CLARE GATHOGA

KENYATTA UNIVERSITY
INSTITUTE OF OPEN DISTANCE & e-LEARNING
IN COLLABORATION WITH

SCHOOL OF ECONOMICS
DEPARTMENT: ECONOMETRICS AND STATISTICS

UNIT CODE: EES 300

MATHEMATICS FOR ECONOMISTS III

WRITTEN BY: DR.AFLONIA MBUTHIA EDITED BY: DR. DIANAH MUCHAI

Copyright © Kenyatta University, 2009
All Rights Reserved
Published By:
KENYATTA UNIVERSITY PRESS
INTRODUCTION

This course is an advanced course in mathematics for economics. A student

should have learnt mathematics for economics two for her/him to be able to understand this
course. The course introduces the student to more optimization techniques, both for
constrained and unconstrained optimization problems.

OBJECTIVES

The learner should be able to:

1. Formulate linear programming problems in a standard way.
2. Solve linear programming problems using the graphical and simplex
method.
3. Formulate the dual of a linear programming problem.
4. Test linear dependence using the Jacobian Determinant
5. Solve simultaneous equations using Gaussian and Gauss Jordan
elimination methods.
6. Test the convexity and concavity of a non-linear function using Hessian
matrices.
7. State the Kuhn Tucker conditions for a maximization and minimization
problem.
8. Solve a non-linear Programming problem.
9. Integrate logarithmic and non-algebraic functions.
10. Integrate functions using Integration by parts
11. Get solutions to First order differential equations used in Economics
12. Get solutions to First order difference equations used in Economics
13. Determine stability of economic variables.
TABLE OF CONTENTS

Lecture One: Introduction to linear programming.

 Formulation of LPP
Lecture Two: Graphical method of solving LP problems
Lecture Three: Simplex method for maximization problems, Duality in LPP
Lecture Four: Tests of linear dependence using the Jacobian Determinant.
 Solution of simultaneous equations using Gaussian and Gauss
Jordan elimination methods.
Lecture Five: Tests of convexity and concavity using Hessian matrices and
determinants
Lecture six: Non-linear Programming
 Kuhn Tucker problem for maximisation
 Kuhn Tucker problem for minimization
 Economic applications of non-linear programming
Lecture Seven: Integration of logarithmic and non-algebraic
functions,Integration by parts
 Economic applications of integration of non-algebraic functions.
Lecture Eight: First order differential equations
 FODE with constant coefficient and term.
 FODE with variable coefficient and term
 Exact differential equations
 Economic applications of FODE
Lecture Nine: First order difference equations
 Homogenous and non-homogenous difference equations.
 Solution to FO difference equations
Lecture Ten: Stability of an equilibrium.
 Economic applications in cobweb model and market
model.
LECTURE ONE: LINEAR PROGRAMMING

1.1 INTRODUCTION

This lecture introduces the student to the method of formulating economic problems with
linear objective functions and also linear constraints/limitations. Standard steps of
formulating an LPP will be introduced.

1.2 LECTURE OBJECTIVES

By the end of this lecture the learner should be:

1. Able to define key terms used in a linear programming problem.

2. Able to formulate a linear programming problem in the standard way.

 1.3 Introduction to linear programming

• In most business organizations, allocation of limited resources to unlimited needs is a

major problem. Resources have to be allocated to the alternative that makes an economic

agent most well off.

• The management of a firm makes many allocations decisions within any given time.

• Linear programming is a mathematical technique concerned with allocation of scarce

resources. It involves optimization of an objective function subject to one or several

constraints faced by the economic agent.

• Optimization could either be maximization or minimization problem.

For a problem to be solved using linear programming technique, a number of conditions must

hold.

 Problem can be stated in numeric terms.

 All factors involved must have linear relationships

 Problem must permit a choice/s between alternative courses of action

 There must be one/more constraints involved

Definition of key terms

Programming – process of getting optimum values.

Linearity- In all relevant relationship, there are linear relationships between the

variables. For example, in production kind of problems, we have constant input prices,

output prices and we have constant returns to scale, constant profit per unit of output sold
 and factors of production are used in fixed ratios. Although linearity may not always

apply, the relationship can be linear within a given range

1.4 Expressing Linear programming problems in a standard manner

We need to express all linear programming problems in a standardized manner. This

helps in calculations required in getting solutions. In addition, it ensures that no important

element is overlooked.

Steps involved

 Decide on the objective of firm

 Decide on decision variables

 Formulate the objective function

 Formulate the constraints

 Set up the non-negative constraints

 Solve by graphical or simplex method

i) Objective

What results are required/what do we want to achieve. e.g. maximise profit,

minimize cost/time or other. Once the objective is decided upon, the elements in

achieving it are stated mathematically.

ii) Decision variables

In order to achieve the objective, the decision maker has to decide on something.

For example, number of units of a product to produce.

iii) Objective function – This is the function to be maximized or minimized.

Example1

A firm produces 2 goods, x and y. The profit per unit of x = 100/= and y is 300/=.

The firm wants to maximize the profits.

 Objective: maximize profits

 Decision variables

x = number of units x to produce

y = number of units y to produce

 Objective function: Maximize Profits = 100x + 300y

iv) Constraints/limitations

These are the factors that could hinder the decision maker from achieving the

objective set. These are conditions that need to be met when achieving the

objective.For example, as a student, the objective is to maximise the marks

obtained in a given course. But this objective is limited by the time available for

reading, resources available to buy relevant books among others.

The limitations in any problem must be clearly identified, quantified and

expressed mathematically.

If a problem is concerned with scarce resources, constraints take the form of:

Resources used≤Resources available

In a profit maximization problem, the firm would want to maximize the profits by

producing as much as possible. However, the amount of raw materials and

machine hours available in the firm could act as limitations. A consumer would

want to maximize the utility derived from consuming two products, but the

income/budget constraints him/her.

v) Non-negative constraints

Most of the decision variables in LP problems cannot take negative values. To

ensure that decision varibles are non-negative, we have the non-negativity

constraints.

For example, the number of units of a product manufactured should be greater or

equal to zero but cannot be negative. i.e. X≥0

vi) Solve using either graphical or simplex method.

1.5 Examples of Formulating linear programming problems in a standard way:

We will look at several examples that will assist us to formulate the linear

programming problems in a standard way.

PROBLEM 1.1

A firm makes two types of furniture: chairs and tables. The contribution for each product

as calculated by the accounting department is Kshs. 2000 per chair and Kshs. 3000 per

table. Both products are processed on three machines M 1 , M2 and M3. The time

required by each product and total time available per week on each machine is as

follows:

Machine Chair Table Available Hours

M1 3 3 36

M2 5 2 50

M3 2 6 60

Formulate the linear programming problem in the standard way given that the

manufacturer wants to maximize contribution.

SOLUTION

Let x1 = Number of chairs to be produced

X2 = Number of tables to be produced

Since the objective is to maximize the profit, the objective function is given by:

Maximise Z= 20x1+ 30x2

Subject to constraints:

3x1 + 3x2 ≤ 36 ( Total time of machine M 1)

5x1 + 2x2 ≤ 50 ( Total time of machine M2)

2x1 + 6x2 ≤ 60 ( Total time of machine M 3)

X1 , X2≥ 0 [ Non-Negativity constraint)

PROBLEM 1.2

The ABC manufacturing company can make two products P1 and P2 . Each of the

products requires time on a cutting machine and a finishing machine. Relevant data are:

Products

P1 p2

Cutting Hours ( per unit) 2 1

Finishing Hours ( per Unit) 3 3

Profit ( $ per week) 6 4

Maximum Sales ( Unit per week) 200

The number of cutting hours available per week is 390 and the number of finishing hours

available per week is 810. Formulate the linear programming problem in the standard

way given that the manufacturer wants to maximize profits.

 SOLUTION

Let X1 =Number of product p1 to be produced

X2 = Number of product p2 to be produced

Since the objective is to maximize profit, the objective function is given by:

Maximise Z = 6x1 +4x2

Subject to constraints

2x1 +x2 ≤ 390 ( Availability of cutting hours)

3x1 +3x1 ≤ 810 (Availability of finishing hours)

X2 ≤ 200 ( Maximum sales)

X1 , X2≥ 0 ( Non-Negativity constraint)

• PROBLEM 1.3

• An animal feed company must produce 100 kg of a mixture consisting of ingredients X1 ,

and X2 daily. X1 cost Kshs. 3 per kg and X2 Kshs. 8 per kg. Not more than 80 kg of X1

can be used, and at least 60kg of X2 must be used. Formulate the problem in a standard

way to enable the company minimize cost.

Solution

Let X1 = Number of kg of ingredient X1 to be used

X2 = Number of kg of ingredient X2 to be used

Since the objective is to minimize the cost, the objective function is given as:

minimize Z = 3x1 +8x2

Subject to constraints:

X1 +X2 = 100 (Total mixture to be produced)

X1 ≤ 80 ( Maximum use of X1)

 X2≥ 60 (Minimum use of x2)

X1 ≥ 0 (Non-negativity constraint

Note: There is no need for a non-negativity constraint for X2 since it is already supposed

to be greater or equal to 60.

1.6 SUMMARY

In this lecture we have learnt how to formulate minimization and

maximization problems in a standard way. In our next lecture we will look at how

we solve the linear programming problems using the graphical method.

NOTE

Linear programming problems can only be used if all relationships have been

expressed as linear functions

1.7 ACTIVITIES

Activity

As a distant learning student, try to think of the limitations you encounter when

trying to maximize your study time.

1.8 FURTHER READING

References
1. Gupta S.K. Cozzolino J.M. (1975) Fundamentals of Operations Research
for Management
2. Lucey T. Quantitative techniques
3. Monga G.S: Mathematics and statistics for Economists, second revised
edition
4. Tulsian and Pandey V. Quantitative Techniques, Theory and Problems.
1.9 SELF-TEST QUESTIONS 1

• The Tusome Printing Company is facing a tight financial squeeze and is attempting to

cut costs wherever possible. At present it has only one printing contract and, luckily,

the book is selling well in both the hardcover and paperback editions. It has just

received a request to print more copies of this book in either the hardcover or

paperback form. Printing cost for hardcover books is Kshs. 700 per 100 while

printing cost for paperback is only Kshs. 600 per 100. Although the company is

attempting to economise, it does not wish to lay off any employees. Therefore, it feels

obliged to run its two printing presses at least 90 and 70 hours per week, respectively.

Press I can produce 100 hardcover books in 2 hours or 100 paperback books in 1 hour,

while Press I can produce 100 hardcover books in 1 hours or 100 paperback books in

2 hours. Formulate the problem in a standard way, in order to minimize cost.

LECTURE TWO: GRAPHICAL SOLUTION TO LINEAR

PROGRAMMING

2.1 INTRODUCTION

This lecture discusses the method of solving linear programming problems (LPP) using the

graphical method.

2.2 LECTURE OBJECTIVES

By the end of this lecture the learner should be able to solve maximization and minimization
linear programming problem using the graphical method
 2.3 Graphical solution to linear programming.

In our lecture one, we were able to see how a LPP can be formulated in a standard way. In this

section we will see how a LPP can be solved using the graphical method.

Graphical methods of solving Linear programming can only be used for problems with TWO

unknowns or decision variables. Problems with THREE OR MORE unknowns must be solved

by techniques such as the simplex method. Graphical methods are the simplest to use and should

be used wherever possible. Graphical method can be used to solve both maximization and

minimization problems. It can deal with limitations of the form ‘greater than or equal to’ type

(≥) , the ‘less than or equal to’ type. (≤), and even constraints of ‘equal to’( =) type.

The following is a step by step procedure for solving LP maximisation problems graphically.

Example 1

A manufacturer produces two products, toothpaste and detergents. Toothpaste has a contribution

of Kshs. 3000 per carton and Detergent Kshs. 4000 per carton. The manufacturer wishes to

establish the weekly production plan which maximizes contribution.

The machine hours, labor hours and raw materials (in kgs) required to produce a carton of

toothpaste and detergent respectively are as follows:

PER UNIT REQUIREMENTS

Machine (Hours) Labour (Hours) Material (kgs)
Toothpaste 4 4 1
Detergent 2 6 1
Total Available 100 180 40
Per week
Because of a trade agreement, sales of Toothpaste are limited to a weekly maximum of 20 units

and to honor an agreement with an old established customer at least 10 units of Detergent must

be sold per week.

Solution

STEP 1. Formulate the linear programming model in the standardized manner.

• Decision variables

Let X1 be the number of units of toothpaste to be made and:

X2 be no. of cartons of detergents to be made

Objective function

Maximise: Contribution = 3000x1 + 4000x2

Constraint

• A: 4X1+2X2≤100 machine hours constraint

• B: 4X1+6X2 ≤180 Labour hours constraint

• C: X1+X2≤ 40 Materials constraint

• D: X1≤ 20 maximum toothpaste sales

• E: X2≥10 minimum detergent sales

• X1 ≥ 0 non-negativity constraint
• (No need for having a non-negativity constraint for X2 since it is already supposed to be

greater than 10 which is non-negative).

• The source and sales constraints include both types of restrictions (i.e. ≥ and ≤).

• The problem has only 2 decision variables ( X 1 and X2) and can it can be solved

graphically.

• The number of limitations does not exclude a graphical solution.

• STEP 2. Draw the axes of the graph to represent the X1 and X2

X2

0
X1

You should determine the scale once you know your limitations. Each axis must start at

zero and the scale must be constant (i.e. linear) along the axis but it is not necessary for

the scales on both axes to be the same.

STEP 3. Draw each limitation as a separate line on the graph.

(For illustration purposes each of these lines is drawn separately but they should all be

combined and drawn on the same graph).

• Since the constraints are all linear, they will be represented by straight lines.

• In order to draw a straight line, we only need two points on a graph.

• Each constraint will be drawn as follows:

• 4X1+2X2 ≤100 machine hours constraint

• We assume there is no inequality to enable us get the two points, to aid us in drawing the

graph.

• i.e. Assume 4X1+2X2=100

• When X1 = 0, X2=100/2=50

• When X2=0, X1=100/4=25

• We have two points : (X1,X2)=(0,50)and (25,0)

• Drawing and shading of constraint 1

• Shade off the region that is not required. In our case it is area to the right of the line.

X2

Equation of the line:4X1+2X2≤100

50

25
0 X1

• Constraint two and three

• 4X1+6X2 ≤180 Labor hours constraint

• Assume 4X1+6X2=180

• When X1=0, X2=30 and when X2=0, X1= 45

• We can plot the constraint using the coordinates (X1,X2 )=(0,30)and (45,0)

• X1+X2 ≤ 40 Materials constraint

• Assume X1+X2= 40

• When X1=0, X2=40, and when X2=0, X1=40

• We can plot the constraint using the coordinates (X1,X2 )=(0,40)and (40,0)
 • Drawing and shading of constraint 2

• Shade off the region that is not required. In our case it is the area to the right of the line.

X2

Equation of the line: 4X1+6X2≤180

30

0 45
X1

Constraints four and five

• Sales limitations : this only affects one of the products at a time and in our case, the sales

restrictions were

X1 ≤ 20 and X2 ≥ 10

If we assume X1= 20 and X2=10, we can represent each of them by one line.

The vertical line represents X1 = 20 and the shaded area to the right of the line represents the area

containing the values greater than 20. The horizontal line represents X2 ≥ 10 and the shaded line

below the line represents values of X2 which are less than 10.
 X2

X2

10

0 20 X1 0 X1

Graphing and shading of the non-negativity constraint

X2

0 X1
Step 4: Identify the feasible region(the region satisfying all the constraints)

• All the constraints are supposed to be drawn on the same graph to help in identifying the

area that satisfies all the constraints.

X2
50

40

30

20
R
10

0 10 20 30 40 50 X1

The region labeled R represents the feasible region i.e. the area that satisfies all the 5 constraints.

Note : The constraint X1+X2≤ 40 does not touch the feasible region, it is overshadowed

by the others. It is called a redundant constraint in that it is not affecting the decision

made by the producer. It is non-binding.

Step 5: Identify the point within the feasible region that ensures the contribution is maximum.

• This can be identified by marking the corner points within the feasible region.

X2
A=(0,30), B=(15,20), C=(0,10), D=(20,10)
50

40

30
A
B
20
R
C D
10

0 10 20 30 40 50 X1

Once the corner points are identified, the one that maximizes the contribution is identified

by evaluating the value of the objective function at each of the points . This has been

done in the next slide.

• Contribution = 3000x1 + 4000x2

• At point A, contribution = 3000(0) + 4000(30)

• = kshs.120,000
 • At point B, contribution = 3000(15) + 4000(20)

• = kshs.125,000

• At point C, contribution = 3000(0) + 4000(10)

• = kshs.40,000

• At point D, contribution = 3000(20) + 4000(10)

• = kshs.100,000

• Since it is a maximization problem, we choose the point giving us the maximum

value. Point B= Kshs.125,000 and it is the maximum contribution

Example Two: Minimisation problem

• A chemical manufacturer uses 2 chemicals X and Y in different proportions to produce 3

products A, B, and C. The manufacturer wants to produce at least 150 units of A, 200 units of

B and 60 units of C. Each ton of X yields 3 units of A, 5 of B and 3 of C. Each ton of Y

yields 5 of A, 5 of B and 1 of C. If chemical X costs $2000 per ton and Y costs $2500 per

ton. Question: Formulate the linear programming problem and solve it using the graphical

method.

• Solution

• Decide on the decision variables:

• Let X=number of tons of chemical X to be used

• Y=number of tons of chemical Y to be used

• Objective:

• To minimize the total cost of purchasing the chemicals

• Cost=2000X+2500Y

1. At least 150 units of A should be produced

3X+5Y≥150

2. At least 200 of B should be produced

5X+5Y≥200

3. At least 60 of B should be produced

3X + Y ≥60

4. Non-negativity constraints

X ≥0, Y ≥0

Graph and shade the constraints on the same graph:

3X+5Y=150

when X=0, Y=30,when Y=0, X=50

(X,Y) = (0,30)(50,0)

5X+5Y=200

When X=0, Y=40, whenY=0, X=40

 (X,Y)= (0,40), (40,0)

3X + Y =60

When X=0, Y=60, when Y=0, X=20

60
A
50

40

30 B

20 C

10
D
0 10 20 30 40 50 X

A=0,60, cost=150,000

• B=10,30 cost=20,000+75000=95,000

• C=20,25 cost=102,500

Point B gives the lowest cost, Kshs. 95,000. We should therefore buy 20 units of chemical X and

25 units of chemical Y.
 2.4 SUMMARY

In this lecture we have learnt how to solve LPP using

graphical method. It can only deal with problems involving 2 decision
variables. In our next lecture we will look at how to solve LPP using
simplex method.

NOTE

When using the graphical method, ensure your

graph is drawn to scale.

2.5 ACTIVITIES

Ensure you do more practice questions from your reference

texts.
2.6 FURTHER READING

References
1. Gupta S.K. Cozzolino J.M. (1975) Fundamentals of
Operations Research for Management.
2. Lucey T. Quantitative techniques
3. Monga G.S: Mathematics and statistics for Economists,
second revised edition.
4. Tulsian and Pandey V. Quantitative Techniques, Theory
and Problems.
2.7 SELF-TEST QUESTIONS 2

A manufacturer produces two products, toothpaste and detergents. The

contribution per unit is $3 and $4 for toothpaste and detergents respectively. The manufacturer

wishes to establish the weekly production plan which maximizes contribution. Both products are

processed on three machines M 1, M2 and M3. The time required by each product and total time

available per week on each machine is as follows:

Machine chairs tables Available hours

M1 3 3 36
M2 5 2 50
M3 2 6 60

Use the graphical method to determine how the manufacturer should schedule his production in

order to maximize contribution.

LECTURE 3: SOLUTION OF LPP USING SIMPLEX METHOD

3.1 INTRODUCTION

We have already looked at how we solve an LPP using the graphical method.
We noted that the graphical method can not be used if we have more than 2 decision
variables. In this lecture we will discuss the method of solving (LPP) using the simplex
method.In addition, we will look at how a maximization/minimisation problem can be turned
into a dual.

3.2 LECTURE OBJECTIVES

By the end of this lecture the learner should be:

1. Able to solve a linear programming problem using the simplex method.

2. Able to formulate the dual of any linear programming problem.

 3.2 Solution of maximisation LPP using the simplex method.

In our lecture two, we were able to see how a LPP can be solved using the graphical method. In

this lecture we will learn how to a LPP can be solved using the simplex method.

Steps

• Formulate the problem in the standard way.

• Convert the inequalities into equations through the addition of a slack variable if the

inequality is of a less than type (≤).

• A slack variable represents unused capacity in the constraint.

E.g. 4X1+6X2≤180 Labour hours constraint

4X1+6X2 + S1 =180

S1= unused capacity

• It will have the coefficient of 1 if both X1 and X2=0 or can be 0 if the labour hours are

fully utilised.

• Each constraint has its own slack variable.

• Once the slack variable is incorporated, the simplex method assigns it a value during the

step by step procedure.

A company can produce 2 products A and B. The profit per unit of A produced is 6 dollars while

the profits per unit of B produced is 8 dollars. To produce a unit of A the company requires 30

labour hours and 20 labour hours. The machine hours required are 5 hours and 10 hours for A
and B respectively. The company has a maximum of 300 and 110 labour and machine hours at its

disposal. Solve the problem using the simplex method.

Step 1

Formulate the problem in the standard way.

• Let X=no. of units of A produced and

Y= no. of units of B produced.

Objective function

Profits = 6X+8Y

Constraints:

6X+8Y≤300 labour hours constraint

5X+10Y≤110 Machine hours constraint

X≥0, Y≥0, non-negativity constraints

Step 2:

Change the constraints into equations by introducing slack variables

Profits = 6X+8Y+0S1+0S2

Constraints:

6X+8Y+S1=300 labour hours constraint

5X+10Y+S2=110 Machine hours constraint

X, Y, S1, S2 ≥0, non-negativity constraints

Step 3: Prepare the initial simplex table

Solution Products Slack variables Solution

variable X Y S1 S2 Quantity

S1 30 20 1 0 300

S2 5 10 0 1 110

Z=Profit 6 8 0 0 0

Note:

• The values from the table are the values from the objective function and the constraints in

step 2.

• Z is used to represent total profits.

• The table shows that S1=300 and S2=110 and profits (Z)=0.

• Nothing is produced hence the slacks are at their maximum values.

Step Four

Profits can be improved by producing the products A and B in varying quantities.

• Improve the profits by producing as much as possible of the product with highest profit

per unit.

• i.e. highest value in Z row.

• The number of the units to be produced will be limited by one or more constraints

becoming operative.

• E.g. in our case, product B has the highest profit.

• It becomes the incoming variable.

 • Determine the row it should be come into by dividing the solution quantity column by the

corresponding values in the Y column.

• 300/20=15

• 110/10=11

• Select the row with the lowest quotient. (Why: it is the one that will limit the maximum

amount of product B that can be produced.

• The selected row is known as the key row. It gives us the outgoing variable.

• The column with the highest profit per unit is called the key column and gives us the

incoming variable.

• The value at the intersection of the key row and the key element is known as the pivot

element/key element.

This is shown in the table below:

Solution Products Products Slack variables Solution

variable Quantity
X Y S1 S2
S1 30 20 1 0 300
S2 5 0 1 110
Z=Profit 6 8 0 0 0

Divide all elements in the pivot row by the pivot element.

• Old row

5 10 0 1 110

New row

0.5 1 0 0.1 11

Enter this new row in a new simplex table

Second simplex table

Row no. Solution Products Slack variables Solution

variable Quantity
X Y S1 S2
1 S1 30 20 1 0 300
2 Y 0.5 0 0.1 11
3 Z=Profit 6 8 0 0 0

• The entire table is similar to the initial simplex table except that row 2 is replaced with

new row 2.

• This implies that we will now produce 11 units of product B.

• As a result of producing product B, the labour and machine hours available have reduced.

In addition, the profits have also been improved. We therefore need to adjust the other

two rows to reflect this change in resources and also the profits.

• This is done by an iterative procedure that ensures that all the other values in the pivot

column become 0. In order to ensure the equality of the row, we ensure that the entire

row is altered.

• New Row 1=Row 1-20row 2

• Old row 1

• 20 1 0 300

• New row 1

• 20 0 1 -2 80

• Old row 3

• 8 0 0 0

• New row 3=row 3-8row 1

• 2 0 0 -0.8 88

Third Simplex table

Row Solution Products Slack variables Solution

no. variable Quantity
X Y S1 S2
1 S1 20 0 1 -2 80
2 Y 0.5 0 0.1 11
3 Z=Profit 2 0 0 -0.8 -88

• If in the Z row, in the products columns we still have a positive number, the profits can

be improved further. So we need to introduce the product whose profits can still be

positive.

The above steps are repeated as follows:

• If we introduce a unit of X, the profit will increase by 2 per unit.

• X becomes the incoming variable.

• X column is the key column.

• To identify the key row and pivot element, we divide solution quantity with

corresponding values in the X column.

80/20=4

11/0.5=22

We choose the row with smallest quotient.

Row 1 is chosen, 20 becomes the pivot element.

• Divide the key row values by 20.

New row 1:

1 0 1/20 -0.1 4
 Fourth Simplex table

Row Solution Products Slack variables Solution

no. variable Quantity
X Y S1 S2
1 S1 1 0 1/20 -0.1 4
2 Y 0.5 0 0.1 11
3 Z=Profit 2 0 0 -0.8 -88

• Adjust the other rows in the table to ensure that all the other elements in key column are

zero.

• New Row 2=old row 2 - 0.5New Row1

• Old row 2: 0.5 1 0 1/10 11

• New row 1: 1 0 1/20 -2/20 4

• New row 2: 0 1 -1⁄40 3/20 9

• New row 3=old row 3-2newrow 1

• Old row 3: 2 0 0 -8/10 -88

• New row 1: 1 0 1/20 -2/20 4

New row 3: 0 0 -2/20 -12/20 -96

Fifth Simplex table

Row Solution Products Slack variables Solution

no. variable Quantity
X Y S1 S2
1 S1 1 0 1/20 -0.1 4
2 Y 0 1 -1/40 3/20 9
3 Z=Profit 0 0 -2/20 -12/20 -96

• The optimal solution is producing 4 units of A and 9 units of product B.

 • The total profits will be: Z=6X+8Y=6*4+8*9=96

The values in the Z row and in the columns of the slack variables S 1 and S2 are the shadow prices

of the labour hours and machine hours respectively. If we have an extra unit of labour hours, we

will be able to increase the profits by 1/10 dollars while an extra unit of machine hours will

increase the profits by 6/10 dollars.

3.4 DUALITY IN LPP

Every maximization or minimization problem in a linear programming has a corresponding

minimization or maximization problem.

The original problem is referred to as a PRIMAL while the corresponding one is the DUAL.

The relationship between the two can best be expressed through the use of parameters that they

share in common.

Consider a primal maximization problem generally expressed as.

𝑀𝑎𝑥 𝑓(𝑥 ) = 𝐶1 𝑋1 + 𝐶2 𝑋2 + 𝐶3 𝑋3 +. . . +𝐶𝑛 𝑋𝑛

𝑠. 𝑡

𝑎11 𝑥1 + 𝑎12 𝑥2 + ⋯ + 𝑎1𝑛 𝑥𝑛 ≤ 𝑏1

𝑎21 𝑥1 + 𝑎22 𝑥2 + ⋯ + 𝑎2𝑛 𝑥𝑛 ≤ 𝑏2

𝑎𝑚1 𝑥1 + 𝑎𝑚2 𝑥2 + ⋯ + 𝑎𝑚𝑛 𝑥𝑛 ≤ 𝑏𝑚

𝑥1 , 𝑥2 … 𝑥𝑛 ≥ 0

Rewriting the above in matrix form

 𝑥1
Max 𝑓(𝑥) = [𝐶1 𝐶2 … 𝐶𝑛 ] ⌈𝑥2 ⌉
𝑥𝑛

𝑠. 𝑡

𝑎11 𝑎12 𝑎1𝑛 𝑥1 𝑏1

⌊ 𝑎12 𝑎22 𝑎2𝑛 ⌋ ⌈𝑥2 ⌉ ≤ ⌈ 𝑏2 ⌉
𝑎𝑚1 𝑎𝑚2 𝑎𝑚𝑛 𝑥𝑛 𝑏𝑛

𝑋1 , 𝑋2 … 𝑋𝑛 ≥ 0

RULES OF TRANSFORMATION
The direction of optimization changes i.e maximization becomes minimization and vice versa.

e.g. min Q

 The row vector of coefficient of the objective function in the primal problem is

transposed into a column vector of constants in the dual.

 The column vector of constants in the constants in the primal problem is transposed

into a row vector of coefficients in the objective function of the dual.

 The matrix of coefficients in the constraint equation of the primal problem is

transposed into a matrix of coefficients in the matrix of the dual problem.

 The inequality sign in the primal problem is reversed, i.e. ≤ in a primal maximization

problem becomes in the dual problem.

Note. 1. The non-negativity constraints remain unchanged.

2. The decision variables will change say from Xi to Zj.

The dual problem therefore becomes, in matrix form:

𝑧1
𝑀𝑖𝑛 𝑄 = 𝑏1 𝑏2 … 𝑏𝑛 ⌈𝑧2 ⌉
[ ]
𝑧𝑛

𝑠. 𝑡

𝑎11 𝑎21 𝑎31 𝑧1 𝑐1

⌊ 𝑎12 𝑎22 𝑎32 ⌋ ⌈𝑧2 ⌉ ≥ ⌈𝑐2 ⌉
𝑎1𝑛 𝑎2𝑛 𝑎3𝑛 𝑧𝑛 𝑐𝑛

𝑧1 , 𝑧2 … 𝑧𝑛 ≥ 0

𝑀𝑖𝑛 𝑄 = 𝑏1 𝑧1 + 𝑏2 𝑧2 + 𝑏3 𝑧3

𝑠. 𝑡

𝑎11 𝑧1 + 𝑎21 𝑧2 + 𝑎31 𝑧3 ≥ 𝑐1

𝑎21 𝑧1 + 𝑎22 𝑧2 + 𝑎32 𝑧3 ≥ 𝑐2

𝑎31 𝑧1 + 𝑎23 𝑧2 + 𝑎33 𝑧3 ≥ 𝑐3

𝑧1 , 𝑧2 … 𝑧𝑛 ≥ 0

Example

Let the primal problem be

𝑀𝑎𝑥 𝜋 = 5𝑋1 + 3𝑋2

𝑠. 𝑡

6𝑋1 + 2𝑋2 ≤ 36

5𝑋1 + 5𝑋2 ≤ 40
 2𝑋1 + 4𝑋2 ≤ 28

𝑋1 𝑋2 ≥ 0

Rewriting into matrix

𝑥1
𝑚𝑎𝑥𝜋 = ⌊5 3⌋ [ ]
𝑥2

𝑠. 𝑡

6 2 𝑥 36
1
[5 5] [ ] ≤ [40]
𝑥2
2 4 28

𝑋1 𝑋2 ≥ 0
The dual will be,

𝑧1
𝑀𝑖𝑛 𝑄 = [36 40 28] [𝑧2 ]
𝑧3

𝑠. 𝑡

𝑧1
6 5 2 𝑧 5
[ ] [ 2] ≥ [ ]
2 5 4 𝑧 3
3

𝑧1 𝑧2 𝑧3 ≥ 0

In the linear form,

min 𝑄 = 36𝑧1 + 40𝑧2 + 28𝑧3

𝑠. 𝑡

6 𝑧1 + 5𝑧2 + 2𝑧3 ≥ 5
 2 𝑧1 + 5𝑧2 + 4𝑧3 ≥ 5

𝑧1 𝑧2 𝑧3 ≥ 0

3.5 SUMMARY

This lecture has looked at how you can solve LPP using
the simplex method. It can be used for problems involving more than
2 decision variables. .Formulation of dual from the primal has also
been learnt.

NOTE

Shadow prices are obtained only for the completely utilized

resources.
Duality in LPP is important especially if you want to use
maximization instead of minimization problem.

3.6 ACTIVITIES

Solve the LPP problems in lecture 2 using the simplex

method and check whether the answers obtained are similar.
 3.7 FURTHER READING

References
1. Gupta S.K. Cozzolino J.M. (1975) Fundamentals of
Operations Research for Management.
2. Lucey T. Quantitative techniques
3. Monga G.S: Mathematics and statistics for Economists,
second revised edition.
4. Tulsian and Pandey V. Quantitative Techniques, Theory
and Problems.

3.8 SELF-TEST QUESTIONS 3

Use the simplex method to solve the following:

• Z=6X+8Y

• Subject to:

• 2X+3Y≤16

• 4X+2Y≤16

X, Y≥0
 LECTURE FOUR: FUNCTIONAL DEPENDENCE AND SOLUTION TO
SIMULATNEOUS EQUATIONS

4.1 INTRODUCTION

In this lecture you will learn how to test for functional

dependence in non-linear functions. In addition, the solution to
simultaneous equations using the Gaussian and Gauss Jordan
eliminitain methods will be learnt.

4.2 LECTURE OBJECTIVES

By the end of this lecture, the learner should be able to:

1. Determine if functions of two and three independent variables
are functionally dependent using the Jacobian determinant.
2. Solve three simultaneous equations using the gauss Jordan
elimination method.
4.3 Jacobian Determinant
The Jacobian matrix can help to test whether there exists functional linear dependence in a set of
𝑛 functions in 𝑛 variables. It is a matrix of the first order partial derivatives. If the functions are
linearly dependent, the determinant of the Jacobian matrix is 0.
E.g. 𝑦1 = 2𝑥1 + 3𝑥2
𝑦2 = 4𝑥1 + 12𝑥1 𝑥2 + 9𝑥2

𝜕𝑦1 𝜕𝑦
=2 ⁄𝜕𝑥 = 3
𝜕𝑥1 2

𝜕𝑦2 𝜕𝑦2
= 8𝑥1 + 12𝑥2 = 12𝑥2 + 18𝑥2
𝜕𝑥1 𝜕𝑥2

𝜕𝑦1 𝜕𝑦2
𝜕𝑥1 𝜕𝑥2
𝐽=
𝑑𝑦2 𝜕𝑦2
[𝑑𝑥1 𝜕𝑥2 ]

2 3
(𝐽 ) = | |=0
8𝑥1 + 12𝑥2 12𝑥2 + 18𝑥2
The functions are linearly dependent.

Example 2
𝑦1 = 3𝑥12 + 2𝑥22
𝑦2 = 5𝑥1 + 1
 6𝑥1 4𝑥2
𝐽=[ ]
5 0
6𝑥 4𝑥2
|𝐽 | = | 1 |=−20𝑥2 unless X2
5 0
Is zero, the determinant is not zero, hence the 2 functions are functionally independent.
If we have several functions,
𝑓1′ 𝑓2′ … 𝑓𝑛′
𝑦1 = 𝑓 (𝑥1 ⋯ 𝑥𝑛 ) |𝐽 | = | ⋮ |
𝑓1𝑛 𝑓𝑛𝑛
𝑦2 = 𝑓(𝑥1 ⋯ 𝑥𝑛 )
𝑦𝑛 = 𝑓 (𝑥1 ⋯ 𝑥𝑛 )

A Jacobian |𝐽 |is identically zero for all values of 𝑥1 , 𝑥2 … , 𝑥𝑛 if the n functions are functionally
dependant.

4.4 Solution of simultaneous equations using the Gaussian method

Gaussian method involves solving for simultaneous equations using matrix algebra by getting the
lower echelon matrix for the matrix of coefficients.
If we have a 3𝑋3 matrix, given as:
𝑎11 𝑋 + 𝑎12 𝑌 + 𝑎13 𝑍 = 𝑏1
𝑎21 𝑋 + 𝑎22 𝑌 + 𝑎23 𝑍 = 𝑏2
𝑎31 𝑋 + 𝑎32 𝑌 + 𝑎33 𝑍 = 𝑏3
We can get the values of X, Y and Z by rewriting the functions in matrix form such that:
𝑎11 𝑎12 𝑎13 𝑋 𝑏1
[𝑎21 𝑎22 𝑎23 ] [ 𝑌 ] = [𝑏2 ]
𝑎31 𝑎32 𝑎33 𝑍 𝑏3
The lower echelon matrix for matrix of coefficients is given as:

1 𝑘 𝑚
[0 1 𝑛]
0 0 1
This is done through a step by step iteration procedure.
Let us use a practical example.
 2𝑋 + 12𝑌 − 2𝑍 = 20
2𝑋 + 3𝑌 + 3𝑍 = 17
3𝑋 − 3𝑌 − 2𝑍 = −9
If we rewrite the equations in matrix form, we get:
2 12 −2 𝑋 20
2 3 3 [ 𝑌 ] = [ 17 ]
3 −3 −2 𝑍 −9
As we do the iterations, we need to maintain the row equality. This will be done by some row
operations, similar to those we used when we were looking at the simplex method in LPP. We
will only affect the matrix of the coefficients and the constants matrix. We can join them
together into one augmented matrix as we undertake the step by step procedure.

2 12 −2 20
2 3 3 | 17
3 −3 −2 −9

We want to make the first element in the first row to be 1. This can be accomplished by dividing
the entire row 1 by 2. This gives us:
1 6 −1 10
2 3 3 | 17
3 −3 −2 −9
The elements 2 and 3 in column 1 Row 2 and 3 respectively will be turned into 0 using the
following row procedure:
𝑁𝑒𝑤 𝑅𝑜𝑤 2 = 𝑂𝑙𝑑 𝑅𝑜𝑤2 − 2𝑛𝑒𝑤 𝑅𝑜𝑤 1
𝑁𝑒𝑤 𝑅𝑜𝑤 3 = 𝑂𝑙𝑑 𝑅𝑜𝑤3 − 3𝑛𝑒𝑤 𝑅𝑜𝑤 1
This gives us the following matrix:
1 6 −1 10
0 −9 5 | −3
0 −21 1 −39

We then change the −9 in the second row , second column into 1 by dividing the entire row by
−9. This gives us the following matrix.
1 6 −1 10
5 1
0 1 − | 3
9
0 −21 1 −39
We change the value -21 in the second column into a 0 by the following procedure:
𝑁𝑒𝑤 𝑅𝑜𝑤 3 = 𝑂𝑙𝑑 𝑅𝑜𝑤3 − (−21)𝑛𝑒𝑤 𝑅𝑜𝑤 2
It will give us:

1 6 −1 10
5
0 1 − | 1
9
32 3
0 0 − −32
3

The next step involves changing the element in 3 rd row, 3rd column into a 1. This can be done by
32
dividing the entire row 3 by − giving us the following matrix
3

1 6 −1 10
5 1
0 1 − |
9 3
0 0 1 3
If we bring back our variables, X, Y and Z we have the following matrices:

1 6 −1 10
5 𝑋 1
[0 1 − ] [ 𝑌 ] = [ ⁄3]
9 𝑍
0 0 1 3

𝑋 + 6𝑌 − 𝑍 = 10
5 1
𝑌− 𝑍=
9 3
𝑍=3

If we replace Z=3 in the second equation, it will give us the following

5 5 1 1 5 6
𝑌 − (3) = 𝑌 − = → 𝑌 = + = = 2
9 3 3 3 3 3

𝑍 = 3𝑌 = 2
By replacing the values of Y and Z in the first equation, we get:

𝑋 + 6(2) − 3 = 10
𝑋 = 10 − 9 = 1
 𝑍=3 𝑌=2 𝑋=1
These will be the values that will satisfy the three equations.

4.5 Gauss Jordan Elimination Method

This method is just an extension of the Gaussian method. It involves ensuring that the matrix of
coefficients is turned into an identity matrix.

1 6 −1
5
Using our previous example, the matrix [0 1 − ] will be changed to an identity matrix
9
0 0 1
through the same row procedures.
We will start by changing the elements −1 and −5/9 in the 3rd column, 1st and 2 nd row
respectively into 0 using the 3 rd row. This will be done by the following process:
𝑁𝑒𝑤 𝑅𝑜𝑤 1 = 𝑂𝑙𝑑 𝑅𝑜𝑤1 − (−1)𝑛𝑒𝑤 𝑅𝑜𝑤 3
5
𝑁𝑒𝑤 𝑅𝑜𝑤 2 = 𝑂𝑙𝑑 𝑅𝑜𝑤2 − (− )𝑛𝑒𝑤 𝑅𝑜𝑤 3
9

1 6 0 𝑋 13
[0 1 ]
0 𝑌[ ] = [ 2]
0 0 1 𝑍 3
Next step involves changing the 6 in row 1 column 2 into a 0. This will be done by using row 2.
𝑁𝑒𝑤 𝑅𝑜𝑤 1 = 𝑂𝑙𝑑 𝑅𝑜𝑤1 − 6𝑛𝑒𝑤 𝑅𝑜𝑤 2
It will give us:

1 0 0 𝑋 1
[0 1 0] [ 𝑌 ] = [2]
0 0 1 𝑍 3

We can then rewrite our matrix into equations and we will have:
𝑋=1
𝑌=2
𝑍=3
Example 2
After using the Gaussian method to solve for three simultaneous equations, the following three
equations were obtained:
Determine the values of 𝑋1 , 𝑋2 , 𝑎𝑛𝑑 𝑋3 using the Gauss Jordan elimination method.
𝑥1 − 0.4𝑥2 − 0.3𝑥3 = 130
𝑥2 − 0.25𝑥3 = 125
𝑥3 = 300

1 −0.4 −0.3 𝑥1 130

[0 1 −0.25] [𝑥2 ] = [125 ]
0 0 1 𝑥3 300

Work from the last column to eliminate the coefficients of 𝑥3 in the other rows
i.e.
𝑅1 = 𝑅1 + 0.3𝑅3
𝑅2 = 𝑅2 + 0.25𝑅3

1 −0.4 0 220
0 1 0 |200
0 0 1 300
Finally change −0.4 into a 0 by using the second row. This will be done through the following
operation:
𝑅1 = 𝑅1 + 0.4𝑅2

1 0 0 300
0 1 0 |200
0 0 1 300

𝑥1 = 300 𝑥2 = 200 𝑥3 = 300

 4.6 SUMMARY

In this lecture, you have learnt about the Jacobian matrix which is a
matrix of second order partial derivatives. It is very helpful when
testing for functional dependence. You have also learnt an additional
method for solving three simultaneous equations.

NOTE

In order to use the gauss Jordan elimination method, you need to

understand how the Gaussian elimination method is done.

4.7 ACTIVITIES

Determine if the following functions are functionally

dependent
 4.8 FURTHER READING

References
1. Main Text: Chiang A.C. and Wainwright K:
Fundamental methods of mathematical Economics.
2. Monga G.S: Mathematics and statistics for Economists,
second revised edition.
3. A cook-book of Mathematics, Viatcheslav
VINOGRADOV, June 1999

4.9 SELF-TEST QUESTIONS 4

Solve the following three simultaneous equations using

the gauss Jordan elimination method.
𝑋−𝑍 =4
2𝑌 − 𝑍 = 6
𝑋 + 𝑌 = 10
 LECTURE FIVE: TESTS OF CONVEXITY AND CONCAVITY USING HESSIAN
DETERMINANTS

5.1 INTRODUCTION

In this lecture we will learn how to test for convexity and concavity
of a function using the hessian matrix. In order to understand better,
you may need to remind yourself about the rules of partial
differentiation.

5.2 LECTURE OBJECTIVES

By the end of this lecture the learner should be able to:

1. Determine the convexity of two and three independent
variables.
2. Get the determinants of a hessian matrix.
 5.3 CONCAVE AND CONVEX FUNCTIONS

A concave function has an absolute maximum when plotted on a graph, while a convex function
gives an absolute minimum when plotted on a graph.

It is easy to check for concavity and convexity in functions of single independent variable by just
checking the second order derivative.

If the second order derivative is negative, the function is concave and if positive, the function is
said to be convex.
In functions of two independent variables, there is a corresponding test which also uses the
second order derivatives.

If we have a function Z given as 𝑍 = 𝑓(𝑋, 𝑌 )

𝜕𝑓(𝑋, 𝑌 ) 𝜕𝑓 (𝑋, 𝑌 )
𝑑𝑍 = 𝑑𝑋 + 𝑑𝑌
𝜕𝑋 𝜕𝑌
It can be written as:

𝑑𝑍 = 𝑓𝑥 𝑑𝑋 + 𝑓𝑦 𝑑𝑋

The second order partial derivative will be given as:

𝑑2 𝑍 = 𝑓𝑥𝑥 𝑑𝑋 2 + 𝑓𝑦𝑦 𝑑𝑌 2 + 2𝑓𝑥𝑦 𝑑𝑋𝑑𝑌

a) The function is convex if: 𝑓𝑥𝑥 ≥ 0, 𝑓𝑦𝑦 ≥ 0 𝑎𝑛𝑑 𝑓𝑥𝑥 . 𝑓𝑦𝑦 − (𝑓𝑥𝑦 )2 ≥ 0
b) The function is concave if: 𝑓𝑥𝑥 ≤ 0, 𝑓𝑦𝑦 ≤ 0 𝑎𝑛𝑑 𝑓𝑥𝑥 . 𝑓𝑦𝑦 − (𝑓𝑥𝑦 )2 ≥ 0
c) The function is strictly convex if: 𝑓𝑥𝑥 > 0, 𝑎𝑛𝑑 𝑓𝑥𝑥 . 𝑓𝑦𝑦 − (𝑓𝑥𝑦 )2 > 0
d) The function is strictly concave if: 𝑓𝑥𝑥 < 0, 𝑎𝑛𝑑 𝑓𝑥𝑥 . 𝑓𝑦𝑦 − (𝑓𝑥𝑦 )2 > 0

Example one

𝑍 = 2𝑥 − 𝑦 − 𝑥 2 + 2𝑥𝑦 − 𝑦 2 Is the function convex or concave

𝑓𝑥 = 2 − 2𝑥 + 2𝑦 𝑓𝑥𝑥 = −2 𝑓𝑦 = −1 + 2𝑥 − 2𝑦 𝑓𝑦𝑦 = −2 𝑓𝑥𝑦 = 2

𝑓𝑥𝑥 = −2 < 0, 𝑓𝑥𝑥 𝑓𝑦𝑦 = 4 (𝑓𝑥𝑦 ) = 4

ℎ𝑒𝑛𝑐𝑒
2
𝑓𝑥𝑥 𝑓𝑦𝑦 − (𝑓𝑥𝑦 ) = 0
The function Z is concave but not strictly so.

Determinantal test for sign definiteness

𝑓𝑥𝑥 𝑓𝑥𝑦
The matrix of second order partial derivatives: H= [ ] is called a hessian matrix.
𝑓𝑥𝑦 𝑓𝑦𝑦

A function Z is positive definite if the first and second principal leading minors of matrix H are
positive.

i.e. |H1 | = |fxx | > 0

𝑓𝑥𝑥 𝑓𝑥𝑦
|H2 |= | | >0 If we get the determinant, it will be given as: 𝑓𝑥𝑥 𝑓𝑦𝑦 − 𝑓𝑥𝑦 𝑓𝑦𝑥 >0
𝑓𝑥𝑦 𝑓𝑦𝑦

A function Z is negative definite if the first leading minor is negative while the second principal
leading minors of matrix H are positive.

i.e. |H1 | = |fxx | < 0

𝑓𝑥𝑥 𝑓𝑥𝑦
|H2 |= | | >0
𝑓𝑥𝑦 𝑓𝑦𝑦

Example Two
Determine if the following function is positive definite or negative definite u sing the Hessian
determinants.

𝑍 = 2𝑋𝑌 − 𝑋 2 + 5𝑌 2
2 2
𝐻=[ ]
2 10
|H1 | = 2 > 0

|H2 |= |2 2
| = 20 − 4 = 16 > 0
2 10
Since the determinants of both H1 and H2 are greater than 0, we say the function is positive
definite and by extension the function is convex.
The determinantal test can be extended to the case of more than 2 independent variables.

𝑍 = 𝑓(𝑋, 𝑌, 𝑊)
𝑓𝑥𝑥 𝑓𝑥𝑦 𝑓𝑥𝑤
𝐻 = [ 𝑓𝑥𝑦 𝑓𝑦𝑦 𝑓𝑦𝑤 ]
𝑓𝑤𝑥 𝑓𝑤𝑦 𝑓𝑤𝑤
The function Z is positive definite (convex) iff
|H1 | = |fxx| > 0

𝑓𝑥𝑥 𝑓𝑥𝑦
|H2 |= | | >0
𝑓𝑥𝑦 𝑓𝑦𝑦

𝑓𝑥𝑥 𝑓𝑥𝑦 𝑓𝑥𝑤

|𝐻3 | = | 𝑓𝑥𝑦 𝑓𝑦𝑦 𝑓𝑦𝑤 | >0
𝑓𝑤𝑥 𝑓𝑤𝑦 𝑓𝑤𝑤

It is negative definite (concave iff

|H1 | = |fxx | ≤ 0
𝑓𝑥𝑥 𝑓𝑥𝑦
|H2 |= | |≥0
𝑓𝑥𝑦 𝑓𝑦𝑦
𝑓𝑥𝑥 𝑓𝑥𝑦 𝑓𝑥𝑤
|𝐻3 | = | 𝑓𝑥𝑦 𝑓𝑦𝑦 𝑓𝑦𝑤 | ≤ 0
𝑓𝑤𝑥 𝑓𝑤𝑦 𝑓𝑤𝑤

If Z is a function of n independent variables such that:

𝑍 = 𝑓 (𝑋1 , 𝑋2 … … . , 𝑋𝑛 )
The function is negative definite (convex) iff:

|H1 | < 0, |H2 | > 0 … … … … … (−1)n|Hn| > 0

Its positive definite iff:

|H1 | > 0, |H2 | > 0 … … … … … |Hn| > 0

Example three

Determine if function Z is convex or concave

𝑍 = 200 − 2𝑥 2 − 𝑦 2 − 3𝑤 − 𝑥𝑦 − 𝑒 𝑥+𝑦+𝑤
The matrix of second order derivatives will be given as:
−4 − 𝑒 𝑢 −1 − 𝑒 𝑢 −𝑒 𝑢
𝐻 = [−1 − 𝑒 𝑢 −2 − 𝑒 𝑢 −𝑒 𝑢 ]
−𝑒 𝑢 −𝑒 𝑢 −𝑒 𝑢
|H1 | = |−4 − 𝑒 𝑢 | < 0
𝑢 −1 − 𝑒 𝑢
|H2 |= |−4 − 𝑒 𝑢 | = 7 + 4𝑒 𝑢 > 0
−1 − 𝑒 −2 − 𝑒 𝑢
 −4 − 𝑒 𝑢 −1 − 𝑒 𝑢 −𝑒 𝑢
|𝐻3 | = [−1 − 𝑒 𝑢 −2 − 𝑒 𝑢 −𝑒 𝑢 ] = −7𝑒 𝑢 <0
−𝑒 𝑢 −𝑒 𝑢 −𝑒 𝑢
The function is therefore strictly concave.

5.4 SUMMARY

We have learned how to check for concavity and convexity using the
hessian determinant.

NOTE

Convex functions have a minimum as an extreme point while

concave functions have a maximum as the extreme point.
5.5 ACTIVITIES

Determine if a cobb-douglas production function given

as 𝑄 = 𝐴𝐾𝛽 𝐿𝛼 is concave or convex given that β+α≤1

5.6 FURTHER READING

References
1. Main Text: Chiang A.C. and Wainwright K:
Fundamental methods of mathematical Economics.
2. Monga G.S: Mathematics and statistics for Economists,
second revised edition.
3. A cook-book of Mathematics, Viatcheslav
VINOGRADOV, June 1999
1. CERGE-
5.7 SELF-TEST QUESTIONS FIVE

• Practice Question Five

• Determine if the following function is concave or convex
𝑍 = −𝑥 2 + 4𝑥𝑦 − 9𝑦 2 − 𝑤 2
LECTURE SIX: NON-LINEAR PROGRAMMING

6.1 INTRODUCTION

This lecture is an advancement of constrained optimization involving

inequality and non-negative constraints.

6.2 LECTURE OBJECTIVES

By the end of this lecture the learner should be able to:

1. Formulate non-linear programming problems.
2. State the Kuhn tucker necessary conditions for a minimum and for a
maximum.
3. Solve for optimal values for both maximization and minimization
problems.
 6.3 Non Linear programming

This is a constrained optimization technique that makes it possible to handle problems that
involve non-linear inequalities and non-negativity constraints.

In normal optimization problems, first order necessary conditions for a relative extremum require
that the first order partial derivatives of the function with respect to each choice variable be zero.

In non-linear programming the first order necessary conditions are called Kuhn Tucker
conditions. In normal optimization first order conditions are necessary but not sufficient for an
extreme. The Kuhn Tucker conditions are necessary and sufficient conditions as well.

Effect of Non-negativity constraints

1. Maximisation case
Let’s assume a firm wants to maximize the profits subject to ensuring that the quantity X is non-
negative. This can be written as:
𝑀𝑎𝑥 𝐼𝐼 = 𝑓(𝑋)
𝑠𝑡. 𝑋 ≥ 0
We can have three cases:

A
𝑓(𝑋) 𝑓(𝑋) 𝑓(𝑋)
B C

𝑋 𝑋 𝑋
In case A, the profits are maximized (the maximum point on curve) when X is greater than 0. It
is an interior solution. At the point where profits are maximum, X>0 and the first order
derivative of the profit function with respect to X is 0 i.e. 𝑋 > 0, 𝑓 ′ (𝑥 ) = 0.
In case B, the profits are maximized when X is equal to 0. It is a boundary solution. At the point
where profits are maximum, X=0 and the first order derivative of the profit function with respect
to X is 0 i.e. 𝑋 = 0, 𝑓 ′ (𝑥 ) = 0.

In case C, the profits are maximized when X is less than 0. Since X is not allowed to take
negative values, the firm will not produce. The firm will therefore operate at the point where
X=0, and first order derivative of the profit function with respect to X is negative i.e. 𝑋 =
0, 𝑓 ′ (𝑥 ) < 0.
These conditions can be summarized as:
1. 𝑓 ′ (𝑋) = 0 2. 𝑓 ′ (𝑋) = 0 3. 𝑓 ′ (𝑋) < 0
𝑋 >0 𝑋=0 𝑋=0
If the three conditions are combined, maximization will occur at the point where:
𝑋≥0
𝑓′ (𝑋) ≤ 0
and 𝑋. 𝑓 ′ (𝑋) = 0 This is known as the Complementary slackness condition between 𝑋
and 𝑓 ′ (𝑋).

The three conditions are the Kuhn Tucker first order necessary conditions for a maximum.

If a problem has got 𝑛 choice variables,

i.e.
𝑀𝑎𝑥 𝐼𝐼 = 𝑓(𝑥1 , 𝑥2 , ⋯ 𝑥𝑛 )
𝑥1 ⋯ 𝑥𝑛 ≥ 0
𝑥𝑗 ≥ 0(𝑗 = 1,2 ⋯ 𝑛 )
Then
𝑓𝑗 ′ ≤ 0 𝑥𝑗 ≥ 0 𝑥𝑗 𝑓𝑗 ′ = 0 (𝐽 = 1, 2 ⋯ 𝑛 )

6.4 Effect of non equality constraints

𝑀𝑎𝑥 𝐼𝐼 = 𝑓(𝑥1 , 𝑥2 , 𝑥3 )
𝑠𝑡
𝑔1 (𝑥1 , 𝑥2 , 𝑥3 ) ≤ 𝑟1
𝑔2 (𝑥1 , 𝑥2 , 𝑥3 ) ≤ 𝑟2
𝑥1 , 𝑥2 , 𝑥3 ≥ 0

Can be written as:

𝑀𝑎𝑥 𝐼𝐼 = 𝑓(𝑥1 , 𝑥2 , 𝑥3 )
𝑠𝑡
𝑔1 (𝑥1 , 𝑥2 , 𝑥3 ) + 𝑠1 = 𝑟1
𝑔2 (𝑥1 𝑥2 𝑥3 ) + 𝑠2 = 𝑟2
𝑥1 , 𝑥2 , 𝑥3 , 𝑠1 , 𝑠2 ≥ 0

Using classical approach:

𝑍 = (𝑓(𝑥1 , 𝑥2 , 𝑥3 ) + 𝜆1 (𝑟1 − 𝑔1 (𝑥1 𝑥2 𝑥3 )+𝑠1 ) + 𝜆2 (𝑟2 – 𝑔2 (𝑥1 𝑥2 , 𝑥3 ) + 𝑠2 )

In normal constrained optimization, first order conditions are:

𝜕𝐿 𝜕𝐿 𝜕𝐿 𝜕𝐿 𝜕𝐿 𝜕𝐿 𝜕𝐿
= = = = = = =0
𝜕𝑥1 𝜕𝑥2 𝜕𝑥3 𝜕𝑑1 𝜕𝑑2 𝜕𝑠1 𝜕𝑠2

But now, due to the presence of non-negativity constraints, we modify them to become:
𝜕𝑍 𝜕𝐿
1. ≤ 0 𝑋𝑗 ≥ 0 𝑎𝑛𝑑 𝑋𝑗 . =0
𝜕𝑋𝑗 𝜕𝑋𝑗

𝜕𝑍 𝜕𝐿
2. ≤ 0 𝑆𝑗 ≥ 0 𝑆𝑗 . =0
𝜕𝑆𝑗 𝑑𝑠𝑗

𝜕𝑍
3. =0
𝜕𝜆𝑖

𝑑𝑍
= 𝜆𝑖 ≤ 0
𝜕𝑆𝑖
−𝜆𝑖 ≤ 0 𝑠𝑗 ≥ 0 𝜆𝑖 . 𝑠𝑗 = 0
𝑖𝑓: −𝜆𝑖 ≤ 0 𝑖𝑡 𝑖𝑚𝑝𝑙𝑖𝑒𝑠 𝑡ℎ𝑎𝑡 𝜆𝑖 ≥ 0

If we ignore the non-negativity constraints and inequality signs in constraints we can write it as:
𝑍 = 𝑓(𝑥1 , 𝑥2 , 𝑥3 ) + 𝜆1 (𝑟1 − 𝑔1 (𝑥1 , 𝑥2 , 𝑥3 ) + 𝜆2 (𝑟2 − 𝑔2 (𝑥1 , 𝑥2 , 𝑥3 )
 𝜕𝑍 𝜕𝑍
a. ≤0 𝑥𝑗 ≥ 0 𝑥𝑗 . =0
𝜕𝑥𝑗 𝜕𝑥𝑗

𝜕𝑍 𝜕𝑍
≥0 𝜆𝑖 ≥ 0 𝜆𝑖 . =0
𝜕𝜆𝑖 𝜕𝜆𝑖

How?
𝑔1 (𝑥1 , 𝑥2 , 𝑥3 ) ≤ 𝑟1
𝑟1 ≥ 𝑔1 (𝑥1 , 𝑥2 , 𝑥3 )
𝜕𝑍
= (𝑟1 − 𝑔1 𝑥1 𝑥2 𝑥3 ) ≥ 0
𝜕𝜆𝑖

Example
𝑀𝑎𝑥
𝑢 = 𝑥𝑦
𝑠𝑡. 𝑥 + 𝑦 ≤ 100
𝑥 ≤ 40
𝑥, 𝑦 ≥ 0
𝑍 = 𝑥𝑦 + 𝜆1 (100 − 𝑥 − 𝑦) + 𝜆2 (40 − 𝑥 )

K.T.C.
𝜕𝑧
= 𝑦 − 𝜆1 − 𝜆2 ≤ 0 − − − 1 𝑋≥0 (𝑦 − 𝜆1 − 𝜆2 ) (𝑥 ) = 0
𝜕𝑥

𝜕𝑍
= 𝑥 − 𝜆1 ≤ 0 − − − 2 𝑦≥0 (𝑥 − 𝜆1 ) (−1) = 0
𝜕𝑌

𝜕𝑍
= 100 − 𝑥 − 𝑦 ≥ 0 − − − 3 𝜆1 ≥ 0 (100 − 𝑥 − 𝑦) (𝜆2 ) = 0
𝜕𝜆1

𝜕𝑍
= 40 − 𝑥 ≥ 0 − − − 4 𝜆2 ≥ 0 (40 − 𝑥 ) (𝜆2 ) = 0
𝜕𝜆2

Use trial and error: but ensure all inequalities hold. If they do not, try another solution
We can try: 𝑥=0 𝑦=0 → Not possible for 𝑢 = 0 ignore this and assume that both
𝑥 𝑎𝑛𝑑 𝑦 are non zero.
𝑦 − 𝜆1 − 𝜆2 ≤ 0 𝑥≥0 𝑖𝑓 𝑥 > 0 𝑡ℎ𝑒𝑛 𝑖𝑡 ℎ𝑎𝑠 𝑡𝑜 𝑏𝑒 𝑡ℎ𝑎𝑡
(𝑦 − 𝜆1 − 𝜆2 ) = 0 → 𝑓𝑜𝑟 𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑠𝑙𝑎𝑐𝑘𝑛𝑒𝑠𝑠 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 to hold.

𝑌 = 𝜆1 + 𝜆2 𝑥 − 𝜆1 = 0 𝑥 = 𝜆1

→ 𝑌 = 𝑥 + 𝜆2 → 𝑌 − 𝜆2 = 𝑋

𝐼𝑓 𝜆2 = 0 𝑡ℎ𝑒𝑛 𝑥 = 𝑦
𝐼𝑓 100 − 𝑥 − 𝑦 = 0
𝑡ℎ𝑒𝑛
100 − 2𝑥 = 0
2𝑥 = 100
𝑥 = 50
𝑇ℎ𝑖𝑠 𝑣𝑖𝑜𝑙𝑎𝑡𝑒𝑠 𝑡ℎ𝑒 𝑐𝑜𝑛𝑠𝑡𝑟𝑎𝑖𝑛𝑡 𝑥 ≤ 40 𝐹𝑜𝑟 𝑡ℎ𝑖𝑠 𝑡𝑜 ℎ𝑜𝑙𝑑 𝜆2 ≠ 0

Other alternative
𝜆2 > 0 ℎ𝑒𝑛𝑐𝑒:
𝑚𝑒𝑎𝑛𝑖𝑛𝑔 𝑥 = 40 𝑦 ∗ = 60 𝜆1 = 𝑥 = 40

100 − 𝑥 − 𝑦 ≥ 0
𝑦 = 𝑥 + 𝜆2
60 = 40 + 𝜆2 𝜆∗2 = 20

6.5 Kuhn Tucker conditions for a minimization case.

Let’s assume a firm wants to minimize the costs subject to ensuring that the quantity X is non-
negative. This can be written as:
𝑀𝑖𝑛 𝐶 = 𝑓(𝑋)
𝑠𝑡. 𝑋 ≥ 0
We can have three cases:
Y
f(x)

C
A B

In case A, the costs are minimized (the minimum point on curve) when X is greater than 0. It is
an interior solution. At the point where costs are minimum, X>0 and the first order derivative of
the cost function with respect to X is 0 i.e. 𝑋 > 0, 𝑓 ′ (𝑥 ) = 0.

In case B, the costs are minimized when X is equal to 0. It is a boundary solution. At the point
where costs are minimized, X=0 and the first order derivative of the cost function with respect to
X is 0 i.e. 𝑋 = 0, 𝑓 ′ (𝑥 ) = 0.

In case C, the costs are minimized when X is less than 0. Since X is not allowed to take negative
values, the firm will not produce. The firm will therefore operate at the point where X=0, and
first order derivative of the cost function with respect to X is positive i.e. 𝑋 = 0, 𝑓 ′ (𝑥 ) > 0.
These conditions can be summarized as:

a. 𝑓 ′ (𝑋) = 0 𝑓 ′ (𝑋 ) = 0 𝑓 ′ (𝑋 ) > 0
𝑋 >0 (𝑋 ) = 0 𝑋=0
In each of the cases,
𝑓 ′ (𝑋 ) ≥ 0 𝑋 ≥0 𝑓 ′ (𝑥 )(𝑋) = 0
Combined, these are the Kuhn tucker necessary conditions for a minimum
𝑓 ′ (𝑋) ≥ 0 𝑋 ≥0 𝑓 ′ (𝑋)(𝑋) = 0

Example two
𝑀𝑖𝑛:
𝐶 = (𝑥 − 4)2 + (𝑥2 − 4)2
𝑠𝑡.
2𝑥1 + 3𝑥2 ≥ 6
3𝑥1 − 2𝑥2 ≥ −12
𝑥1 , 𝑥2 ≥ 0

𝑍 = (𝑥1 − 4) 2 + (𝑥2 − 4) 2 + 𝑑1 (6 − 2𝑥1 − 3𝑥2 ) + 𝑑2 (−12 + 3𝑥1 + 2𝑥2 )

𝑑𝑧
= 2(𝑥1 − 4) − 2𝑑1 + 3𝑑2 ≥ 0 𝑥1 ≥ 0
𝜕𝑥1
𝜕𝑧
− 2(𝑥2 − 4) − 3𝑑1 + 2𝑑2 ≥ 0 𝑥2 ≥ 0
𝜕𝑑1
𝜕𝑧
= 6 − 2𝑥1 − 3𝑥2 ≤ 0
𝜕𝑑1
𝜕𝑧
= 12 + 3𝑥1 + 2𝑥2 ≤ 0
𝜕𝑑2

𝐴𝑠𝑠𝑢𝑚𝑒 𝑑1 > 0, 𝑑2 > 0

4⁄ 1⁄
2𝑥1 + 3𝑥2 = 6 𝑥1 = 4 5 𝑥2 = −1 5

3𝑥1 + 2𝑥2 = 12
This violates the non-negativity condition that
𝑥2 > 0
Assume both X1 and X2 are greater than 0
𝑥1 > 0 𝑥2 > 0

𝜕𝑍 𝜕𝑍
=0= =0
𝜕𝑥1 𝜕𝑥2

2(𝑥1 − 4) − 2𝑑1 + 3𝑑2 = 0 (2)

2𝑥2 − 4 − 3𝑑1 + 2𝑑2 = 0 (3)

4𝑥1 − 16 − 4𝑑1 + 6𝑑2 = 0

6𝑥2 − 24 = 9𝑑1 + 6𝑑2
4𝑥1 − 6𝑥2 + 5𝑑1 = 0
𝑑1 = 0
𝑥1 = 3⁄2 𝑥2 = −2

𝐴𝑠𝑠𝑢𝑚𝑒 𝑑2 ≠ 0

𝜕𝑍
=0 3𝑥1 + 2𝑥2 = 12
𝜕𝑑2

28 2 36 10⁄
𝑥1 = = 2 ⁄13 𝑥2 = 2 13 > 0
13 13

𝑑1 = 0 𝑑2 = 16⁄13

→ all choice variables are non-negative and all constraints are satisfied.
6.6 SUMMARY

Kuhn tucker first order conditions are important when

dealing with inequality constraints and non-negativity constraints in
non-linear programming.

NOTE

Complementary slackness conditions are very

important when determining the solutions that satisfy all the
constraints in non-linear programming.
6.7 ACTIVITIES

What are the Kuhn-tucker necessary conditions for a

minimum and a maximum?

6.8 FURTHER READING

References
1. Main Text: Chiang A.C. and Wainwright K:
Fundamental methods of mathematical Economics
2. Monga G.S: Mathematics and statistics for Economists,
second revised edition.
3. A cook-book of Mathematics, Viatcheslav
VINOGRADOV, June 1999
6.9 SELF-TEST QUESTIONS SIX

Max REVENUE =R = X1(10 -X1) + X2(20 - X2)

subject to 5X 1 + 3X 2 ≤40
X1 ≤ 5
X2 ≤10
X1 ≥ 0; X2 ≥0:
 LECTURE SEVEN: INTEGRATION

7.1 INTRODUCTION

This lecture is an advanced topic in integration that was learnt in

mathematics for economics two. It will help the learner to be able to
integrate exponential and logarithmic functions. It also introduces the
concept of integration by parts especially when one is dealing with
complex functions.

7.2 LECTURE OBJECTIVES

By the end of the lecture, the learner should be able to:

1. Integrate non-algebraic functions
2. Integrate economic functions by parts.
 7.3 Integration of non-algebraic functions.
Recall from the differentiation of non-algebraic functions:
If 𝑦 = 𝑒 𝑥
𝑑𝑦
= 𝑒𝑥
𝑑𝑥

And
if 𝑦 = 𝐿𝑛𝑥 then
𝑑 1
𝐿𝑛 (𝑥 ) =
𝑑𝑥 𝑥

7.1.1 Exponential Rule

∫ 𝑒 𝑥 . 𝑑𝑥 = 𝑒 𝑥 + 𝑐
∫ 𝑓 ′ (𝑥 )𝑒 𝑓(𝑥) 𝑑𝑥 = 𝑒 𝑓(𝑥) + 𝑐
7.1.2 Logarithmic rule
1
∫ 𝑑𝑥 = 𝐿𝑛𝑥 + 𝑐
𝑥
𝑓′ (𝑥)
∫ 𝑓(𝑥) 𝑑𝑥 = 𝐿𝑛 𝑓(𝑥 ) + 𝑐 𝑓 (𝑥 ) > 0

𝑜𝑟
𝐿𝑛 |𝑓 (𝑥 )| + 𝑐 𝑖𝑓 [𝑓(𝑥 ) = 0]

Example 1

∫(2𝑒 2𝑥 )𝑑𝑥

Using the exponential rule,

∫ 2𝑒 2𝑥 𝑑𝑥 = 𝑒 2𝑥 + 𝐶

Example 2
14𝑥
∫ 7𝑥2+5 𝑑𝑥
 14𝑥 𝑓 1 (𝑥 )
∫ . 𝑑𝑥 = ∫ . 𝑑𝑥 = 𝐿𝑛 𝑓(𝑥 ) + 𝑐
(7𝑥 2 + 5) 𝑓((𝑥 ))

= 𝐿𝑛 (7𝑥 2 + 5) + 𝐶

Example 3

∫(5𝑒 𝑥 − 𝑥 −2 + 3⁄𝑥 )𝑑𝑥

= ∫ 5𝑒 𝑥 . 𝑑𝑥 − ∫ 𝑥 −2 . 𝑑𝑥 + ∫ 3⁄𝑥 . 𝑑𝑥

= 5 ∫ 𝑒 𝑥 . 𝑑𝑥 = 5𝑒 𝑥 + 𝑐. +𝑥 −1 + 3𝐿𝑛𝑥 + 𝑐

7.4 Integration by substitution

Used if appropriate

𝑑𝑣
∫ 𝑓(𝑢). . 𝑑𝑥 = ∫ 𝑓(𝑢). 𝑑𝑣 = 𝑓(𝑢) + 𝑐
𝑑𝑥

Comes from chain rule

𝑑 𝑓 (𝑣 ) 𝑑 𝑑𝑣 𝑑𝑣 𝑑𝑣
= 𝑓 (𝑢 ). = 𝑓 1 (𝑢 ). = 𝑓(𝑢)
𝑑𝑥 𝑑𝑢 𝑑𝑥 𝑑𝑥 𝑑𝑥

𝑑𝑣
∫ [𝑓(𝑣 ). ] 𝑑𝑥 = 𝑓(𝑢 ) + 𝑐
𝑑𝑥

e.g.
∫ 2𝑥. (𝑥 2 + 1)𝑑𝑥 = 𝑙𝑒𝑡 𝑢 = 𝑥 2 + 1
𝑑𝑢
= 2𝑥
𝑑𝑥
 𝑑𝑢
𝑑𝑥 =
2𝑥

𝑑𝑢
∫𝑢 = ∫ 𝑢𝑑𝑢 =
𝑥
𝑢2
= + 𝑐1
2
(𝑥 2 + 1) 2 𝑥4 1
= + 𝑐1 = + 𝑥 2 + [ + 𝑐1 ]
2 1 2
𝑥4
+ 𝑥2 + 𝑐
2

Definite integral after substitution

2
2𝑥
∫ . 𝑑𝑥
𝑥2 + 3
1

𝐿𝑒𝑡 𝑢 = 𝑥 2 + 3

𝑑𝑢 𝑑𝑣
= 2𝑥 = 𝑑𝑥
𝑑𝑥 2𝑥

2
2𝑥 𝑑𝑣
∫
𝑢 2𝑥
1

2
𝑑𝑣
∫ = [𝐿𝑛 𝑢 + 𝑐]
𝑢
1

𝑥 = 2.
𝑢 = 22 + 3 = 4 + 3 = 7

𝑥=1
𝑢=1+3=4
[Lnv + 𝑐]74 = 𝐿𝑛 |7| − Ln|4|

*Ensure you change the limits.

Definite Integral as on area under the curve

𝑦 = 𝑓 (𝑥 )

𝑦 = 𝑓 (𝑥 )
𝑦 = 𝑓 (𝑥 )

0 2

∫ 𝑓1 (𝑥 ). 𝑑𝑥 = [𝑓(𝑥 ) + 𝑐]20
0
9 𝑏

∫ 𝑓(𝑥 ). 𝑑𝑥 = − ∫ 𝑓 (𝑥 ) . 𝑑 (𝑥 )
𝑏 𝑎
9

∫ 𝑓(𝑥 ) 𝑑𝑥 = 0
𝑎

𝑏 𝑏

∫ 𝐾 𝑓(𝑥 ) 𝑑𝑥 = 𝐾 ∫ 𝑓(𝑥 ). 𝑑𝑥
𝑎 𝑎
𝑏 𝑏

∫ −𝑓(𝑥 ) 𝑑𝑥 = − ∫ 𝑓 (𝑥 )𝑑𝑥
𝑎 𝛼

∫ 6𝑥 2 (𝑥 3 + 2) 99𝑑𝑥 = 2 ∫ 3𝑥 2 (𝑥 3 + 2)99 . 𝑑𝑥
 𝑑𝑣
𝑢 = 𝑥3 + 2 𝑑𝑥 =
3𝑥 2
𝑑𝑢
= 3𝑥 2
𝑑𝑥

= 2 ∫ 𝑢 99 . 𝑑𝑢

2. 𝑢100 1 100
+𝑐 = 𝑢 +𝑐
100 50

1
= (𝑥 3 + 2)100
50

∫ 8𝑒 2𝑥+3 .dx 𝑢 = 2𝑥 + 3
𝑑𝑣 𝑑𝑣
=2 𝑑𝑥 =
𝑑𝑥 2

𝑑𝑢
= 4𝑒 𝑢 = ∫ 4𝑒 𝑢 . 𝑑𝑣
1

4𝑒 𝑢 + 𝑐
4𝑒 2𝑥+3 +c

𝑑𝑣
This rule applies whenever it is possible to express the integral as 𝑓 𝑢 .
𝑑𝑥

∫ 𝑑 (𝑥 ) = 𝑥

∫(29𝑋 + 𝑏 )(𝛼𝑥 2 + 𝑏𝑋)7

Substitution 𝑢 = 𝛼𝑥 2 + 𝑏𝑥
𝑑𝑣
= 29𝑥 + 𝑏
𝑑𝑥
𝑑𝑣
𝑑𝑥 = (29𝑥+𝑏)

= ∫ 𝑢 7 . 𝑑𝑢
 8
𝑢8 (𝑎𝑥 2 +𝑏4)
+ 𝑐. = +𝑐
8 8

7.5 Integration by Parts

The rule applied when integrating by parts is given as:

∫ 𝑣. 𝑑𝑣 = 𝑢𝑣 − ∫ 𝑢. 𝑑𝑣

Comes from the product rule:

i.e.
𝑑(𝑢𝑣 ) = 𝑣. 𝑑𝑢 + 𝑢𝑑𝑣

∫ 𝑑 (𝑢𝑣 )

∫ 𝑣. 𝑑𝑢 + ∫ 𝑢𝑑𝑣

𝑢𝑣 = ∫ 𝑣. 𝑑𝑢 + ∫ 𝑢. 𝑑𝑣

𝑢𝑣 − ∫ 𝑢 𝑑𝑣 = ∫ 𝑣. 𝑑𝑢

Example 1
1⁄
∫ 𝑥 (𝑥 + 1) 2 . 𝑑𝑥 → Substitution cannot be applied.

1⁄
𝑑𝑣 = (𝑥 + 1) 2 𝑑𝑥
1
(𝑥 + 1) ⁄2+1
𝑢=
3⁄
2
2 3
(𝑥 + 1) ⁄2
3

𝐿𝑒𝑡 𝑣 = 𝑥
1⁄
𝑑𝑣 = (𝑥 + 1) 2 . 𝑑𝑥

𝑑𝑣 1
∫ = (𝑥 + 1) ⁄2
𝑑𝑥
 1⁄
∫ 𝑥 (𝑥 + 1) 2 𝑑𝑥

𝑑𝑣
= 1 𝑑𝑣 = 𝑑𝑥
𝑑𝑥

2 3 3
= (𝑥 + 1) ⁄2 . 𝑥 − ∫ 2⁄3 (𝑥 + 1) ⁄2 . 𝑑𝑥
3
4 5
= − (𝑥 + 1) ⁄2 + 𝑐
15

Find
1⁄
∫(𝑥 + 3)(𝑥 + 1) 2 𝑑𝑥

𝑣 = 𝑥+3
𝑑𝑣
=1 𝑑𝑣 = 𝑑𝑥
𝑑𝑥
1⁄
𝑑𝑣 = (𝑥 + 1) 2

3⁄
2(𝑥 + 1) 2
𝑢=
3

∫ 𝑣. 𝑑𝑢 = 𝑢𝑣 − ∫ 𝑢. 𝑑𝑣
3⁄
2(𝑥 + 1) 2 4 5
= (𝑥 + 3) − (𝑥 + 1) ⁄2 + 𝑐
3 15

Example 2

∫ 𝑥 3 . (𝐿𝑛𝑥 2 ) 𝑑𝑥

= 𝑢𝑣 − ∫ 𝑣. 𝑑𝑢

𝑥4 𝑥4 2
𝐿𝑛𝑥 2 − ∫ . 2 . 𝑑𝑥
4 4 𝑥
𝑥4 𝑥4
= 𝐿𝑛𝑥 2 − +𝐶
4 4𝑥2
𝑥4 𝑥4
= 𝐿𝑛𝑥 2 − + 𝐶
4 8
7.6 SUMMARY

In this lecture, we have learnt how to integrate non-algebraic

functions and also integrating by parts. The knowledge of integrating
the non-algebraic functions will be very helpful when looking at the
next topic of differential functions.

NOTE

When integrating logarithmic functions by parts you should always

make the log function the part that will be differentiated.

7.7 ACTIVITIES

Integrate the following functions:

1. ∫ 𝐿𝑛 𝑋. 𝑑𝑥 2. ∫ 𝑒 𝑡−1 . 2𝑡 + 5𝑡 2 . 𝑑𝑡
7.8 FURTHER READING

References
4. Main Text: Chiang A.C. and Wainwright K:
Fundamental methods of mathematical Economics
5. Monga G.S: Mathematics and statistics for Economists,
second revised edition.
6. A cook-book of Mathematics, Viatcheslav
VINOGRADOV, June 1999

7.9 SELF-TEST QUESTIONS SEVEN

Problem question seven

Integrate the following function by parts
1⁄
∫(𝑥 + 3)(𝑥 + 1) 2 𝑑𝑥
LECTURE EIGHT: FIRST ORDER DIFFERENTIAL EQUATIONS

8.1 INTRODUCTION

Differential Equations are very important in Economics. In order to

be able to solve differential equations, the knowledge derived from
integral calculus will be very important.

8.2 LECTURE OBJECTIVES

By the end of this lecture, the learner should be able to:

1. Formulate a differential equation.
2. Solve homogenous and non-homogenous first differential
equations.
3. Solve exact differential equations.
 8.3 Definition of Differential Equations
A differential equation helps in expressing the relationship between a variable with its past
value/s if it changes continuously with time. A differential equation is an equation involving
derivatives. It is an implicit functional relationship between a variable and its differentials.
For example
𝑑𝑦
= −3𝑦
𝑑𝑡

- The order of a Differential Equation is the highest order of any of the derivatives
contained in it. It’s said to be of order 𝑛 if the 𝑛 𝑡ℎ derivative is the highest order
derivative in it.
𝑑𝑦
𝑦= + 𝑡 2 = 𝑡 This is a 𝐷𝐸 of order 1
𝑑𝑡

- Highest power to which the derivative of the highest order occurs is the degree of a
differential equation. For example:

𝑑2 𝑦 3 𝑑𝑦 2
( 2
) − 6 ( ) + 𝑥𝑦 → Order 2, degree 3
𝑑𝑡 𝑑𝑡

A differential equation is linear if the dependent variable is of 1 st degree e.g.

𝑑𝑛 𝑦 𝑑𝑛−1 𝑦
𝛼0 𝑛 + 𝛼1 + 𝛼𝑛 𝑦 = 𝑓 ( 𝑡 )
𝑑𝑡 𝑑𝑡 𝑛−1
-Linear 𝑑. 𝐸 of order 𝑛

Assuming 𝛼0 , 𝛼1 and 𝛼𝑛 are constants and 𝑓 (𝑥 ) = 0 the equation is said to be a linear

homogenous differential equation of order 𝑛.

𝑑𝑦⁄ st
If 𝑑𝑡 appears only in the 1 degree and so does the dependent variable 𝑦 and we do not
have any product of the form:
 𝑑𝑦
𝑦 ( ⁄𝑑𝑡) the equation is linear.

8.4 First order linear differential equation (FOLDE)

If
𝑑𝑦
+ 𝑢 (𝑡 )𝑦 = 𝑤 (𝑡 )
𝑑𝑡

If : 𝑢 (𝑡) = constant ie. Coefficient of 𝑦 is constant and 𝑤 is a constant additive term then we
have a FOLDE with constant coefficient and constant term.
Given a differential equation of the form:
𝑑𝑦
+ 𝛼𝑦 = 𝑂
𝑑𝑡
Since 𝑤(𝑡 ) = 𝑂, we say it is a homogenous differential equation.

Solution to a differential equation

Involves getting all solutions and if it does not have one, to show it doesn’t have a solution.
A differential equation of order 𝑛 has a general solution containing 𝑛 independent
constants.
A particular solution is obtained from the general solution, by assigning specific values to the
arbitrary constants which are obtained from the information given as initial conditions.

Example
𝑑𝑦
1. If: = 𝑓 (𝑡 )
𝑑𝑡
𝑑𝑦
=𝑦
𝑑(𝑡)

In order to get back Y, we need to integrate the function

∫ 𝑑𝑦 = ∫ 𝑦 . 𝑑𝑥 𝑦 = ∫ 𝑓(𝑡). 𝑑𝑡 + 𝐶
𝑦 = 𝐹 (𝑡) + 𝐶.

𝑑𝑦
2. . 𝑎𝑦=0
𝑑𝑡
𝑑𝑦
∫ = ∫ 𝑎𝑑𝑡
𝑦
 𝐿𝑛 𝑌 = 𝑎𝑡 + 𝐶
𝑦 = 𝑒 𝑎𝑡+𝐶
= 𝑒 𝛼𝑡 . 𝑒 𝐶 𝑙𝑒𝑡 𝑒 𝐶 = 𝐶
𝑦 = 𝐶 𝑒 𝑎𝑡

𝑑𝑦
3. = 𝑡𝑒 𝑡
𝑑𝑥

∫ 𝑑𝑦 = ∫ 𝑡𝑒 𝑡 . 𝑑𝑡

𝑦 = (𝑡 − 1) (𝑒 𝑡 ) + 𝐶 → 𝐺𝑒𝑛𝑒𝑟𝑎𝑙 𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑦 = 2 𝑤ℎ𝑒𝑛 𝑡 = 0
2 = (𝑡 − 1) (𝑒 0 ) + 𝐶
2 = −1.1 + 𝐶 𝐶=3
𝑦 = (𝑡 − 1) 𝑒 𝑡 + 3

8.5 Non Homogenous differential functions

𝑑𝑦
+ 𝛼𝑦 = 𝑏 − 𝑟𝑒𝑑𝑢𝑐𝑒𝑑 𝑓𝑜𝑟𝑚
𝑑𝑡

The Solution is composed of 2 parts:

1. Complementary function 𝑌𝑐 and
2. Particular integral (𝑌𝑝 )
If 𝑦 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡, then,
𝑑𝑦
=𝑂
𝑑𝑡
Then
𝑎𝑦 = 𝑏
𝑦 = 𝑏⁄𝑎
𝑌𝑝 = 𝑏⁄𝑎 𝑎≠𝑜

𝑌 (𝑡) = 𝑌𝑐 + 𝑌𝑝 = 𝐴𝑒 −𝑎𝑡 + 𝑏⁄𝑎

𝑑𝑦
To solve, assume = 0. Whats 𝑌?
𝑑𝑡

The solution is the particular integral that ensures that A has a definite value,
 When 𝑌 (𝑡 ) = 𝐴𝑒 −𝑎𝑡 + 𝑏⁄𝑎

𝑌 (𝑜 ) = 𝐴 + 𝑏⁄𝑎 𝐴 = 𝑦𝑜 − (𝑏⁄𝑎)

𝑌 (𝑡) = [𝑦 (𝑜 ) − 𝑏⁄𝑎] 𝑒 −𝑎𝑡 + 𝑏⁄𝑎

8.6 Separable differential equations

These are equations that can be specifically separated into two parts, one a derivative of one
variable and the other a derivative of another variable. For example:
𝑑𝑦
𝑀 (𝑡 ) + 𝑁 (𝑦 ) =𝑂
𝑑𝑡
𝑀 (𝑡) . 𝑑𝑡 + 𝑁 (𝑦) 𝑑𝑦 = 𝑂

∫ 𝑀 (𝑡) . 𝑑𝑡 + ∫ 𝑁 (𝑦) 𝑑𝑦 = 𝑂

Example
𝑑𝑦 1
. = −𝛼
𝑑𝑡 𝑦
𝑑𝑦
∫ = ∫ 𝑎𝑑𝑡
𝑦
𝐿𝑛𝑦 = −𝑎𝑡 + 𝐶
𝑦 = 𝑒 −𝑎𝑡+𝑐
= 𝑒 −𝑎𝑡 𝑒 𝑐 𝐿𝑒𝑡 𝐴 = 𝑒 𝑐
Then
Y = Αe−at − General solution
At t𝑖𝑚𝑒 𝑡 = 𝑂
𝑌 = Αe−𝑜
= Α
𝑦 = 𝑦(𝑜 ) 𝑒 −𝑎𝑡 − 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛

Definite solution of a differential equation

𝑦 (𝑜 ) – Is the only one that make solution satisfy initial condition.
The solution is not numerical rather it is a function of 𝑡.

𝑦𝑡 is free of any derivative or differential expressions

e.g.
𝑑𝑦 𝑦
=
𝑑𝑡 𝑡
𝑑𝑦 𝑑𝑡
=∫ = ∫
𝑦 𝑡
𝐿𝑛 𝑌 = 𝐿𝑛𝑡 + 𝐿𝑛𝐶 𝑌 = 𝑡𝑐
𝑌 = 𝑒 𝐿𝑛𝑡+𝐿𝑛𝑐 𝑌=𝑥
If 𝑌 = 1 𝑤ℎ𝑒𝑛 𝑡 = 1
=
Then
1 = 1 + 𝐿𝑛𝑐
𝐿𝑛𝑐 = 𝑂. 𝐶 = 1

𝑑𝑦
3𝑡 2 + 2𝑡 − 3𝑦 =0
𝑑𝑡
3𝑡 2 . 𝑑𝑡 + 2𝑡. 𝑑𝑥 − 3𝑦 𝑑𝑦 = 0
3𝑡 3 2𝑡 2 3𝑦 2
+ + 𝐶= + 𝐶
3 2 2
𝑡𝑥 3 + 𝑡 2 = 3⁄2 𝑦 2 + 𝐶

2 +2 2
√ (𝑡 3 + 𝑡 2 ) 𝐶 = √𝑦 2 𝑦 = √ (𝑡 3 + 𝑡 2 ) + 𝐶
3 3 3
𝑑𝑦
= 𝑒 𝑡+𝑦
𝑑𝑡
𝑑𝑦
= 𝑒𝑡 . 𝑒𝑦
𝑑𝑡
𝑑𝑦
∫ 𝑦 = ∫ 𝑒 𝑡 . 𝑑𝑡
𝑒
 = ∫ 𝑒 −𝑦 𝑑𝑦 = ∫ 𝑒 𝑡 . 𝑑𝑡

−𝑒 −𝑦 = 𝑒 𝑡 + 𝐶
𝑒 𝑡 + 𝑒 −𝑦 = 𝐶
Example
𝑑𝑦 𝑦
=
𝑑𝑡 𝑡
𝑑𝑦 𝑑𝑡
=
−𝑦 𝑡
𝑑𝑦 𝑑𝑡
−∫ = ∫
𝑦 𝑡
−𝐿𝑛 𝑦 = 𝐿𝑛𝑡 + 𝐶
𝐿𝑛𝑡 + 𝐿𝑛𝑦 = 𝐶
𝐿𝑛 (𝑡𝑦) = 𝐶

8.7 Exact Differential Equations

𝑑𝑦 𝑑𝑚 𝑑𝑛
If 𝑀(𝑡, 𝑦) + 𝑁 (𝑡, 𝑦) . = 0 We say the function is exact if =
𝑑𝑡 𝑑𝑦 𝑑𝑡
𝑑𝑚 𝑑𝑛
= is a necessary and sufficient condition for the equation to be exact.
𝑑𝑦 𝑑𝑡

Solution of E.DE

𝐹 (𝑦, 𝑡 ) = ∫ 𝑀. 𝑑𝑦 + 𝜇(𝑡)

𝐼𝑓 𝑀. 𝑑𝑦 + 𝑁𝑑𝑡 = 0
𝑡ℎ𝑒𝑛

𝐹 (𝑦, 𝑡 ) = ∫ 𝑚𝑑𝑦 + 𝜇(𝑡)

𝐸. 𝑔. 𝜇(𝑡) Takes care of any terms containing (𝑡) that dropped out in the process of
differentiating
2𝑦𝑡. 𝑑𝑦 + 𝑦 2 . 𝑑𝑡 = 0
𝑀 = 2𝑦𝑡
 2𝑦 2
∫ 𝑚. 𝑑𝑦 = 𝑡 = 𝑦2 𝑡
2

Steps
1. 𝐹 (𝑦, 𝑡 ) = 𝑦 2 𝑡 + 𝜇 (𝑡 ) − 𝑝𝑟𝑒𝑙𝑖𝑚𝑖𝑛𝑎𝑟𝑦 𝑟𝑒𝑠𝑢𝑙𝑡

𝑑𝑓 (𝑦,𝑡)
2. = 𝑁 = 𝑦 2 + 𝑢 ′ (𝑡 ) = 𝑦 2
𝑑𝑡

3. 𝑢 ′ (𝑡 ) = 0
𝐼𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑒 𝑢 ′ (𝑡) 𝑡𝑜 𝑔𝑒𝑡 𝑢 (𝑡)

𝜇(𝑡) = ∫ 𝑢 ′ (𝑡) = ∫ 0. 𝑑𝑡 = 𝐾

4. 𝐹 (𝑦, 𝑡 ) = 𝑦 2 𝑡 + 𝐾 = 𝐶
𝑦2 𝑡 = 𝐶
𝑐
𝑦2 =
𝑡
𝑐
𝑦 = ( ) 1⁄2
𝑡
1⁄
= (𝑐𝑡 −1 ) 2

2𝑦𝑡 3 3𝑦 2 𝑡 2
𝑑𝑦 + 𝑑𝑡 = 0
𝑚 𝑁
𝑑𝑚 𝑑𝑁
=
𝑑𝑡 𝑑𝑦
6𝑦𝑡 2 6𝑦𝑡 2

𝐹 (𝑦, 𝑡 ) = ∫ 𝑀. 𝑑𝑦 + 𝜇(𝑡)

= ∫ 2𝑦𝑡 3 . 𝑑𝑦 + 𝑢 (𝑡)

2𝑦 2 𝑡 3
= + 𝑢 (𝑡 )
2

𝐹 (𝑦, 𝑡 ) = 𝑦 2 𝑡 3 + 𝑢 (𝑡)
𝑑𝑓
= 𝑁 = 3𝑦 2 𝑡 2 + 𝑢 ′ (𝑡) = 3𝑦 2 𝑡 2
𝑑𝑡
𝑢1 ( 𝑡 ) = 0

𝐹 (𝑦, 𝑡 ) = 3𝑦 2 𝑡 2 + 𝐾 = 𝐶
3𝑦 2 𝑡 2 = 𝐶
𝑡2
𝑦2 = − +𝑐
3
1 2 1⁄ 1
𝑦=− (𝑡 ) 2 + 𝐶 = − 𝑡 + 𝐶
3 3
𝑀 = 2𝑦𝑡 3
2𝑦 2 𝑡 3
∫ 2𝑦𝑡 3 . 𝑑𝑦 = + 𝑢 (𝑡 )
2
𝐹 (𝑦1 𝑡) = 𝑦 2 𝑡 3 + 𝑢 (𝑡)
𝑑𝐹
= 3𝑦 2 𝑡 2 + 𝑢1 (𝑡 ) = 𝑁
𝑑𝑡
3𝑦 2 𝑡 2 + 𝑢1 (𝑡) = 3𝑦 2 𝑡 2
3𝑦 2 𝑡 2 − 3𝑦 2 𝑡 2 = 𝑢1 𝑡 = 0
𝑢1 ( 𝑡 ) = 0
𝑢 (𝑡 ) = 𝐾

𝐹 (𝑦1 𝑡) = 𝑦 2 𝑡 3 + 𝐾 = 𝐶
𝑦2 𝑡3 = 𝐶
𝐶
𝑦2 =
𝑡3
1⁄
𝐶 2
𝑦 = ( 3)
𝑡

Homogenous differential equation (HDE)

If 𝑀(𝑡, 𝑦) 𝑑𝑥 + 𝑁 (𝑡, 𝑦) 𝑑𝑦 = 0 is homogenous first differential equation of degree 𝑛.degree
A 𝐻. 𝐷. 𝐸 is solved by reducing it to a separable equation after making necessary
substitution. If
𝑦 = 𝑡𝑣 𝑡ℎ𝑒𝑛 𝑑𝑦 = 𝑡. 𝑑𝑣 + 𝑣. 𝑑𝑡
𝑥𝑣 = 0

𝑑𝑦
= 𝑦2
𝑑𝑡
𝑑𝑦
= 𝑑𝑡
𝑦2
𝑑𝑦
∫ = ∫ 𝑑𝑡
𝑦2
𝐿𝑛 𝑦 2
=𝑡
2
𝐿𝑛 𝑦 2 = 2𝑡 + 𝐶
𝑦 2 = 𝑒 2𝑡+𝐶
𝑦 = 𝑒 𝑡+𝐶

𝑦 −2 𝑑𝑦 = 𝑑𝑡
2𝑦 −1
=𝑡
+1
𝑦 −1 = 𝑡 + 𝑐
(𝑦)−1𝑥−1 = (𝑡)−1
−1
𝑦=
𝑡
𝑦 (𝑡 ) = 𝑡 2 𝑦 (𝑡 )
𝑑𝑦
= 𝑡 2 𝑦 (𝑡 )
𝑑𝑡
𝑑𝑦
= 𝑡 2 . 𝑑𝑡
𝑦
𝑡3
𝐿𝑛𝑦 = + 𝐶
3
𝑡 3⁄ 𝑡 3⁄
𝑌 (𝑡) = 𝑐𝑒 3 𝑌 (𝑡) = 𝐾𝑒 3
 𝑑𝑦
= 𝑡2
𝑑𝑡
[𝑑𝑦 =]𝑡 2 . 𝑑𝑡
𝑡3
𝑌= +𝐾
3

𝑑𝑦
𝑌 = 𝐹 (𝑦) 𝑖. 𝑒. = 𝐹 (𝑦 ) → 𝑎𝑢𝑡𝑜𝑛𝑜𝑚𝑜𝑢𝑠 𝑜𝑟 𝑡𝑖𝑚𝑒 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑑. 𝐸
𝑑𝑡
e.g.
𝑑𝑦 𝑑𝑦
= 2𝑦 = 𝑦2
𝑑𝑡 𝑑𝑡
𝑑𝑦
= 𝑡 2 𝑦 → 𝑇𝑖𝑚𝑒 𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑛𝑜𝑛 − 𝑎𝑢𝑡𝑜𝑛𝑜𝑚𝑜𝑢𝑠 𝐷. 𝐸.
𝑑𝑡
→ 𝐴𝑙𝑙 𝑡ℎ𝑒𝑠𝑒 𝑎𝑟𝑒 1𝑠𝑡 𝑜𝑟𝑑𝑒𝑟 𝑑. 𝐸
8.8 Non linear D.E of 1st order 1st degree
𝑌 appears in a higher power or lower power not equal to 1 e.g.
𝑑𝑦
+ 𝑦2 𝑡 = 𝑡3
𝑑𝑡
Separable function e.g.
3𝑦 2 𝑑𝑦 − 𝑡 𝑑𝑡 = 0
𝑑𝑦
+ 3𝑦 2 𝑡. = 0 ∫ 3𝑦 2 𝑑𝑦 = − ∫ 𝑡 𝑑𝑡
𝑑𝑡
𝑑𝑦 3𝑦 3 𝑡2
+ 𝑡 𝑦 = 3𝑡𝑦 2 + 𝐶= − + 𝐶
𝑑𝑡 3 2
𝑑𝑦 + (𝑡𝑦 − 3𝑡𝑦 2 ) 𝑑𝑡 = 0 𝑡2
𝑦3 = − + 𝐶
𝑑𝑦 𝑦 − 3𝑦 2 2
+ 𝑡 ( ) 𝑑𝑡 = 0 1⁄
𝑦 − 3𝑦 2 𝑦 − 3𝑦 2 𝑡2 3
𝑦 3 = (− + 𝐶)
𝑑𝑦 2
∫ + ∫ 𝑡 𝑑𝑡 = 0
𝑦 − 3𝑦 2
 8.9 SUMMARY

In this lecture we have learnt how to get the general solution to first
order homogenous and non-homogenous differential equations.

NOTE

A differential equation has a general and a particular solution.

8.10 ACTIVITIES

Differentiate between a homogenous and non-

homogenous differential equation.?

8.11 FURTHER READING

 References
1. Main Text: Chiang A.C. and Wainwright K:
Fundamental methods of mathematical Economics
2. Monga G.S: Mathematics and statistics for Economists,
second revised edition.
3. A cook-book of Mathematics, Viatcheslav
VINOGRADOV, June 1999

8.12 SELF-TEST QUESTIONS EIGHT

1. Solve for Y in the following differential equation

𝑑𝑦 𝑡
=
𝑑𝑡 𝑦
𝑑𝑦
2. + 2𝑡𝑦 = 𝑡
𝑑𝑡
LECTURE NINE: FIRST ORDER DIFFERENCE EQUATIONS
9.1 INTRODUCTION

This lecture introduces the student to first order difference equations.

These are equations used to model; the relationship between a
variable over time, if the variable changes over discrete time
intervals. We will look at the applications of these equations in
economic relationships.

9.2 LECTURE OBJECTIVES

At this end of this lecture, the student should be able to:

1. Form difference equations from economic relationships.
2. Get general and particular solutions to first order homogenous
difference equations.
3. Get general and particular solutions to first order non-
homogenous difference equations
 9.3 Definition of a Difference Equation
A difference equation describes the relationship between a variable overtime. It describes the
relationship between dependent variables (for example investments) and the independent
variable time.

When we denote 𝑌 as ‘𝑌𝑡 ’ it means the 𝑌 is a discrete function. i.e. 𝑌 changes at fixed points
in time only.
On the other hand when we denote 𝑌 as 𝑌 (𝑡), it implies 𝑌is a continuous function which
changes continuously overtime.

If the following functions:

𝑌𝑡 = 5𝑡 2 + 10 → 𝑑𝑖𝑠𝑐𝑟𝑒𝑡𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
𝑦𝑡+1 = 5(𝑡 + 1) 2 + 15
𝑦(𝑡) = 20𝑡 2 + 18 → 𝑐𝑜𝑛𝑡𝑖𝑛𝑜𝑢𝑠 𝑓𝑢𝑛𝑡𝑖𝑜𝑛
A difference equation represents the relationship between a variable with its value in the
previous period e.g.
𝑦𝑡 = 𝑓 (𝑦𝑡−1 )
𝑦𝑡+1 = 𝑓 (𝑦𝑡 )

A difference equation is used to model dynamic systems in which changes occur at fixed
intervals, which are equally spaced.
For example:
1. If salary increments are 20% per annum, then in any given year, the salary will be 120%
of the salary in the previous year.
120 120
e.g.𝑦𝑡+1 = 𝑦𝑡 𝑜𝑟 𝑦𝑡 = 𝑦𝑡−1
100 100

100𝑦𝑡 − 120𝑦𝑡−1 = 0
𝑦𝑡 − 1.2𝑦𝑡−1 = 0
2. If a machine depreciates at the rate of 10% per annum, then the value of the machine in
year 𝑡 will be 90% of its value in year 𝑡 − 1 i.e.
90
𝑣𝑡 = 𝑣
100 𝑡−1
100𝑣 𝑡 − 90 𝑣𝑡−1 = 0
𝑣𝑡 − 0.9𝑣𝑡−1 = 0

Order of difference equations

This is the number of time intervals spanned by the difference equation.
e.g .𝑌𝑡 = 1.2𝑌𝑡−1 + 0.9𝑦𝑡−2 + 0.1𝑌𝑡−3 − 𝑜𝑟𝑑𝑒𝑟 3

𝑌𝑡 = 1.2𝑌𝑡−1 + 0.9𝑦𝑡−2 = − 𝑜𝑟𝑑𝑒𝑟 2

𝑃𝑡 = 0.9𝑃𝑡−1 = − 𝑜𝑟𝑑𝑒𝑟 1

Homogeneity of a difference equation

A difference equation said to be homogenous if the 𝑅𝐻𝑆 = 0
I.e. 𝑦𝑡 − 1.2𝑦𝑡−1 = 0
𝑦𝑡 − 1.2𝑦𝑡−1 − 0.9𝑦𝑡−2 = 0

On the other hand, if the 𝑅𝐻𝑆 is not equal to zero, it is classified as non-homogenous.
e.g. 𝑃𝑡+2 − 𝑃𝑡+1 − 𝑃𝑡 = 240

9.4 Solution to difference equations

The solution can be obtained using the iteration method: i.e.
If 𝑌𝑡+1 − 1.2𝑌𝑡 = 0
Get the values when 𝑡 = 1 ,2, 3, 4, 5 by assuming 𝑌 = 10,000 in year 1.
𝑌𝑡+1 − 1.2𝑌𝑡 = 0
𝑌𝑡+1 = 1.2𝑌𝑡 . 𝑡=1
𝑌1+1 = 1.2𝑌1 = 1.2 (10,000)
 𝑌2 = 12,000
𝑌3 = 1.2(𝑌2 )
= 1.2(12,000) = 14,400
𝑌4 = 1.2(14,400) = 17,280
𝑌5 = 1.2(17,280) = 20,736

9.5 General solution for a Homogenous first order difference equation

𝑌𝑡+1 − 1.2𝑌𝑡 = 0

If we get the income (𝑌 ) for years 1, 2, 3, 4, 5 in terms of 𝑌1 we will have:

𝑌𝑡+1 = 1.2𝑌𝑡 𝑡 = 1
𝑌2 = 1.2𝑌1
𝑌3 = 1.2[𝑌2 ] = 1.2[1.2𝑌1 ]1.22 𝑌1
𝑌4 = 1.2[𝑌3 ] = 1.2[1.22 ]𝑌1
= 1.23 𝑌1
In general
𝑌𝑡 = 1.2𝑡−1 𝑌1

Assume 𝑡 = 10 𝑌1 = 10,000
𝑌10 = 1.2(10−1) 𝑌1
= 1.29 (10,000)
= 5.15978 𝑥 10,000
= 51,597.8

General and particular solutions of difference equations

We have shown that:

𝑦𝑡 − 1.2𝑦𝑡−1 = 0 can be generalized as:

𝑌𝑡 = 1.2𝑡−1 𝑌1 …………………………1
This can be rewritten as:
𝑌𝑡 = 1.2𝑡 (1.2)−1 (𝑌1 )……………........2
= 1.2−1 . 𝑌1 . 1.2𝑡 ………………………3

Since 𝑌1 is a constant, (1.2)−1 is also a constant, we can combine them into one constant to be
represented as 𝐴 so that equation 3 becomes:

𝑌𝑡 = 𝐴 (1.2)𝑡 = 𝐴. 1.2𝑡 ……………………4

1.2 is also a constant. If we represent it with a small ′𝛼′ we can have equation 4 written as:
𝑌𝑡 = Α. 𝛼 𝑡 …………………………………5
This is called a general solution where ′𝛼′ will be determined from the difference equation.

When we evaluate equation 5 at the given values of 𝑡, and specific values of ′Α′we get a
particular solution.

Example:
Assume 𝑌𝑡+1 = 0.8𝑌𝑡
a) Find the general solution.
b) If 𝑌2 = 80, find the particular solution.
c) Evaluate 𝑌1 , 𝑌20 , 𝑌50 .

Solution
a) 𝑌𝑡 = Α(𝛼 )𝑡
𝑌𝑡+1 = Α(𝑎)𝑡+1 if 𝑌𝑡+1 − 0.8𝑌𝑡 = 0 then we can substitute this into equation of 𝑌𝑡 & 𝑌𝑡+1 to
get:
Α(𝛼 )𝑡+1 − 0.8 Α(α)t = 0
Α 𝛼 𝑡 𝛼 1 − 0.8 Α 𝛼 𝑡 = 0
Α αt α1 − 0.8 Ααt = 0
Α at [α1 − 0.8] = 0
 Either Α 𝛼 𝑡 = 0 or (𝛼 1 − 0.8) = 0
𝛼 = 0.8
If Α = 0 then
𝑌𝑡 = 0
If on the other hand 𝛼 = 0 then:
𝑌𝑡 = Α(0)𝑡 = 0 - This is called a trivial solution.
Therefore:
𝛼 = 0.8
𝑌𝑡 = Α. 0.8𝑡 General solution

b) If 𝑌2 = 80, find the particular solution

𝑌𝑡 = Α(0.8)t
𝑌2 = 80 = Α. 0.82
80 80
→ Α= = = 125
0.82 0.64

The particular solution is therefore:𝑌𝑡 = 125(0.8)𝑡

c) Evaluate 𝑌1 , 𝑌20 , 𝑌50

When 𝑡 = 1 𝑌1 = (125)(0.8) = 100
𝑡 = 20 𝑌20 = (125)(0.8)20 = 100
𝑡 = 50 𝑌50 = (125)(0.8)50 = 0.00178

9.6 Solution to Non-homogenous difference equations

If
𝑌𝑡+1 − 0.8𝑌𝑡 = 40 , the equation is called a non-homogenous difference equation.
The solution will consist of 2 parts:
1. Complementary function (CF)
2. Particular Integral (P.I)
i.e. 𝑌𝑡 = 𝐶𝐹 + 𝑃. 𝐼
Complementary Function
Solution to the homogenous part of the difference equation e.g.
If
𝑌𝑡+1 − 0.95𝑌𝑡 = 1000
then the complementary function is the solution to:
𝑌𝑡+1 − 0.95𝑌𝑡 = 0

Particular Integral
It is a function that satisfies the full difference equation. If the 𝑅𝐻𝑆 is fully a constant then, the
particular integral will be given as: 𝑌𝑡𝑝 = 𝐾(another constant).

If 𝑅𝐻𝑆 is constant 𝑥𝑏 𝑡 , then 𝑃𝐼 will be given as

𝑌𝑡,𝑝 = 𝐾 (𝑏 )𝑡

To show how we get the particular integral lets use the following example:

a) Find the General solution of:

𝑌𝑡+1 − 0.95𝑌𝑡 = 1000

b) Find the particular solution if 𝑌5 = 20,950

Complementary Function
𝑌𝑡+1 − 0.95𝑌𝑡 = 0
𝑌𝑡 = Α 𝛼 𝑡
𝑌𝑡+1 = Α 𝛼 𝑡+1
Α 𝛼 𝑡+1 − 0.95 Α𝛼 𝑡 = 0
Α 𝛼 𝑡 . 𝛼 − 0.95 Α𝛼 𝑡 = 0
Α𝛼 𝑡 (𝛼 − 0.95) = 0 𝛼 = 0.95
𝑌𝑡,𝐶 = Α (0.95) 𝑡 → This is the complementary function.
Particular Integral
𝑌𝑡,𝑝 = 𝐾
𝑌𝑡+1,𝑝 = 𝐾
Constant does not change irrespective of time
𝑌𝑡+1 − 0.95 𝑌𝑡 = 1000
(𝐾 − 0.95 𝐾) = 1000
0.05 𝐾 = 1000 𝐾 = 20,000
Therefore
𝑌𝑡,𝑃 = 20,000
General solution is
𝑌𝑡 = 𝑌𝑡𝐶 + 𝑌𝑡𝑃 = Α(0.95)𝑡 + 20,000
If:
𝑌5 = 20,950, then:
𝑌5 = 𝑌5,𝐶 + 𝑌5𝑝 = Α (0.95) 5 + 20,000 = 20,950
Α(0.95) 5 = 20,950 − 20000 = 950
950
Α= = 1228
(0.95) 5

The particular solution is therefore:

𝑌𝑡 = 1228(0.95)𝑡 + 20,000

9.7 SUMMARY

In this lecture we have learnt about first order difference

equations which are used to model variables which change discretely
with time. Solution to difference equations has also been learnt.
 NOTE

Most economic variables can be modeled using difference

equations. Difference equations are important when forecasting
variables in the future.

9.8 ACTIVITIES

1. Solve for 𝑌𝑡 by iteration method:

a) 𝑌𝑡+1 − 0.6 𝑌𝑡−1 = 0 Given 𝑌1 = 5 find 𝑌5

2. Find the general and particular solutions for”

𝑌𝑡+1 = 10𝑌𝑡 + 900 Given 𝑌0 = 20

3. Ι𝑡 − 0.9 Ι𝑡−1 = 60
Get the general solution
Given that Ι0 = 180 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 , get the particular solution
9.9 FURTHER READING

References
1. Main Text: Chiang A.C. and Wainwright K:
Fundamental methods of mathematical Economics
2. Monga G.S: Mathematics and statistics for Economists,
second revised edition.
3. A cook-book of Mathematics, Viatcheslav
VINOGRADOV, June 1999
9.10 SELF-TEST QUESTIONS

Practice question nine

Determine the general and particular solution of the following
difference equation
3𝑌𝑡+1 + 2𝑌𝑡 = 44 (0.8)𝑡 𝑌0 = 900
LECTURE TEN: STABILITY AND APPLICATIONS IN ECONOMICS

10.1 INTRODUCTION

The lecture discusses how to determine whether a difference

equation is stable.

10.2 LECTURE OBJECTIVES

By the end of the lecture the learner should be able to:

1. Determine the conditions necessary for a variable to be stable
2. Solve difference equations used in economics.
 10.3 Stability and the time path to stability
The general solution of a difference equation

𝑌𝑡 = Α(𝛼 )𝑡

 The expression can be used to forecast the dependent variable Y. To determine if Y will
stabilize i.e. approach some fixed value with time, we can trace how income changes each
year until stability is reached (trace the time path to stability).
The stability of the solution 𝑌𝑡 as ′𝑡′ increases are deduced from the exponential term (𝛼 )𝑡 .
𝛼 𝑡 as 𝑡 increases i.e. 𝑡 > 0.
If:
−∞ < 𝛼 < −1 then (−𝛼 )𝑡 → ±∞,
Solution is unstable e.g. (−2)1 = −2(−2)3 = −8
Time path alternates (−2)2 = 4(−2)4 = 64
If:
−1 < 𝛼 < 0
(−𝛼 )𝑡 → 0 e.g.(−0.5)1 = −0.5
(−0.5)2 = 0.25
(−0.5)3 − 0.125
Value decreases as 𝑡 → ∞
If
0<𝛼<1 e.g. 𝛼 = 0.2
(2)1 = 0.2
𝛼𝑡 → 0 (0.2)2 = 0.04
Solution is stable (0.2)3 = 0.008
Time path tends to zero (0.2)4 = 0.0064
1<𝛼<∞ e.g. 𝛼 = 2
Solution unstable 21 = 2
Time path tends to infinity 22 = 4
24 = 64
Examples
Find the general and particular solutions of the difference equations given as:

𝑌𝑡+1 − 2𝑌𝑡 = 0 𝑌1 = 900

𝑌𝑡 = Α(𝛼 )𝑡
𝑌𝑡+1 = Α𝛼 𝑡+1
Α 𝛼 𝑡+1 − 2 Α at = 0
Α 𝛼 𝑡 . 𝑎 − 2 Α at = 0
Α 𝑎𝑡 (𝛼 − 2) = 0 (𝛼 − 2) = 0 𝛼=2

General solution
𝑌𝑡 = Α(2)𝑡 Since 𝛼 = 2, the solution is unstable.

Particular solution
𝑌𝑡 = Α(2)t 𝑌1 = 900
900
900 = Α(2)1 Α= = 450
2

𝑌𝑡 = (450)(2) 𝑡
= 450 (2)t - Particular solution

Example 2:
𝑌𝑡+1 + 0.7𝑌𝑡 = 0 …. (1) 𝑌1 = 49
𝑌𝑡 = Α 𝑎𝑡
𝑌𝑡+1 = Α 𝛼 𝑡+1

Substituting this in equation 1:

Α 𝛼 𝑡+1 + 0.7 Α αt = 0
Α 𝛼 𝑡 . 𝛼 + 0.7 Α 𝛼 𝑡 = 0
Α 𝛼 𝑡 (𝛼 + 0.7) = 0 (𝛼 + 0.7) = 0
𝛼 = −0.7
Since
−1 < 𝛼 < 0
Solution is stable and time path alternates

𝑌𝑡 = Α (−0.7)𝑡 → General solution

𝑌1 = 49
49
𝑌1 = Α (−0.7)1 = 49 Α= = −70
−0.7

𝑌𝑡 = (−70)(−0.7)𝑡 - Particular solution

Applications of Difference equations:

Cobweb Model:
This is a microeconomic application of difference equations.
Consider the following demand and supply functions over time:
𝑄𝑑, 𝑡 = 𝛼 − 𝑏 𝑃𝑡 - Quality demanded is a function of the price at time𝑡.
𝑄𝑠, 𝑡 = 𝐶 + 𝑑 𝑃𝑡−1 - Quality supplied at time 𝑡 is a function of the price prevailing in past
period i.e. 𝑃𝑡−1 .

At equilibrium:
𝑄𝑑𝑡 = 𝑄𝑠, 𝑡
𝛼 − 𝑏 𝑃𝑡 = 𝐶 + 𝑑 𝑃𝑡−1
−𝑏 𝑃𝑡 − 𝑑 𝑃𝑡−1 = (𝐶 − 𝛼 ) This is a difference equation

Example:
𝑄𝑑𝑡 = 125 − 2𝑃𝑡
𝑄𝑠𝑡 = −50 + 1.5 𝑃𝑡−1

At equilibrium
125 − 2 𝑃𝑡 = −50 + 1.5 𝑃𝑡−1
125 + 50 = 105 𝑃𝑡−1 + 2 𝑃𝑡
→ 1.5 𝑃𝑡−1 + 2 𝑃𝑡 = 175
Since this is a non-homogenous difference equation, the solution will have two parts.
𝑃𝑡 = 𝐶. 𝐹 + 𝑃Ι
𝑃𝑡 = 𝑃𝑡,𝐶 + 𝑃𝑡,𝑃

Find the Complementary functions:

2 𝑃𝑡 + 1.5 𝑃𝑡−1 = 0
𝑃𝑡 = Α (𝛼 )𝑡
𝑃𝑡−1 = Α (𝛼 )𝑡−1
2 Α (𝛼 )𝑡 + 1.5 Α (𝛼 )𝑡−1 = 0
2 Α (𝛼 )𝑡 + 1.5 Α (𝛼 )𝑡 . 𝛼 −1 = 0
Α (𝛼 )𝑡 [2 − 1.5 𝛼 −1 ] = 0
1.5
2+ =0
𝛼
1.5
2=−
𝛼
1.5
𝛼=− = −0.75
2
Therefore:
𝑃𝑡,𝐶 = Α (−0.75)𝑡
Particular Integral
Since the right hand side is a constant, the
𝑃𝑡,𝑃 = 𝐾
𝑃𝑡−1,𝑃 = 𝐾

Hence: in 2𝑃𝑡 + 1.5𝑃𝑡−1 = 175

2(𝐾 ) + 1.5𝐾 = 175
3.5𝐾 = 175
175
𝐾= = 50
345
𝑃𝑡,𝑝 = 50
General solution
𝑃𝑡 = 𝑃𝑡,𝐶 + 𝑃𝑡,𝑃
𝑃𝑡 = Α (−0.75) 𝑡 + 50
Particular solution
If 𝑃0 = 60
Then
𝑃0 = Α (−0.75) 0 + 50
= Α + 50 = 60
Α = 10
𝑃𝑡 = 10(−0.75) 𝑡 + 50 → Particular solution
Is the price stable?
𝛼 = (−0.75) Since 𝛼 is between −1 and 𝑜, the price is stable.

1.2 SUMMARY

In this lecture we have learnt how to check for stability in

a difference equation, especially in the cobweb models.

NOTE

The stability of a difference equation is determined

by the value of ′𝑎′ in the following expression:
𝐴(𝑎)𝑡
1.3 ACTIVITIES

Practice question
1. 𝑌𝑡+1 + 𝑌𝑡 = 0 𝑌1 = 10
Find:

i. The general solution

ii. Determine if solution is stable
iii. Find the particular solution
2. Find the general and particular solution given that
Dt=10-2Pt St= -5+4Pt-1
P(0) = 10

1.7 FURTHER READING

References
1. Main Text: Chiang A.C. and Wainwright K:
Fundamental methods of mathematical Economics
2. Monga G.S: Mathematics and statistics for Economists,
second revised edition.
3. A cook-book of Mathematics, Viatcheslav
VINOGRADOV, June 1999
1.8 SELF-TEST QUESTIONS

Practice question 10
The supply and demand function of cabbage is given as:
𝑄𝑠𝑡 = −4 + 2 𝑃𝑡−1 𝑄𝑑𝑡 = 80 − 2 𝑃𝑡
Required:
i) Determine the equilibrium price

ii) Is it stable?

Place here the self-test questions for this chapter.

 Answers to self-test questions

Solution to self-test one

Let x1 = Number of hardcover books ( per 100) to be produced

X2 = Number of paperback books ( per 100) to be produced

Since the objective is to minimize the cost, the objective function is given by-

Minimise Z = 700x1 + 600x2

Subject to constraints:

2x1 + x2 ≥ 90 ( Minimum running of press I)

x1 +2x2 ≥ 70 ( Minimum running of press II)

x1 ,x2 ≥ 0 ( Non-negativity constraint)

Solution to self test two

• Let x1 = Number of chairs to be produced

• X2 = Number of tables to be produced

• Z=contribution

• Since the objective is to maximize the contribution, the objective function is given by

Maximise Z= 20x1+ 30x2

• Subject to constraints:

• 3x1 + 3x2 ≤ 36 ( Total time of machine M1)

 • 5x1 + 2x2 ≤ 50 ( Total time of machine M2)

• 2x1 + 6x2 ≤ 60 ( Total time of machine M3)

• X1 , x2≥ 0 [ Non-Negativity constraint)

• X1= X2 =

Solution to self test three

• Y=4

• X=2

• Z==44

Solution to self test four

 X=6 Y=4 Z=2

Solution to self test five

−2 4 0
• 𝐻 = [ 4 −18 0]
0 0 −2
• |H1 | < 0
• |H2 | > 0
• |H3 | < 0
• Z=Concave
Solution to self test six
• X1 =2; X2 = 10:
•
Solution to self test seven

• 𝐿𝑒𝑡 𝑣 = 𝑥 + 3
𝑑𝑣
• =1 𝑑𝑣 = 𝑑𝑥
𝑑𝑥
1⁄
• 𝑑𝑈 = (𝑥 + 1) 2
 3⁄
2(𝑥+1) 2
• 𝑢=
3

• ∫ 𝑣. 𝑑𝑢 = 𝑢𝑣 − ∫ 𝑢. 𝑑𝑣
3⁄
2(𝑥+1) 2 4 5⁄
• = (𝑥 + 3) − (𝑥 + 1) 2 +𝑐
3 15

Solution to self test eight

1. ∫ 𝑦. 𝑑𝑦 = ∫ 𝑡𝑑𝑡.
𝑦2 𝑡2
+ 𝐶= + 𝐶
2 2
𝑦2 𝑡2
− =𝐶
2 2
𝑦 2 − 𝑡 2 = 2𝐶
𝑦 2 = 2𝐶 + 𝑡 2
±
𝑦 = √2𝐶 + 𝑡 2
2 1
𝑌(𝑡) = 𝐴𝑒 −𝑡 +
2

Solution to self test nine

Complementary function
𝑡
𝑌𝑡,𝐶 = Α (− 2⁄3)

Particular Integral
Since the right hand side = 44(0.8)𝑡 , the general form of
𝑌𝑡,𝑝 = 𝐾 (0.8)𝑡
𝑌𝑡+1,𝑃 = 𝐾 (0.8)𝑡+1 = 𝐾 (0.8)𝑡 0.8
Substituting this in the original difference equation it will be:
3 𝑌𝑡+1 + 2 𝑌𝑡 = 44 (0.8)𝑡
3 𝐾. 0.8𝑡+1 + 2 𝐾 (0.8)𝑡 = 44 (0.8) 𝑡
3 𝐾 0.8𝑡 . 0.8 + 2 𝐾 0.8𝑡 = 44 (0.8)𝑡
𝐾. (0.8)𝑡 [3(0.8)1 + 2] = 44 (0.8)𝑡

Divide all through by (0.8)𝑡 , we get

𝐾 [2.4 + 2] = 44
4.4 𝐾 = 44
44
𝐾= = 10
4.4

Therefore:
𝑌𝑡,𝑃 = 10. (0.8)𝑡

General solution:
𝑌𝑡 = 𝑌𝑡,𝐶 + 𝑌𝑡,𝑃
𝑡
= Α (−2⁄3) + 10 (0.8)𝑡
Particular solution
𝑌0 = 900
0
𝑌0 = Α (−2⁄3) + 10 (0.8)0 = 900

Α + 10 = 900 Α = 890
t
Yt = 890 (−2⁄3) + 10 (0.8)𝑡

Solution to self-test ten

 Pt = A (−1)t + 21
 Solution is stable
 REFERENCES
1. Main Text: Chiang A.C. and Wainwright K: Fundamental
methods of mathematical Economics
2. Monga G.S: Mathematics and statistics for Economists, second
revised edition.
3. A cook-book of Mathematics, Viatcheslav VINOGRADOV, June
1999.
4. Tulsian and Pandey V. Quantitative Techniques, Theory and
Problems.
5. Gupta S.K. Cozzolino J.M. (1975) Fundamentals of Operations
Research for Management
6. Kothari C.R. (2004) Quantitative Techniques, 3 rd revised edition
7. Kothari C.R. An introduction to Operations Research, Vikas
Publishing Press.
8. Lucey T. (2004) Quantitative Techniques, 4th edition