x2) . – 2x   = 2m y
dx dx 2  dx  dx

d2y dy dy
Or, (1 – x2) 2
–x = m2y [Dividing by 2 ]
dx dx dx
d2y dy
So, (1 – x2) 2
–x – m2 y = 0 (Proved)
dx dx
A1.8 Integration: We know that the integration is the anti-differentiation. Here we shall
learn how to integrate a trigonometric function.
d
Example: We know that, (sin x) = cos x
dx
So, ∫cos x dx = sinx + c; c is the integral constant.

vi
 Annexure – 1

In the similar way we find the following results:

(i) ∫sin x dx = - cosx + c (ii) ∫cos x dx = sin x + c
1 1
(iii) ∫sin mx dx = - cos mx + c (iv) ∫cos mx dx = sin mx + c
m m
2 2
(v) ∫sec x dx = tan x + c (vi) ∫cosec x dx = - cot x + c
(vii) ∫sec x. tan x dx = sec x + c (viii) ∫cosec x. cot x dx = - cosec x + c
1 1 x 1 x
(ix) ∫ 2 dx = tan-1 + c (x) ∫ dx  sin 1 + c
x a 2
a a a x
2 2 a
1 1 x
(xi) ∫ dx  sec 1 + c (xii) ∫tanx dx = log(sec x) + c
x x2  a2 a a
(xiii) ∫cot x dx = log(sin x) + c
1
Example: Evaluate ∫ dx
1  sin x
1
Solution: Let I = ∫ dx
1  sin x
1  sin x 1  sin x 1  sin x
=∫ dx = ∫ dx = ∫ dx
(1  sin x)(1  sin x) 1  sin x
2
cos 2 x
1 sin x
= ∫( 2
 2
) dx = ∫(sec2 x – sec x. tan x)dx
cos x cos x
= ∫sec2 x dx – ∫sec x. tan x dx = tan x – sec x + c (Answer)
1
Example: Evaluate ∫ dx [NU-97]
1 x 2

1
Solution: Let I = ∫ dx
1 x2
1 x
= sin-1 x + c [ . .. ∫ dx  sin 1 + c
a2  x2 a

2
Example: Evaluate  x sin x dx
0
[RU-90]

d
Solution: Here, ∫x sin x dx = x ∫sin x dx - ∫[ (x). ∫sin x dx]dx
dx
= - x cos x - ∫1. (- cos x) dx
= - x cos x + ∫cos x dx = - x cos x + sinx

2 
  
So,  x sin x dx = sin x  x cos x 02 = (sin
0
2
– cos ) – sin 0 – 0.cos 0
2 2

vii
 S. M. Shahidul Islam


= (1 – .0) – 0 = 1 (Answer)
2

A1.9 Exercise:
1. Prove the following relations:
tan tan 1  cos 
(i)   2 cos ec , (ii)  cos ec  cot 
sec  1 sec  1 1  cos 
1  sin 
(iii)  sec  tan
1  sin 
2. Find the value of A when A< 900:
(i) 2 sin2 A = 3cos A [Answer: 600]
(ii) 5 cosec A – 7cot A cosec A – 2 = 0 [Answer: 600]
2

3. Evaluate the following limits:

tan x 5 tan x sin x cos x
(i) lim (ii) lim (iii) lim x sin x (iv) lim (v) lim 
x 0 7 x x 0 x x  0 x   x x  2 ( 2  x)

1 1
[Answer: (i) 7 , (ii) 5, (iii) 0, (iv) 1, (v) 2 ]
4. Differentiate the following functions:
(i) x4 tan x, (ii) sin x .cos x, (iii) sin 6x, (iv) sin ax. e-bx (v) x3 cos x,
(vi) log(sec x + tan x), (vii) (5x3 + sin2 x)1/4, (viii) sin {ln(1 + x2)}
[Answer: (i) x3(4 tan x + x sec2 x), (ii) cos 2x, (iii) 6 cos 6x,
(iv) e-bx(a cos ax – b sin ax), (v) x2(3 cosx – x sin x), (vi) sec x,
15 x 2  sin 2 x 2 x cos{ln(1  x 2 )}
(vii) , (viii) ]
44 (5 x 2  sin 2 x) 3 1 x2
5. Evaluate the following integrals:
1 1
(i) ∫(sin 5x + cos 3x) dx [Answer:
sin 3x - cos 5x + c]
3 5
2
(ii) ∫(sec x. tan x – 3 cosec x) dx [Answer: sec x + 3 cot x + c]
(iii) ∫sec x(sec x – tan x) dx [Answer: tan x – sec x + c]
6. Evaluate the following integrals:

2

(i)  sin 2 x dx [Answer: ]
0
4

2
1
(ii)  1  sin x dx
0
[Answer: 1]

viii
 Annexure – 1

Bibliography

1. Frank S. Budnick – Applied Mathematics for Business, Economics, and the

Social Sciences
2. Abdullah Al Kafi Majumdar & Sayed Sabbir Ahmed – Lecture notes
3. Earl K. Bowen, Gordon D. Prichett & John C. Saber – Mathematics with
Applications in Management and Economics
4. Margaret L. Lial, Charles D. Miller & Raymond N. Greenwell – Finite
Mathematics and Calculus with Applications
5. M.A. Taher – Business Mathematics
6. Seymour Lipschutz – Linear Algebra
7. Dipok Kumar Bishawas & Meer Sajjad Ali – Business Mathematics
8. Professor Md. Abdur Rahaman – College Linear Algebra
9. Dr. S. M. Mahfuzur Rahman – Banijjik Gonit
10. B. C. Das & B. N. Mukerjee – Differential Calculus
11. B. C. Das & B. N. Mukerjee – Integral Calculus
12. D. C. Sancheti & V. K. Kapoor – Business Mathematics
13. Md. Abdur Rahman – College Higher Algebra with Trigonometry & Set
Therory
14. Richard B. Chase, Nicholas J. Aquilano & F. Robert Jacobs – Production and
Operations Management: Manufacturing and Services
15. Lee J. Krajewski & Larry P. Ritzman – Operations Management : Strategy and
Analysis

ix
View publication stats