Mathematical Economics – UNIT-4

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 Mathematical Economics UNIT -4

Introduction to Sequences and Series

Imagine yourself at a pizza hut. You have placed an order and
your order number is 282. So currently they are serving the
order number 275. So how many orders do you think will be
served before your number? Yes, six orders more because you
are in a sequence. To understand this better, let us learn about
the sequences and series. Let us do it right now.

Sequences

A sequence is a list of numbers in a special order. It is a string

of numbers following a particular pattern, and all the elements
of a sequence are called its terms. Let us consider a sequence,

[1, 3, 5, 7, 9, 11……]

We can say this is sequence because we know that they are the
collection of odd natural numbers. Here the number of terms in
the sequence will be infinite. Such a sequence which contains
the infinite number of terms is known as an infinite
sequence. But what if we put end to this.

[1, 3, 5, 7, 9, 11…..131]

If 131 is the last term of this sequence, we can say that the
number of terms in this sequence is countable. So in such
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sequence in which the number of terms is countable, such

sequences are called finite sequences. A finite sequence has the
finite number of terms. So as discussed earlier here 1 is the
term, 3 is the term so is 5, 7…..

Fibonacci Sequence
The special thing about the Fibonacci sequence is that the first
two terms are fixed. When we talk about the terms, there is a
general representation of these terms in sequences and series. A
term is usually denoted as an here ‘ n ‘ is the nth term of a
sequence. For the Fibonacci sequence, the first two terms are
fixed.

The first term is as a1= 1 and a2= 1. Now from the third term
onwards, every term of this Fibonacci sequence will become the
sum of the previous two terms. So a3will be given as a1 + a2

Therefore, 1 + 1 = 2. Similarly,
a4 = a2 + a3
∴1+2=3
a5 = a3+ a4
∴2+3=5

Therefore if we want to write the Fibonacci sequence, we will

write it as, [1 1 2 3 5…]. So, in general, we can say,

an = an-1 + an-2

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where the value of n ≥ 3.

Types of Sequences
 Arithmetic sequence: In an arithmetic (linear) sequence the
difference between any two consecutive terms is constant.
 Quadratic Sequence: A quadratic sequence is a sequence of
numbers in which the second difference between any two
consecutive terms is constant.
 Geometric Sequence: A geometric sequence is a sequence of
numbers where each term after the first is found by multiplying
the previous one by a fixed, non-zero number called the
common ratio.

Series
Series is the sum of sequences. The series is finite or infinite
according to the given sequence is finite or infinite. Series are
represented as sigma, which indicates that the summation is
involved. For example, a series S can be,

S = Sum (1, 3, 5, 7, 9, 11, …)

Solved Examples for You

Question: Identify the sequence of the following function n
(n+1)

A. 4, 10, 18, 28…

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B. 4, 12, 18, 28…

C. 2, 10, 18, 28…
D. 4, 10, 18, 28…
Solution: Correct option is A. The given function is n(n+1),

When n = 1, 1(1+3) = 4
n = 2, 2(2+3) = 10
n=3, 3(3+3) = 27
So, 4, 10, 27…is the function for the sequence n(n+3).

Question: Adding first 100 terms in a sequence is

A. term
B. series
C. constant
D. sequence
Solution: Correct option is B. Adding first 100 terms in a
sequence is series. Also adding the number of some set is a
series.

Arithmetic Progression
Suppose while returning from school, you get into the taxi.
Once you ride a taxi you will be charged an initial rate. But then
the charge will be per mile or per kilometre. This show that the
arithmetic sequence for every kilometre you will be charged a

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certain constant rate plus the initial rate. To understand this let
us study the topic of arithmetic progression in detail.

Arithmetic Progression (AP)

In an arithmetic sequence, the difference between any two

consecutive terms is constant. The constant difference is
commonly known as common difference and is denoted by d.
Let us understand this with one example.

Let’s check whether the given sequence is A.P: 1, 3, 5, 7, 9, 1.

To check if the given sequence is A.P or not, we must first
prove that the difference between the consecutive terms is
constant. So, d = a2– a1 should be equal to a3– a2 and so on…
Here,

d = 3 – 1 = 2 equal to 5 – 3 = 2

Here we can see that the difference is common to both the

terms. So we can say that the given sequences are in arithmetic
progression.

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The nth term of an AP

We can find the nth term of an AP by using the formula,

an = (a1 + (n – 1) d)

Here a1 is the first term and d is the common difference of an

AP. The constant difference between any two terms of an AP is
commonly known as common difference and is denoted by d.

Properties of Arithmetic Progression

Before understanding the properties let us first write the general
form of an A.P: a1 a2 a3……an and if its common difference is d.
If we add or subtract any constant, to every term of this
arithmetic progression, the resulting sequence will also be an
arithmetic progression. So,

(a1 + (n – 1) d) is the general (nth) term of an AP

S = [(n/2) * (2a1 + (n – 1) d)] is the sum of first n terms of an

AP

 If a constant number is added or subtracted from each term of

an A.P, then the resulting term in the sequence are also in A.P
with the same common difference.
 If each term in an A.P is divided or multiply with a constant
non-zero number, then the resulting sequence is also in an A.P

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 Three number x, y and z are in an A.P if 2y = x + z

 A sequence is an A.P if its nth term is a linear expression.
 If we select terms in the regular interval from an A.P, these
selected terms will also be in AP

Solved Examples for You

Question: If l = 20, d = -1 and n = 17, then the first term is?

A. 30
B. 32
C. 34
D. 36
Solution: The correct option is D
Given: l = 20, d = -1, n = 17
We need to find a. By using l = a + (n – 1)d,
∴ 20 = a + (17 – 1) -1
20 = a – 17
a = 36

Question: 30 trees are planted in a straight line at intervals of 55

metres. To water them, the gardener needs to bring water to
each tree, separately from a well, which is 10 metres from the
first tree in line with the trees. How far will he have to walk in
order to water all the trees beginning with the first tree? Assume
that he starts from the first well, and he can carry enough water
to water only one tree at a time.
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A. 4785 meters
B. 4795 meters
C. 4800 meters
D. None of these
Solution: Correct option is B. For the first tree, he has to walk =
10 meters. For the second tree = 25 meters. third tree =
35 meters and fourth tree = 45 meters. So the
total distance travelled = 10 + 25+ 35…..30 terms

= 10 + ( 25 + 35+ …29 terms)

= 10 + 29/2 (2 × 25 + 28 × 10) [S = n/2 * (2a + (n-1)d)

]

= 4795

Continuity
The property of continuity is exhibited by various aspects
of nature. The water flow in the rivers is continuous. The flow
of time in human life is continuous i.e. you are getting older
continuously. And so on. Similarly, in mathematics, we have
the notion of the continuity of a function.

What it simply means is that a function is said to be continuous

if you can sketch its curve on a graph without lifting your pen
even once (provided that you can draw good). It is a very

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straightforward and close to accurate definition actually. But for

the sake of higher mathematics, we must define it in a more
precise way. That’s what we are going to do in this section. So
let’s jump into it!

Definition of Continuity
A function f(x) is said to be continuous at a point x = a, in its
domain if the following three conditions are satisfied:

1. f(a) exists (i.e. the value of f(a) is finite)

2. Limx→a f(x) exists (i.e. the right-hand limit = left-hand limit, and
both are finite)
3. Limx→a f(x) = f(a)
The function f(x) is said to be continuous in the interval I =
[x1,x2] if the three conditions mentioned above are satisfied for
every point in the interval I.

However, note that at the end-points of the interval I, we need

not consider both the right-hand and the left-hand limits for the
calculation of Limx→a f(x). For a = x1, only the right-hand limit
need be considered, and for a = x2, only the left-hand limit needs
to be considered.

Some Typical Continuous Functions

 Trigonometric Functions in certain periodic intervals (sin x,
cos x, tan x etc.)

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 Polynomial Functions (x2 +x +1, x4 + 2….etc.)

 Exponential Functions (e2x, 5ex etc.)
 Logarithmic Functions in their domain (log10x, ln x2 etc.)

Discontinuity
If any one of the three conditions for a function to be continuous
fails; then the function is said to be discontinuous at that point.
On the basis of the failure of which specific condition leads to
discontinuity, we can define different types of discontinuities.

Jump Discontinuity

In this type of discontinuity, the right-hand limit and the left-

hand limit for the function at x = a exists; but the two are not
equal to each other. It can be shown as:

Limx→a+f(x)≠Limx→a−f(x)

Infinite Discontinuity
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The function diverges at x = a to give it a discontinuous nature

here. That is to say,

f(a) is not defined

Since the value of the function at x = a tends to infinity or doesn’t
approach a particular finite value, the limits of the function as x → a
are also not defined.

Point Discontinuity

This is a category of discontinuity in which the function has a

well defined two-sided limit at x = a, but either f(a) is not
defined or f(a) is not equal to its limit. The discrepancy can be
shown as:

Limx→af(x)≠f(a)

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This type of discontinuity is also known as a Removable Discontinuity

since it can be easily eliminated by redefining the function in such a
way that,
f(a)=Limx→af(x)

Solved Examples for You

Question: Let a function be defined as f(x) =

5 – 2x for x < 1
3 for x = 1
x + 2 for x > 1

Is this function continuous for all x?

Solution: Since for x < 1 and x > 1, the function f(x) is defined
by straight lines (that can be drawn continuously on a graph),
the function will be continuous for all x ≠ 1. Now for x = 1, let
us check all the three conditions:

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–> f(1) = 3 (given)

–> Left-Hand Limit:

=Limx→1−f(x)
=Limx→1−(5–2x)
=5–2×1
=3
–> Right-Hand Limit:
=Limx→1+f(x)
=Limx→1+(x+2)
=1+2
=3
–> Limx→1−f(x)=Limx→1+f(x)=3=f(1)
Thus all the three conditions are satisfied and the function f(x)
is found out to be continuous at x = 1. Therefore, f(x) is
continuous for all x.

This concludes our discussion on the topic of continuity of

functions. Continuous functions are very important as they are
necessarily differentiable at every point on which they are
continuous, and hence very simple to work upon.

Differential calculus
Differential calculus, Branch of mathematical analysis, devised by Isaac
Newton and G.W. Leibniz, and concerned with the problem of finding
the rate of change of a function with respect to the variable on which it
depends. Thus it involves calculating derivatives and using them to solve
problems involving nonconstant rates of change. Typical applications

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include finding maximum and minimum values of functions in order to

solve practical problems in optimization.

Applications Of Differential
Calculus
Optimisation problems
We have seen that differential calculus can be used to determine the
stationary points of functions, in order to sketch their graphs. Calculating
stationary points also lends itself to the solving of problems that require some
variable to be maximised or minimised. These are referred to as optimisation
problems.

The fuel used by a car is defined by f(v)=380v2−6v+245f(v)=380v2−6v+245,

where vv is the travelling speed in km/hkm/h.
What is the most economical speed of the car? In other words, determine the
speed of the car which uses the least amount of fuel.

If we draw the graph of this function we find that the graph has a minimum.
The speed at the minimum would then give the most economical speed.

Linear Algebra: Vectors, Matrices and their

properties
Introduction
Large datasets are often comprised of hundreds
to millions of individual data items. It is easier to
work with this data and operate on it when it is
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represented in the form of vectors and matrices.

Linear algebra is a branch of mathematics that
deals with vectors and operations on vectors.
Linear algebra is thus an important prerequisite
for machine learning and data processing
algorithms.

This tutorial covers the basics of vectors and

matrices, as well as the concepts that are
required for data science and machine learning.
It also introduces you terminology, such as "dot
product", "trace of a matrix", etc.

Vectors and their properties

What is a Vector?
Vectors can be thought of as an array of
numbers where the order of the numbers also
matters. They are typically represented by a
lowercase bold letter such as x. The individual
numbers are denoted by writing the vector

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name with subscript indicating the position of

the individual member. For example, x1 is the
first number, x2 is the second number and so
on. If we want to write the vector with the
members explicitly then we enclose the
individual elements in square brackets as,

We have seen that the coordinates of the turning point can be calculated by
differentiating the function and finding the xx-coordinate (speed in the case of
the example) for which the derivative is 00.
f′(v)=340v−6f′(v)=340v−6

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If we set f′(v)=0f′(v)=0 we can calculate the speed that corresponds to the

turning point:
f′(v)0v=340v−6=340v−6=6×403=80f′(v)=340v−60=340v−6v=6×403=80
This means that the most economical speed is 80 km/h80 km/h.
Finding the optimum point:

Let f′(x)=0f′(x)=0 and solve for xx to find the optimum point.

To check whether the optimum point at x=ax=a is a local minimum or a local
maximum, we find f′′(x)f″(x):
 If f′′(a)<0f″(a)<0, then the point is a local maximum.
 If f′′(a)>0f″(a)>0, then the point is a local minimum.
WORKED EXAMPLE 21: OPTIMISATION PROBLEMS

The sum of two positive numbers is 1010. One of the numbers is multiplied by the
square of the other. If each number is greater than 00, find the numbers that make this
product a maximum.
Draw a graph to illustrate the answer.

Analyse the problem and formulate the equations that are required

Let the two numbers be aa and bb and the product be PP.

a+bP=10……(1)=a×b2……(2)a+b=10……(1)P=a×b2……(2)
Make bb the subject of equation (11) and substitute into equation (22):
P∴P(a)=a(10−a)2=a(100−20a+a2)=100a−20a2+a3P=a(10−a)2=a(100−20a+a
2)∴P(a)=100a−20a2+a3

Differentiate with respect to aa

P′(a)=100−40a+3a2P′(a)=100−40a+3a2
Determine the stationary points by letting P′(a)=0P′(a)=0
We find the value of aa which makes PP a maximum:
P′(a)0∴a=10=3a2−40a+100=(3a−10)(a−10) or a=103P′(a)=3a2−40a+1000
=(3a−10)(a−10)∴a=10 or a=103

Substitute into the equation (11) to solve for bb:

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If a=10:b∴no
solutionIf a=103:b=10−10=0(but b>0)=10−103=203If a=10:b=10−10=0(b
ut b>0)∴no solutionIf a=103:b=10−103=203

Determine the second derivative P′′(a)P″(a)

We check that the point (103;203)(103;203) is a local maximum by showing
that P′′(103)<0P″(103)<0:
P′′(a)∴P′′(103)=6a−40=6(103)−40=20−40=−20P″(a)=6a−40∴P″(103)=6(103
)−40=20−40=−20
Write the final answer

The product is maximised when the two numbers are 103103 and 203203.

Draw a graph

To draw a rough sketch of the graph we need to calculate where the graph intersects
with the axes and the maximum and minimum function values of the turning points:

Intercepts:

P(a)Let P(a)=0:=a3−20a2+100a=a(a−10)2(0;0) and (10;0)P(a)=a3−20a2+1

00a=a(a−10)2Let P(a)=0:(0;0) and (10;0)

Turning points:

P′(a)∴a=103=0 or a=10P′(a)=0∴a=103 or a=10

Maximum and minimum function values:

Substitute (103;203):PSubstitute (0;10):P=ab2=(103)(203)2=400027≈1

48(Maximum turning point)=ab2=(10)(0)2=0(Minimum turning
point)Substitute (103;203):P=ab2=(103)(203)2=400027≈148(Maximum turning
point)Substitute (0;10):P=ab2=(10)(0)2=0(Minimum turning point)

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Note: the above diagram is not drawn to scale.

WORKED EXAMPLE 22: OPTIMISATION PROBLEMS

Michael wants to start a vegetable garden, which he decides to fence off in the shape of
a rectangle from the rest of the garden. Michael has only 160 m160 m of fencing, so
he decides to use a wall as one border of the vegetable garden. Calculate the width and
length of the garden that corresponds to the largest possible area that Michael can
fence off.

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Examine the problem and formulate the equations that are required

The important pieces of information given are related to the area and modified perimeter
of the garden. We know that the area of the garden is given by the formula:

Area =w×lArea =w×l

The fencing is only required for 33 sides and the three sides must add up
to 160 m160 m.
160=w+l+l160=w+l+l
Rearrange the formula to make ww the subject of the formula:
w=160−2lw=160−2l
Substitute the expression for ww into the formula for the area of the garden. Notice that
this formula now contains only one unknown variable.
Area =l(160−2l)=160l−2l2Area =l(160−2l)=160l−2l2
Differentiate with respect to ll
We are interested in maximising the area of the garden, so we differentiate to get the
following:

dAdl=A′=160−4ldAdl=A′=160−4l
Calculate the stationary point

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To find the stationary point, we set A′(l)=0A′(l)=0 and solve for the value(s) of ll that
maximises the area:
A′(l)04l∴l=160−4l=160−4l=160=40A′(l)=160−4l0=160−4l4l=160∴l=40
Therefore, the length of the garden is 40 m40 m.
Substitute to solve for the width:

w=160−2l=160−2(40)=160−80=80w=160−2l=160−2(40)=160−80=80
Therefore, the width of the garden is 80 m80 m.
Determine the second derivative A′′(l)A″(l)
We can check that this gives a maximum area by showing that A′′(l)<0A″(l)<0:
A′′(l)=−4A″(l)=−4

Linear Algebra/Introduction to Matrices and

Determinants
Matrices

Organization of a matrix

Informally an m×n matrix (plural matrices) is a rectangular table of entries from

a field (that is to say that each entry is an element of a field). Here m is the
number of rows and n the number of the columns in the table. Those
unfamiliar with the concept of a field, can for now assume that by a field of
characteristic 0 (which we will denote by F) we are referring to a particular
subset of the set of complex numbers.

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An m×n matrix (read as m by n matrix), is usually written as:

The row is an element of , showing the n components . Similary

the column is an element of showing the m components .

Here m and n are called the dimensions of the matrix. The dimensions of a
matrix are always given with the number of rows first, then the number of
columns. It is also said that an m by n matrix has an order of m×n.

Formally, an m×n matrix M is a function where A = {1,2...m} × {1,2...n}

and F is the field under consideration. It is almost always better to visualize a
matrix as a rectangular table (or array) then as a function.

A matrix having only one row is called a row matrix (or a row vector) and a
matrix having only one column is called a column matrix (or a column vector).
Two matrices of the same order whose corresponding entries are equal are

considered equal. The (i,j)-entry of the matrix (often written as or ) is

the element at the intersection of the row (from the top) and the
column (from the left).

For example,

is a 3×3 matrix (said 3 by 3). The 2nd row is and the 3rd column is .
The (2,3) entry is the entry at intersection of the 2nd row and the 3rd column,
that is 11.

Some special kinds of matrices are:

 A square matrix is a matrix which has the same number of rows and
columns. A diagonal matrix is a matrix with non zero entries only on the

main diagonal (ie at positions).

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 The unit matrix or identity matrix In, is the matrix with elements on the
diagonal set to 1 and all other elements set to 0. Mathematically, we

may say that for the identity matrix (which is usually written as

and called Kronecker's delta) is given by:

For example, if n = 3:

 The transpose of an m-by-n matrix A is the n-by-m matrix AT formed by

turning rows into columns and columns into rows, i.e. . An example

is

 A square matrix whose transpose is equal to itself is called a symmetric

matrix; that is, A is symmetric if . An example is

 A square matrix whose transpose is equal to its negative is called skew-

symmetric matrix; that is, A is skew-symmetric if . An example

is

Properties of these matrices are developed in the exercises.

Determinants

To define a determinant of order n, suppose there are n 2 elements of a field

sij where i and j are less than or equal to n. Define the following function (this
function is important in the definition):

S(a1,a2,a3,...,an)=# of reversals, meaning the number of times an1<an2 when

n1>n2, for each possible combination.

Suppose you have a permutation of numbers from 1 to n {a 1,a2,a3,...,an). Then

define a term of the determinant to be equal to (-
1)S(a1,a2,a3,...,an)s1a1,s2a2,s3a3,...,snan. The sum of all possible terms (i. e. through
all possible permutations) is called the determinant.

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Theorem

Definition: The transpose of a matrix A, AT is the matrix resulting when the

columns and rows are interchanged i. e. the matrix sji when A is the matrix
sij A matrix and its transpose have the same determinant:

Proof[

All terms are the same, and the signs of the terms are also unchanged since
all reversals remain reversals. Thus, the sum is the same.

TheoremInterchanging two rows (or columns) changes the sign of the

determinant:

Proof

To show this, suppose two adjacent rows (or columns) are interchanged.
Then any reversals in a term would not be affected except for the reversal of
the elements of that term within that row (or column), in which case adds or
subtracts a reversal, thus changing the signs of all terms, and thus the sign of
the matrix. Now, if two rows, the ath row and the (a+n)th are interchanged, then
interchange successively the ath row and (a+1)th row, and then the (a+1)th row
and (a+2)th row, and continue in this fashion until one reaches the (a+n-
1)th row. Then go backwards until one goes back to the ath row. This has the
same effect as switching the ath row and the (a+n)th rows, and takes n-1
switches for going forwards, and n-2 switches for going backwards, and their
sum must then be an odd number, so it multiplies by -1 an odd number of
times, so that its total effect is to multiply by -1.

Corollary A determinant with two rows (or columns) that are the same has the
value 0. Proof: This determinant would be the additive inverse of itself since
interchanging the rows (or columns) does not change the determinant, but still
changes the sign of the determinant. The only number for which it is possible
is when it is equal to 0.

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Theorem

It is linear on the rows and columns of the matrix.

Proof

The terms are of the form a1... ...an. Using the distributive law of fields, this

comes out to be a1... ...an + a1... ...an, an thus its sum of such terms is
the sum of the two determinants:

Corollary Adding a row (or column) times a number to another row (or
column) does not affect the value of a determinant.

Proof

Suppose you have a determinant A with the kth column added by another

column times a number:

where akb are elements of another column. By the linear property, this is equal
to

The second number is equal to 0 because it has two columns that are the

same. Thus, it is equal to

which is the same as the matrix A.

 It is easy to see that and thus

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for all -by- matrices and all scalars .

 A matrix over a commutative ring R is invertible if and only if its

determinant is a unit in R. In particular, if A is a matrix over a field such
as the real or complex numbers, then A is invertible if and only if det(A)
is not zero. In this case we have

Expressed differently: the vectors v1,...,vn in Rn form a basis if and only if

det(v1,...,vn) is non-zero.

The determinants of a complex matrix and of its conjugate

transpose are conjugate:

Vector Spaces
In this section we present a formal definition of a vector space, which will lead to an
extra increment of abstraction. Once defined, we study its most basic properties.

Here is one of the two most important definitions in the entire course.

Definition VSVector Space

Suppose that VV is a set upon which we have defined two operations: (1) vector
addition, which combines two elements of VV and is denoted by “+”, and (2) scalar
multiplication, which combines a complex number with an element of VV and is
denoted by juxtaposition. Then VV, along with the two operations, is a vector
space over CC if the following ten properties hold.

 AC Additive Closure
If u,v∈Vu,v∈V, then u+v∈Vu+v∈V.

 SC Scalar Closure
If α∈Cα∈C and u∈Vu∈V, then αu∈Vαu∈V.

 C Commutativity
If u,v∈Vu,v∈V, then u+v=v+uu+v=v+u.

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 AA Additive Associativity
If u,v,w∈Vu,v,w∈V, then u+(v+w)=(u+v)+wu+(v+w)=(u+v)+w.

 Z Zero Vector
There is a vector, 00, called the zero vector, such that u+0=uu+0=u for
all u∈Vu∈V.

 AI Additive Inverses
If u∈Vu∈V, then there exists a vector −u∈V−u∈V so that u+(−u)=0u+(−u)=0.

 SMA Scalar Multiplication Associativity

If α,β∈Cα,β∈C and u∈Vu∈V, then α(βu)=(αβ)uα(βu)=(αβ)u.

 DVA Distributivity across Vector Addition

If α∈Cα∈C and u,v∈Vu,v∈V, then α(u+v)=αu+αvα(u+v)=αu+αv.

 DSA Distributivity across Scalar Addition

If α,β∈Cα,β∈C and u∈Vu∈V, then (α+β)u=αu+βu(α+β)u=αu+βu.

 O One
If u∈Vu∈V, then 1u=u1u=u.

The objects in VV are called vectors, no matter what else they might really be,
simply by virtue of being elements of a vector space.

Now, there are several important observations to make. Many of these will be easier
to understand on a second or third reading, and especially after carefully studying
the examples in Subsection VS.EVS.

An axiom is often a “self-evident” truth. Something so fundamental that we all agree

it is true and accept it without proof. Typically, it would be the logical underpinning
that we would begin to build theorems upon. Some might refer to the ten properties
of Definition VS as axioms, implying that a vector space is a very natural object and
the ten properties are the essence of a vector space. We will instead emphasize that
we will begin with a definition of a vector space. After studying the remainder of this
chapter, you might return here and remind yourself how all our forthcoming
theorems and definitions rest on this foundation.

As we will see shortly, the objects in VV can be anything, even though we will call
them vectors. We have been working with vectors frequently, but we should stress
here that these have so far just been column vectors — scalars arranged in a
columnar list of fixed length. In a similar vein, you have used the symbol “+” for
many years to represent the addition of numbers (scalars). We have extended its

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use to the addition of column vectors and to the addition of matrices, and now we
are going to recycle it even further and let it denote vector addition in any possible
vector space. So when describing a new vector space, we will have to define exactly
what “+” is. Similar comments apply to scalar multiplication. Conversely, we
can define our operations any way we like, so long as the ten properties are fulfilled
(see Example CVS).

In Definition VS, the scalars do not have to be complex numbers. They can come
from what are called in more advanced mathematics, “fields”. Examples of fields are
the set of complex numbers, the set of real numbers, the set of rational numbers,
and even the finite set of “binary numbers”, {0,1}{0,1}. There are many, many
others. In this case we would call VV a vector space over (the field) FF.

A vector space is composed of three objects, a set and two operations. Some would
explicitly state in the definition that VV must be a nonempty set, but we can infer this
from Property Z, since the set cannot be empty and contain a vector that behaves as
the zero vector. Also, we usually use the same symbol for both the set and the
vector space itself. Do not let this convenience fool you into thinking the operations
are secondary!

This discussion has either convinced you that we are really embarking on a new
level of abstraction, or it has seemed cryptic, mysterious or nonsensical. You might
want to return to this section in a few days and give it another read then. In any
case, let us look at some concrete examples now.

Our aim in this subsection is to give you a storehouse of examples to work with, to
become comfortable with the ten vector space properties and to convince you that
the multitude of examples justifies (at least initially) making such a broad definition
as Definition VS. Some of our claims will be justified by reference to previous
theorems, we will prove some facts from scratch, and we will do one nontrivial
example completely. In other places, our usual thoroughness will be neglected, so
grab paper and pencil and play along.

Example VSCV The vector space CmCm

Example VSM The vector space of matrices, M mnMmn

So, the set of all matrices of a fixed size forms a vector space. That entitles us to call
a matrix a vector, since a matrix is an element of a vector space. For example,
if A,B∈M34A,B∈M34 then we call AA and BB “vectors,” and we even use our
previous notation for column vectors to refer to AA and BB. So we could legitimately
write expressions likeu+v=A+B=B+A=v+uu+v=A+B=B+A=v+uThis could lead to
some confusion, but it is not too great a danger. But it is worth comment.

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The previous two examples may be less than satisfying. We made all the relevant
definitions long ago. And the required verifications were all handled by quoting old
theorems. However, it is important to consider these two examples first. We have
been studying vectors and matrices carefully (Chapter V, Chapter M), and both
objects, along with their operations, have certain properties in common, as you may
have noticed in comparing Theorem VSPCV with Theorem VSPM. Indeed, it is
these two theorems that motivate us to formulate the abstract definition of a vector
space, Definition VS. Now, if we prove some general theorems about vector spaces
(as we will shortly in Subsection VS.VSP), we can then instantly apply the
conclusions to both CmCm and MmnMmn. Notice too, how we have taken six
definitions and two theorems and reduced them down to two examples. With greater
generalization and abstraction our old ideas get downgraded in stature.

Let us look at some more examples, now considering some new vector spaces.

Example VSP The vector space of polynomials, PnPn

Example VSIS The vector space of infinite sequences
Example VSF The vector space of functions

Here is a unique example.

Example VSS The singleton vector space

Perhaps some of the above definitions and verifications seem obvious or like
splitting hairs, but the next example should convince you that they are necessary.
We will study this one carefully. Ready? Check your preconceptions at the door.

Example CVS The crazy vector space

Subsection VS.EVS has provided us with an abundance of examples of vector
spaces, most of them containing useful and interesting mathematical objects along
with natural operations. In this subsection we will prove some general properties of
vector spaces. Some of these results will again seem obvious, but it is important to
understand why it is necessary to state and prove them. A typical hypothesis will be
“Let VV be a vector space.” From this we may assume the ten properties
of Definition VS, and nothing more. It is like starting over, as we learn about what
can happen in this new algebra we are learning. But the power of this careful
approach is that we can apply these theorems to any vector space we encounter —
those in the previous examples, or new ones we have not yet contemplated. Or
perhaps new ones that nobody has ever contemplated. We will illustrate some of
these results with examples from the crazy vector space (Example CVS), but mostly
we are stating theorems and doing proofs. These proofs do not get too involved, but
are not trivial either, so these are good theorems to try proving yourself before you
study the proof given here.

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First we show that there is just one zero vector. Notice that the properties only
require there to be at least one, and say nothing about there possibly being more.
That is because we can use the ten properties of a vector space (Definition VS) to
learn that there can never be more than one. To require that this extra condition be
stated as an eleventh property would make the definition of a vector space more
complicated than it needs to be.

Theorem ZVUZero Vector is Unique

Suppose that VV is a vector space. The zero vector, 00, is unique.

Proof

Theorem AIUAdditive Inverses are Unique

Suppose that VV is a vector space. For each u∈Vu∈V, the additive inverse, −u−u, is
unique.
Proof

As obvious as the next three theorems appear, nowhere have we guaranteed that
the zero scalar, scalar multiplication and the zero vector all interact this way. Until
we have proved it, anyway.

Theorem ZSSMZero Scalar in Scalar Multiplication

Suppose that VV is a vector space and u∈Vu∈V. Then 0u=00u=0.

Proof

Here is another theorem that looks like it should be obvious, but is still in need of a
proof.

Theorem ZVSMZero Vector in Scalar Multiplication

Suppose that VV is a vector space and α∈Cα∈C. Then α0=0α0=0.

Proof

Here is another one that sure looks obvious. But understand that we have chosen to
use certain notation because it makes the theorem's conclusion look so nice. The
theorem is not true because the notation looks so good; it still needs a proof. If we
had really wanted to make this point, we might have used notation like u♯u♯ for the
additive inverse of uu. Then we would have written the defining property, Property
AI, as u+u♯=0u+u♯=0. This theorem would become u♯=(−1)uu♯=(−1)u. Not really quite
as pretty, is it?

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Theorem AISMAdditive Inverses from Scalar Multiplication

Suppose that VV is a vector space and u∈Vu∈V. Then −u=(−1)u−u=(−1)u.

Proof

Because of this theorem, we can now write linear combinations

like 6u1+(−4)u26u1+(−4)u2 as 6u1−4u26u1−4u2, even though we have not formally
defined an operation called vector subtraction.

Our next theorem is a bit different from several of the others in the list. Rather than
making a declaration (“the zero vector is unique”) it is an implication (“if…, then…”)
and so can be used in proofs to convert a vector equality into two possibilities, one a
scalar equality and the other a vector equality. It should remind you of the situation
for complex numbers. If α,β∈Cα,β∈C and αβ=0αβ=0, then α=0α=0 or β=0β=0. This
critical property is the driving force behind using a factorization to solve a polynomial
equation.

Theorem SMEZVScalar Multiplication Equals the Zero Vector

Suppose that VV is a vector space and α∈Cα∈C. If αu=0αu=0, then

either α=0α=0 or u=0u=0.
Proof

Example PCVS Properties for the Crazy Vector Space

When we say that VV is a vector space, we then know we have a set of objects (the
“vectors”), but we also know we have been provided with two operations (“vector
addition” and “scalar multiplication”) and these operations behave with these objects
according to the ten properties of Definition VS. One combines two vectors and
produces a vector, the other takes a scalar and a vector, producing a vector as the
result. So if u1,u2,u3∈Vu1,u2,u3∈V then an expression
like5u1+7u2−13u35u1+7u2−13u3would be unambiguous in any of the vector spaces
we have discussed in this section. And the resulting object would be another vector
in the vector space. If you were tempted to call the above expression a linear
combination, you would be right. Four of the definitions that were central to our
discussions in Chapter V were stated in the context of vectors being column vectors,
but were purposely kept broad enough that they could be applied in the context of
any vector space. They only rely on the presence of scalars, vectors, vector addition
and scalar multiplication to make sense. We will restate them shortly, unchanged,
except that their titles and acronyms no longer refer to column vectors, and the
hypothesis of being in a vector space has been added. Take the time now to look
forward and review each one, and begin to form some connections to what we have
done earlier and what we will be doing in subsequent sections and chapters.
Specifically, compare the following pairs of definitions:

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STATIC OPTIMIZATION
INTRODUCTION The goal in optimization problems is to find the
minimum or maximum of a cost function l(xI, ... ,xn) that depends on n
variables XI"",Xn' This chapter develops the necessary and sufficient
conditions for a set of variables, XI, ... ,Xn, to be a minimum (or
maximum) of the cost function. Here, the following optimization
problems are considered: (i) the function l(xI, ... , xn) is unconstrained,
(ii) the function l(xI, ... , xn) is subject to equality constraints, and (iii)
the function l(xI, ... , xn) is subject to inequality constraints. Throughout
the chapter it is assumed that the cost function and the constraints are
at least twice continuously differentiable. These types of problems
frequently occur in engineering and are sometimes called parameter or
static optimization problems. The problem is called static because the
variables Xi, i = 1, ... , n, are independent of time. This is in contrast to
dynamic optimization problems where the variables Xi, i = 1, ... , n, are
functions of time. Although this book concentrates on the dynamic
optimization problem, much of the terminology and techniques used in
the static optimization problem provides a foundation for the material
in subsequent chapters. Furthermore, several techniques for the
numerical solution of the dynamic optimization problem involve
transforming the dynamic problem into a static problem.
UNCONSTRAINED OPTIMIZATION

MINIMUM AND MAXIMUM

The function f(Xl, ... , xn} is said to have a global minimum at the point iT =

[Xl X2 ... xn] if

(1.1)

for all h,T = [hl h2 '" hn] # O. This implies that Xl, ... , Xn produces the

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 Mathematical Economics UNIT -4
smallest value of the function f(x1, ... , xn}, and there are no other combinations

of independent variables that will produce a smaller value. The point Xl, ... , Xn

is said to be a local minimum if the condition (1.1) holds for all 71, such that

1171,11 = Jh~ + h~ + ... + h; < p

where p > O. Using the inequality given in the equation above we see that

(Xl + hl , ... ,Xn + hn) defines a sphere in the n-dimensional space centered

at Xl, ... , Xn! with radius p. This spherical region is called a neighborhood of

Xl, ... , Xn. Thus, X1, ... , Xn is a local minimum if there are no other points in

a neighborhood of X1, ... ,Xn for which the function f(Xl,''''Xn} has a smaller

value. Clearly, as p approaches infinity the neighborhood of Xl, ... , Xn becomes

the entire n-dimensional space, and Xl, ... , Xn represents a global minimum.

The point Xl, ... , Xn is said to be a global maximum if

for all hT = [hl h2 ... hn] # O. If the condition (1.2) is satisfied only in some

neighborhood of Xl, ... , Xn , then the maximum is said to be local.

1.2.2 FIRST ORDER OPTIMALITY CONDITIONS

The points at which the function f(Xl, ... ,Xn} become minimum or maximum

are the extrema of the function. This section establishes the conditions that

each extremum must satisfy, i.e., the necessary conditions for optimality. As

in the previous section, let Xl, ""xn denote an extremum of f(Xl, ... ,xn}. Now

consider an increment hi, in the variables Xi, i = 1, ... , n. Then the Taylor

series expansion of f(Xl + hl, ... ,xn + hn} about the point Xl,"',Xn yields

(Oakley, 1971)

f(Xl + h1, ... ,Xn + hn} = f(Xl, ... ,xn} + 6f(Xl,'" ,Xn}

+62 f(Xl, .. . , Xn} + higher-order terms

IMIZATION-BASED MODELS

Problem Formulation and Applications

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The fact that the musculoskeletal system is redundant and that apparently only a
limited number of MAP are used in skilled tasks has led to the development of
optimization-based models for estimating activation and forces of individual muscles.
The main assumption in these models is that the MAP are selected in such a way as to
optimize a specific objective function or a combination of objective functions (7,14).
There are two types of optimization-based models: static and dynamic.
In static models, it is assumed that the MAP at any movement instant are independent
of those at other instants. In other words, the value of the objective function does not
depend on time explicitly. In static optimization, muscle activation and/or forces are
calculated for each time instant of a movement. For example, optimal forces of nine
muscles in a two-dimensional, four-segment model of the leg (Fig. 1) can be found by
solving the following static optimization problem for each time instant:
minimize cost function Z = f(F1, F2, …, F9),
(1)
subject to the equality constraints

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Figure 1
A two-dimensional model of the human leg. It has three DoF and is controlled by nine muscles.
Therefore, the moments (M) at the joints can be produced by an infinite number of muscle force
combinations. TA is tibialis anterior, SO is soleus, GA is gastrocnemius, VA is vastii, RF is
rectus femoris, BFS is short head of biceps femoris, HA is two-joint hamstrings, GLM is gluteus
maximus, and IL is iliacus. For review of model parameters (segment inertial parameters, muscle
moment arms, PCSA, muscle composition, Fmax, Vmax) and their values (10).

M1=d11·F1−d12·F2−d13·F3M2=−d23·F2+d24·F4+d25·F5−d26·F6−d27·F7M3=−d3
7·F7+d38·F8−d39·F9
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(2)
and the inequality constraints
F1, F2, …, F9 ≥ 0,
(3)
where M1, M2, and M3 are the resultant moments at the ankle, knee, and hip joints,
respectively, and dji are the moment arms of the i-th muscle with respect to the j-th
joint (the moment arms are assumed to be known).
A variety of optimization criteria have been used in the literature to solve the above
problem (for example (7,14)). Some of the criteria have been selected arbitrarily,
whereas others have been based on various physiological reasons. Among the latter
are versions of criteria of minimum muscle fatigue, minimum muscle stress, and
minimum metabolic energy expenditure. Two most frequently used versions of
minimum muscle fatigue criteria, Z1 and Z2, are formulated as follows, first (5):
Z1=maxi{1/Ti}→min,i=1,…,9
(4)
where Z1 is muscle fatigue function; Ti = ai · (Fi/Fi max · 100)Pi is endurance time of
the i-th muscle; ai = exp(3.48+0.169 ·Si) and pi = −0.25–0.036 ·Si are the functions of
the percentage of slow-twitch fibers in the i-th muscle, Si, derived from experiments
on cat muscles (5); and Fi max is maximum isometric force of the i-th muscle. The
maximum isometric force of each muscle can be estimated as Fi max = K·PCSAi,
where K = 40 N/cm2 on average.
The second formulation of fatigue criterion is (4):
Z2=∑i=19(Fi/PCSAi)p→min,p=2,3,4
(5)
where Z2 is a muscle fatigue function; PCSAi is the physiological cross-sectional area
of the i-th muscle; and constant p was derived from the experimentally obtained
relationship between muscle stress (F/PCSA) and endurance time (T) of human
muscles (p = 3 on average) (for details see (4)). Mathematically, criterion Z2 is
equivalent to minimizing a norm N = [Σ(Fi/PCSAi)p]1/p of a vector
[F1/PCSA1, F2/PCSA2, …, F9/PCSA9]. When the value of p approaches infinity, the
above norm becomes the maximum norm, and its optimum solution approaches the
equal distribution of stresses among the involved muscles in a one-DOF case (see, for
example, (4)). This property of polynomial criterion Z2 led to the development of the
min/max optimization criterion, max{Fi/PCSAi}→ min, which was first presented in a
bound formulation (2):

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Z3 = σ → min
(6)
where σ > Fi/PCSAi (i = 1, …, 9) is the upper bound for stresses of all muscles. When
the power p in criterion Z2 approaches infinity, the solutions of these criteria converge
to the solution of criterion Z3 (13). The formulations of the above criteria suggest that
muscle stress and muscle fatigue criteria are related.
Criteria from another group minimize energy expenditure. Consider, for example, a
minimum metabolic cost criterion suggested by Alexander (1) and examined in (10):
Z4=∑i=19αip·Fimax·νimax·Φ[νi/νimax]→min,p=1,2,
(7)
where Z4 is the metabolic rate of the i-th muscle and function φ determines the
metabolic cost; Fi max and vimax are known maximum force and maximum velocity of
the i-th muscle, respectively; vi is instantaneous velocity of the i-th muscle; and ai (0
≤ ai ≤ 1) is the unknown normalized activation of the i-th muscle, sought by
minimizing criterion Z4. Moment constraints (equation 2) for this criterion take into
account the muscle force-velocity properties (for details, see (1) and (10)).
Static optimization models are especially well suited for estimating activation and
forces of individual muscles in static tasks, e.g., exerting external forces in different
directions. The major shortcomings of applying static optimization models to dynamic
tasks include the following (3,7): (i) in many dynamic tasks, the dynamics of the
skeleton and muscles at any time instant are likely to depend on the dynamics during
other time instances (e.g., a stretch of the muscle-tendon unit in one movement phase
can modify motor performance in the next phase) and (ii) static optimization models
typically cannot predict kinematics and dynamics of movement when only initial
and/or final states of the system are known.
Dynamic optimization-based models can overcome these limitations (3,7). Dynamic
optimization problems are solved only once for the entire movement time by
minimizing a functional, the value of which often depends on the time-dependent
system state variables (e.g., joint angles and their time derivatives), the control
variables (e.g., muscles activation), and values of the state variables and their time
derivatives at the initial and/or final movement instants. The solution of the dynamic
optimization problem, i.e., optimal MAP, must satisfy the differential constraints,
which include the equations of skeletal motion and the equations describing muscle
activation and muscle-tendon dynamics. When optimal activation patterns are found,
muscle forces, joint moments, and the kinematics of the system can be obtained.
Among limitations of dynamic optimization-based models, probably the most
important is that these models require the knowledge of a large number of model
parameters describing a substantially more sophisticated model of muscle contraction

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dynamics. The exact values of these parameters in humans are essentially unknown.
In dynamic optimization, muscle stress, muscle fatigue, metabolic energy expenditure,
movement performance (time, speed, height), or combinations of those can be
optimized.
Because of the relative simplicity of static optimization models, they have been used
more often than dynamic models to predict patterns of muscle activation and forces in
dynamic tasks (for review, see (10,14)). Therefore in this review, we consider mostly
static optimization models.
In the above formulations of static and dynamic optimization problems it was
assumed that muscles are controlled independently. In reality, this may not be the
case. To explain this idea, let us compare the muscles with the hand fingers. When
people are asked to generate force using one finger, other fingers also involuntarily
produce force. This phenomenon is called enslaving (15). It is not known whether the
enslaving exists at the level of muscle control. If it does, the enslaving may explain
some nonoptimal patterns of muscle activity. Present optimization-based models do
not consider possible enslaving.

COMPARING PREDICTIONS WITH MEASUREMENTS

At present, subject-specific values of muscle activation and forces cannot be predicted
by optimization-based models because many of subject-specific model parameters are
not known. The optimization-based models, however, can predict nominal patterns of
muscles activation in a typical subject with typical model parameters in some motor
tasks (3,7,10,14). Examples of reasonably good correspondence between the
measured MAP and MAP predicted by criteria Z2, Z3, and Z4 in human walking and
cycling are shown in Figures 2 and and3.3. In these tasks, the Pearson correlation
coefficients calculated between predicted and measured patterns are typically between
0.6 and 0.9 for the majority of muscles. However, force predictions and EMG for
some muscles have dissimilar patterns (rectus femoris in walking, Fig. 2; soleus and
gluteus maximus patterns predicted by criterion Z4/p (N = 1) in cycling, Fig. 3).

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Figure 2
Recorded EMG linear envelopes (EMG) and muscle forces and activation predicted by
optimizing minimum fatigue criterion Z2, min/max criterion Z3, and minimum metabolic cost
criteria (Z4, p = 1 and p = 2) during cycle of walking. EMG was obtained from 10 subjects
walking on a treadmill with a speed of 1.82 m·s−1, and muscle forces and activation were
calculated from kinematics and ground reaction forces of one typical subject during over ground
walking at a similar speed (10). The EMG linear envelopes were normalized to the EMG peak in
the cycle and shifted in time by 40 ms to account for the delay between the EMG and the joint
moments. The Pearson correlation coefficients (r) calculated between the EMG and predicted
force/activation patterns are typically between 0.7 and 0.9 (exception: the RF muscle). The best
performance is demonstrated by the minimum fatigue criterion Z2; the linear version of the
metabolic cost criterion (Z4, p = 1) has typically the worst performance. [Adapted from Prilutsky,
B.I., “Coordination of two- and one-joint muscles: functional consequences and implications for
motor control,” Motor Control 4:18, 2000, and from Prilutsky, B.I., “Muscle coordination: the
discussion continues,” Motor Control, 4:103, 2000. Copyright © 2000 Human Kinetics
Publishers, Inc. Used with permission.]

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Recorded EMG linear envelopes (EMG) and muscle forces and activation predicted by
optimizing minimum fatigue criterion Z2, min·max criterion Z3, and minimum metabolic cost
criteria (Z4, p = 1 and p = 2) during cycle of pedaling at a cadence of 60 rpm and a power of 200
W. The EMG was recorded, and the muscle forces and activation were calculated from
kinematics and pedal reaction forces of one typical subject (11). The EMG linear envelopes were
normalized to the peak EMG values recorded in maximum isometric contractions. The EMG
envelopes were shifted in time to account for the delay between the EMG and the joint moments.
The time shift was found for each muscle by cross-correlating the EMG envelopes and the joint
moments (11). The Pearson correlation coefficients (r) calculated between the EMG and
predicted force/activation patterns are typically between 0.6 and 0.9. The best performance is
demonstrated by the minimum fatigue criterion Z2; the linear version of the metabolic cost
criterion (Z4, p = 1) has typically the worst performance.

Three main obstacles make the distribution problem difficult to solve: 1)

incompleteness of the models, 2) insufficient accuracy of the model parameters, and
3) difficulty in validating the models.
1. Muscle forces creating moments about one DoF also induce mechanical effects
about other DoF. Ideally, optimization models should be applied to the entire
body in three-dimensional space. In reality, the majority of the developed
models target either a single joint or a single extremity in two-dimensional
space (for rare exceptions, see (7) and (14)). Hence, numerous secondary and
tertiary moments are neglected. For instance, the trunk is not included in the
model presented in Figure 1 and, hence, the effects of the hip joint moments on
the upper part of the body are disregarded. The inclusion of additional DoF and
body parts increases the model’s complexity. Most importantly, including
many DoF increases the number of assumptions that have to be made and,
hence, may decrease rather than increase the accuracy of the prediction.
2. As seen from the formulation of optimization problem (1–7), there are many
model parameters (dJ,i, PCSAi, Fi max, vi max) and input variables (MJ, vi) that need
to be estimated to calculate muscle forces or activation. A sensitivity analysis
of the optimization-based models to errors in the model and input parameters
suggests that values of muscle forces may be very sensitive to variation in
muscle moment arms, physiological cross-sectional area, muscle fiber
composition, and joint moments (for review, see (10)). Therefore, it is
generally difficult to compare absolute values of predicted and measured
muscle forces. However, published reports on sensitivity analyses suggest that,
in contrast to force magnitudes, time patterns of predicted forces are less
sensitive to errors in model parameters.
Another complication in the comparison of the predicted and measured muscle
forces and activation is the influence of the muscle force-length-velocity
properties on the actual muscle force output. The formulation of the static

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optimization problem presented above does not account for many of these
properties (however they can be accounted for in static optimization to a certain
extent; see, for example (3)). In some tasks, however, the force-length-velocity
properties seem not to significantly affect predictions of the actual force output.
For example, measured forces of the cat soleus, gastrocnemius, and plantaris
during walking and trotting at different speeds were rather successfully
predicted by static optimization and criterion Z1 (Fig. 4; (12)) without
accounting for the force-length-velocity properties of the muscles. In human
walking, including the muscle force-length-velocity properties into the static
optimization problem did not substantially change the obtained optimal
solution (3).

Figure 4

Measured forces (solid lines) and predicted forces (using criterion Z1; dotted lines) of the
cat soleus (SO), plantaris (PL), and gastrocnemius (GA) muscles during seven
consecutive step cycles of trotting of the cat at a speed of 1.5 m·s−1. Because the

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percentage of slow-twitch fibers engaged in the force development during this trial was
unknown, different physiologically reasonable values were examined (see Table 2 in
(12)). The percentage of slow-twitch fibers in SO, GA, and PL used to calculate forces
presented in this figure were 100%, 20%, and 35%, respectively. [Reproduced from
Prilutsky, B.I., W. Herzog, and T.L. Allinger. Forces of individual cat ankle extensor
muscles during locomotion predicted using static optimization. J. Biomech. 30:1029,
1997. Copyright © 1997 Elsevier Science Ltd. Used with permission.]

3. In human studies it is difficult to measure forces in individual muscles.

Therefore, predicted on/off timing or time histories of muscle forces are often
compared against the on/off timing or time histories of EMG linear envelopes.
The comparison of time histories seems more revealing, because predicted
forces and EMG can have similar on/off timing but different time histories. It
should be noted that predicted forces and EMG patterns can be compared if the
force-EMG relation is defined (e.g., linear). This seems to be the case in a
number of isometric tasks and some dynamic tasks like load lifting, cycling,
and walking in humans (3,10). Thus, these activities are convenient for the
application and evaluation of static optimization methods.
Another experimental paradigm, which seems free of the above-mentioned
limitations, is direct measurements of finger forces in multifinger tasks. The model of
hand fingers has been used to study motor redundancy (15). Optimal finger forces
were calculated by solving the following optimization problem:
Minimize
Z5=(∑in(FiFmi)p)1/p
(8)
subject to constraint function:
h=∑i=1nFi=Ftot
(9)
where i stands for a finger involved in a task, Fi are individual finger forces, p is value
of power (p > 1), n is the number of fingers involved in the tasks (n = 2, 3, 4),
and Ftot is the total force level achieved by all the involved fingers.
Go to:

INTERPRETATION AND VALIDATION

Despite the fact that quantitative subject-specific predictions of MAP are very
difficult to achieve at present, some important insights can be made based on
qualitative comparisons of measured and predicted MAP. It seems remarkable that
MAP of such dissimilar human movements as human walking (Fig. 2), cycling (Fig.

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3), exerting isometric forces in different directions (10), and some other tasks have
been qualitatively predicted by minimizing the same cost functions of muscle fatigue
and metabolic cost (Z2 and Z4). These optimization criteria predict: (i) reciprocal
coactivation of one-joint antagonist muscles (e.g., tibialis anterior vs soleus, Figs.
2 and and3);3); (ii) coactivation of one-joint synergists with their two-joint
antagonists (e.g., VA vs HA, Figs. 2 and and3);3); (iii) simultaneous activation of
synergists crossing the same joints (e.g., medial and lateral heads of gastrocnemius;
three heads of vastii; (4); and (iv) a strong relationship between force and activation
of two-joint muscles and moments at the two joints (a two-joint muscle has the
greatest activation when acting as an agonist at the two joints and the smallest
activation when acting as an antagonist at both joints (for review, see (10))). All these
features of muscle coordination can be seen, to a certain extent, in the MAP of a
number of skilled tasks performed with submaximal effort and in a stereotypic
manner.

Which Criterion Is Better?

To select an optimization criterion for interpreting MAP among all tested criteria, one
can rely on at least three factors: (i) quantitative measures of the criterion
performance, i.e., errors of predicted forces (if measured forces are available (12,15)),
or correlation coefficients (in the case of qualitative pattern comparisons; Figs. 2 and
and3);3); (ii) a physiological meaning of the cost function; and (iii) availability of
reasonable physiological mechanisms that could implement a given criterion.
In the case of cat locomotion, criterion Z1 predicts muscle forces with smaller errors
than criterion Z2 (see Fig. 4; see also Table 1 in (12)), whereas
criterion Z2 demonstrates slightly higher correlation between predicted and measured
MAP in studied human tasks (10,11). Both of these criteria minimize muscle fatigue.
The analysis of muscle forces and activation in humans, as predicted by optimizing
linear criteria (see, for example, seemingly logical minimum metabolic cost
criterion Z4 with p = 1 (equation 7; Figs. 2 and and3)),3)), indicates that linear criteria
do not adequately predict activation patterns of some synergist muscles (e.g.,
individual heads of gastrocnemius and vastii (4)). The same criterion Z4 with p = 2 has
a substantially better performance (Figs. 2 and and3).3). However, setting p = 2 has
been done arbitrarily, merely to make the cost function nonlinear and to eliminate the
deficiency of linear criteria that cannot predict more than three simultaneously active
muscles for the given optimization problem (1–3). Of course, there are other
formulations of the metabolic criteria (3), which can have a better performance.
Minimum fatigue criteria Z1 and Z2, min/max criterion Z3, and minimum metabolic
cost criterion Z4 seem reasonable for tasks with submaximum efforts (walking,
running, cycling, load lifting, contact force tasks, etc.), in which the time of sustaining

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a given performance (i.e., endurance time) may be critical and should be maximized.
The minimum fatigue function 1/T = F/PCSA3 (see criterion Z2, equation 5) closely
resembles the function that relates the perceived effort to the exerted muscle force (for
details, see (10)). This leads to speculation that humans might solve some motor
redundancy problems based on the sense of perceived effort. For example, there are
indications that a preferred gait (walking or running) and preferred rate of pedaling in
cycling are selected based on the sense of effort rather than on the metabolic cost (for
references, see (10)).
The sense of effort is thought to originate primarily from motor commands but also
from afferent signals from muscles and joints. Thus, minimizing the sense of effort
may be interpreted, to some extent, as minimizing the central motor commands to
motor neuron pools. These commands can be perceived through corollary
discharges/efference copy of the motor commands from the motor to the sensory
cortex. This minimization of central motor commands is analogous to the principle of
“minimization of interaction” between different motor levels of control, which was
thought to be a central mechanism of mastering expedient behavior by the nervous
system (6) and which is consistent with the idea of lowering a level of control from
higher to lower centers during acquisition of motor skills.
When finger forces, rather than muscle forces, were explored, minimal-norm
criterion Z5 (equation 8) was successful in predicting the individual finger forces
during pressing tasks (15). However, this criterion did not predict finger forces
correctly when the task was to exert a torque on a hand-held object. The reason for the
poor performance was that the criterion neglects the enslaving effects that cause
involuntary activation of the antagonist fingers. In manipulative tasks, the index and
middle fingers and the ring and little fingers exert moments of normal force about a
pivot point, created by the thumb, in opposite directions. The antagonist fingers
generate moments of force opposite to the resultant moment produced by all fingers.
For instance, if the resultant moment is in pronation, an antagonist finger generates a
supination moment. Activation of the antagonist fingers in the torque production tasks
was observed experimentally; however, criterion Z5 neglects it.

How Well Do Optimization Approaches Work in General?

In the literature, reports about the application of optimization methods range from
stories of success (12) to accounts of failure (8). Personal opinions also range from
extremely pessimistic (“… there is no objective function that has been shown to
predict individual muscle forces accurately” and the optimization “has repeatedly
failed to give acceptable results”; (8), p. 187) to very optimistic (“During gait, the
observed muscle activity (as determined by EMG) shows substantial agreement with
that activity pattern predicted when endurance is used as the optimization criterion”;
(4), p. 793). There are several reasons for such diverse opinions. First, there is no
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accepted standard on what to consider a successful prediction of activation patterns.

The predicted and measured patterns presented in Figures 2–4 can be perceived by
some readers as examples of a good correspondence, whereas other readers will
evaluate the same results much more pessimistically. Second, the optimization may
fail or succeed because of the incompleteness of the model and/or errors in parameter
estimation. These deficiencies can be overcome in the future with the advent of new
theoretical and experimental methods. Third, some cost functions may not work well
for some motor tasks. Our bias is that the negative results obtained in some research
projects do not undermine the applicability of the optimization methods in principle.

Input-Output Analysis:
Features, Static and Dynamic
Model
Input-Output Analysis: Features, Static and Dynamic
Model!
Input-output is a novel technique invented by Professor
Wassily W. Leontief in 1951. It is used to analyse inter-
industry relationship in order to understand the inter-
dependencies and complexities of the economy and thus the
conditions for maintaining equilibrium between supply and
demand.

Thus it is a technique to explain the general equilibrium of the

economy. It is also known as “inter-industry analysis”. Before
analysing the input-output method, let us understand the
meaning of the terms, “input” and “output”. According to
Professor J.R. Hicks, an input is “something which is bought

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for the enterprise” while an output is “something which is sold

by it.”

An input is obtained but an output is produced. Thus input

represents the expenditure of the firm, and output its receipts.
The sum of the money values of inputs is the total cost of a
firm and the sum of the money values of the output is its total
revenue.

The input-output analysis tells us that there are industrial

interrelationships and inter-dependencies in the economic
system as a whole. The inputs of one industry are the outputs
of another industry and vice versa, so that ultimately their
mutual relationships lead to equilibrium between supply and
demand in the economy as a whole.

Coal is an input for steel industry and steel is an input for coal
industry, though both are the outputs of their respective
industries. A major part of economic activity consists in
producing intermediate goods (inputs) for further use in
producing final goods (outputs).

There are flows of goods in “whirlpools and cross currents”

between different industries. The supply side consists of large
inter-industry flows of intermediate products and the demand

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side of the final goods. In essence, the input-output analysis

implies that in equilibrium, the money value of aggregate
output of the whole economy must equal the sum of the
money values of inter-industry inputs and the sum of the
money values of inter-industry outputs.

Contents
1. Main Features

2. The Static Input-Output Model

3. The Dynamic Input-output Model

1. Main Features:

The input-output analysis is the finest variant of general

equilibrium. As such, it has three main elements; Firstly, the
input-output analysis concentrates on an economy which is in
equilibrium. Secondly, it does not concern itself with the
demand analysis. It deals exclusively with technical problems
of production. Lastly, it is based on empirical investigation.
The input-output analysis consists of two parts: the
construction of the input-output table and the use of input-
output model.

2. The Static Input-Output Model:

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The input-output model relates to the economy as a whole in a

particular year. It shows the values of the flows of goods and
services between different productive sectors especially inter-
industry flows.
Assumptions:
This analysis is based on the following assumptions:
(i) The whole economy is divided into two sectors—“inter-
industry sectors” and “final-demand sectors,” both being
capable of sub-sectoral division.

(ii) The total output of any inter-industry sector is generally

capable of being used as inputs by other inter-industry
sectors, by itself and by final demand sectors.

(iii) No two products are produced jointly. Each industry

produces only one homogeneous product.

(iv) Prices, consumer demands and factor supplies are given.

(v) There are constant returns to scale.

(vi) There are no external economies and diseconomies of

production.

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(vii) The combinations of inputs are employed in rigidly fixed

proportions. The inputs remain in constant proportion to the
level of output. It implies that there is no substitution between
different materials and no technological progress. There are
fixed input coefficients of production.
Explanation:
For understanding, a three-sector economy is taken in which
there are two inter-industry sectors, agriculture and industry,
and one final demand sector.

Table 1 provides a simplified picture of such economy in

which the total output of the industrial, agricultural and
household sectors is set in rows (to be read horizontally) and
has been divided into the agricultural, industrial and final
demand sectors. The inputs of these sectors are set in
columns. The first row total shows that altogether the
agricultural output is valued at Rs. 300 crores per year.

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Of this total, Rs. 100 crores go directly to final consumption

(demand), that is, household and government, as shown in
the third column of the first row. The remaining output from
agriculture goes as inputs: 50 to itself and 150 to industry.
Similarly, the second row shows the distribution of total
output of the industrial sector valued at Rs. 500 crores per
year. Columns 1, 2 and 3 show that 100 units of manufactured
goods go as inputs to agriculture, 250 to industry itself and
150 for final consumption to the household sector.

Let us take the columns (to be read downwards). The first

column describes the input or cost structure of the
agricultural industry. Agricultural output valued at Rs. 300
crores is produced with the use of agricultural goods worth
Rs. 50, manufactured goods worth Rs. 100 and labour or/and
management services valued at Rs. 150. To put it differently, it
costs Rs. 300 crores to get revenue of Rs. 300 crores from the
agricultural sector. Similarly, the second column explains the
input structure of the industrial sector (i.e., 150 + 250 + 100 =
500).

Thus “a column gives one point on the production function of

the corresponding industry.” The ‘final demand’ column
shows what is available for consumption and government
expenditure. The third row corresponding to this column has
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been shown as zero. This means that the household sector is

simply a spending (consuming) sector that does not sell
anything to itself. In other words, labour is not directly
consumed.

There are two types of relationships which indicate and

determine the manner in which an economy behaves and
assumes a certain pattern of flows of resources.

They are:
(a) The internal stability or balance of each sector of the
economy, and

(b) The external stability of each sector or intersectoral

relationships. Professor Leontief calls them the “fundamental
relationships of balance and structure.” When expressed
mathematically they are known as the “balance equations’ and
the “structural equations”.

If the total output of say X. of the ‘ith’ industry is divided into

various numbers of industries 1, 2, 3, n, then we have the
balance equation:

X1 = xi1 + xi2 + xi3 + xin…… + D1

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and if the amount say У. absorbed by the “outside sector” is

also taken into consideration, the balance equation of the
ith industry becomes

It is to be noted that Yi stands for the sum of the flows of the

products of the ith industry to consumption, investment and
exports net of imports, etc. It is also called the “final bill of
goods” which it is the function of the output to fill. The
balance equation shows the conditions of equilibrium between
demand and supply. It shows the flows of outputs and inputs
to and from one industry to other industries and vice versa.
Since x12 stands for the amount absorbed by industry 2 of the
ith industry, it follows that xij stands for the amount absorbed
by the ith industry of jth industry.
The “technical coefficient” or “input coefficient” of
the ith industry is denoted by:
aij = xij/Xj

where xij is the flow from industry i to industry j, Xj is the

total output of industry aij and aij, as already noted above, is a
constant, called “technical coefficient” or “flow coefficient” in
the ith industry. The technical coefficient shows the number

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of units of one industry’s output that are required to produce

one unit to another industry’s output.

Equation (3) is called a “structural equation.” The structural

equation tells us that the output of one industry is absorbed
by all industries so that the flow structure of the entire
economy is revealed. A number of structural equations give a
summary description of the economy’s existing technological
conditions.

Using equation (3) to calculate the aij for our example of the
two-sector input-output Table 1, we get the following
technology matrix.

These input coefficients have been arrived at by dividing each

item in the first column of Table 1 by first row total, and each
item in the second column by the second row, and so on. Each
column of the technological matrix reveals how much
agricultural and industrial sectors require from each other to
produce a rupee’s worth of output. The first column shows
that a rupee’s worth of agricultural output requires inputs

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worth 33 paise from industries and worth 17 paise from

agriculture itself.

The Leontief Solution:

The table can be utilised to measure the direct and indirect
effects on the entire economy of any sectoral change in total
output of final demand.

Again using equation (3)

aij = xij/Xj

Cross multiplying, xij = aij. Xj

By substituting the value of xij into equation (2) and

transposing terms, we obtain the basic input- output system
of equations

In terms of our two-sector economy, there would be

two linear equations that could be written
symbolically as follows:

The above symbolic relationship can be shown in

matrix form:
X- [A]X= Y
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X [l-A] = Y

where matrix (I – A) is known as the Leontief Matrix

Numerical Solution:
Our technology matrix as per Table 2 is

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3. The Dynamic Input-output Model:

So far we have studied an open static model. “The model

becomes Dynamic when it is closed by the linking of the
investment part of the final bill of goods to output. The
dynamic input-output model extends the concept of inter-
sectoral balancing at a given point of time to that of inter-
sectoral balancing over time.

This necessarily involves the concept of durable capital. The

Leontief dynamic input-output model is г generalization of
the static model and is based on the same assumptions. In a
dynamic model, the output of a given period is supposed to go
into stocks,

i. e., capital goods, and the stocks, in turn, are distributed

among industries.

The balance equation is:

Here Xi (t) represents the total flow of output of ith

industry in period t, which is used for three
purposes:
(i) For production in the economy’s n industries x11 (t), x12 (t),
etc., in that period;

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(ii) As net addition to the stock of capital goods in n industries

i.e. S’t which can also be written as S1 (t) = S1 (t+1) – S1 (t),
where S1 (t) indicates the accumulated stock of capital in the
current period (t), and S1 (t + l) is next year’s stock; and
(iii) As consumption demand for the next period D. (t + 1). If
we ignore depreciation and wear-tear, then S. (t+1) — S1 (t) is
the net addition to capital stock out of current production.
Equation (4) can, therefore, be written as:
Xi (t) t x1i1 + xi2 + xi3 + xin + S. (t +1) – S1 (t) + D2 (t) + Yi (t)
where Yi (t) stands for the amount absorbed by the outside
sector in period t.
Just as the technical co-efficient was derived in the case of the
static model, the capital coefficient can be found out in a
similar manner. Capital co-efficient of the i th product used by
the jth industry is denoted by

bij =Sij /Xj

Cross multiplying, we have Sij = bij. X

where Sij represents the amount of capital stock of the ith

product used by the jth industry. Xj is total output of industry
j, and bij is a constant called capital co-efficient or stock co-
efficient. Equation (5) is known as the structural equation in a
dynamic model.

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If the bij co-efficient is zero, it means that no stock is required

by an industry and the dynamic model becomes a static
model. Moreover, bij can neither be negative nor infinite. If
the capital coefficient is negative, the input is, in fact, an
output of an industry.

Introductory guide on Linear

Programming for (aspiring) data
scientists
Introduction
Optimization is the way of life. We all have finite resources and time and we want to
make the most of them. From using your time productively to solving supply chain
problems for your company – everything uses optimization. It’s a especially
interesting and relevant topic in data science.

It is also a very interesting topic – it starts with simple problems, but can get very
complex. For example, sharing a chocolate between siblings is a simple optimization
problem. We don’t think in mathematical term while solving it. On the other hand
devising inventory and warehousing strategy for an e-tailer can be very complex.
Millions of SKUs with different popularity in different regions to be delivered in
defined time and resources – you see what I mean!

Linear programming (LP) is one of the simplest ways to perform optimization. It

helps you solve some very complex optimization problems by making a few
simplifying assumptions. As an analyst you are bound to come across applications
and problems to be solved by Linear Programming.

For some reason, LP doesn’t get as much attention as it deserves while learning
data science. So, I thought let me do justice to this awesome technique. I decided to
write an article which explains Linear programming in simple English. I have kept the

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content as simple as possible. The idea is to get you started and excited about
Linear Programming.

Table of Content
1. What is Linear Programming?
o Basic Terminologies
o Process to define a LP problem
2. Solve Linear Program by Graphical Method
3. Solve Linear Program using R
4. Solve Linear Program using OpenSolver
5. Simplex Method
6. Northwest Corner Method and Least Cost Method
7. Applications of Linear programming

1.What is Linear Programming?

Now, what is linear programming? Linear programming is a simple technique where
we depict complex relationships through linear functions and then find the optimum
points. The important word in previous sentence is depict. The real relationships
might be much more complex – but we can simplify them to linear relationships.

Applications of linear programming are every where around you. You use linear
programming at personal and professional fronts. You are using linear programming
when you are driving from home to work and want to take the shortest route. Or
when you have a project delivery you make strategies to make your team work
efficiently for on time delivery.

Example of a linear programming problem

Let’s say a FedEx delivery man has 6 packages to deliver in a day. The warehouse
is located at point A. The 6 delivery destinations are given by U, V, W, X, Y and Z.
The numbers on the lines indicate the distance between the cities. To save on fuel
and time the delivery person wants to take the shortest route.

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So, the delivery person will calculate different routes for going to all the 6
destinations and then come up with the shortest route. This technique of choosing
the shortest route is called linear programming.

In this case, the objective of the delivery person is to deliver the parcel on time at all
6 destinations. The process of choosing the best route is called Operation Research.
Operation research is an approach to decision-making, which involves a set of
methods to operate a system. In the above example, my system was the Delivery
model.

Linear programming is used for obtaining the most optimal solution for a problem
with given constraints. In linear programming, we formulate our real life problem into
a mathematical model. It involves an objective function, linear inequalities with
subject to constraints.

Is the linear representation of the 6 points above representative of real world? Yes
and No. It is oversimplification as the real route would not be a straight line. It would
likely have multiple turns, U turns, signals and traffic jams. But with a simple
assumption, we have reduced the complexity of the problem drastically and are
creating a solution which should work in most scenarios.

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Formulating a problem – Let’s manufacture some

chocolates
Example: Consider a chocolate manufacturing company which produces only two
types of chocolate – A and B. Both the chocolates require Milk and Choco only. To
manufacture each unit of A and B, following quantities are required:

 Each unit of A requires 1 unit of Milk and 3 units of Choco

 Each unit of B requires 1 unit of Milk and 2 units of Choco

The company kitchen has a total of 5 units of Milk and 12 units of Choco. On each
sale, the company makes a profit of

 Rs 6 per unit A sold

 Rs 5 per unit B sold.

Now, the company wishes to maximize its profit. How many units of A and B should
it produce respectively?

Solution: The first thing I’m gonna do is represent the problem in a tabular form for
better understanding.

Milk Choco Profit per unit

A 1 3 Rs 6

B 1 2 Rs 5

Total 5 12

Let the total number of units produced of A be = X

Let the total number of units produced of B be = Y

Now, the total profit is represented by Z

The total profit the company makes is given by the total number of units of A and
B produced multiplied by its per unit profit Rs 6 and Rs 5 respectively.

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Profit: Max Z = 6X+5Y

which means we have to maximize Z.

The company will try to produce as many units of A and B to maximize the profit. But
the resources Milk and Choco are available in limited amount.

As per the above table, each unit of A and B requires 1 unit of Milk. The total amount
of Milk available is 5 units. To represent this mathematically,

X+Y ≤ 5

Also, each unit of A and B requires 3 units & 2 units of Choco respectively. The total
amount of Choco available is 12 units. To represent this mathematically,

3X+2Y ≤ 12

Also, the values for units of A can only be integers.

So we have two more constraints, X ≥ 0 & Y ≥ 0

For the company to make maximum profit, the above inequalities have to be
satisfied.

This is called formulating a real-world problem into a mathematical model.

Common terminologies used in Linear Programming

Let us define some terminologies used in Linear Programming using the above
example.

 Decision Variables: The decision variables are the variables which

will decide my output. They represent my ultimate solution. To solve any
problem, we first need to identify the decision variables. For the above
example, the total number of units for A and B denoted by X & Y respectively
are my decision variables.
 Objective Function: It is defined as the objective of making decisions. In the
above example, the company wishes to increase the total profit represented
by Z. So, profit is my objective function.

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 Constraints: The constraints are the restrictions or limitations on the decision

variables. They usually limit the value of the decision variables. In the
above example, the limit on the availability of resources Milk and Choco are
my constraints.
 Non-negativity restriction: For all linear programs, the decision variables
should always take non-negative values. Which means the values for decision
variables should be greater than or equal to 0.

Process to formulate a Linear Programming problem

Let us look at the steps of defining a Linear Programming problem generically:

1. Identify the decision variables

2. Write the objective function
3. Mention the constraints
4. Explicitly state the non-negativity restriction

For a problem to be a linear programming problem, the decision variables, objective

function and constraints all have to be linear functions.

If the all the three conditions are satisfied, it is called a Linear Programming
Problem.

2. Solve Linear Programs by Graphical Method

A linear program can be solved by multiple methods. In this section, we are going to
look at the Graphical method for solving a linear program. This method is used to
solve a two variable linear program. If you have only two decision variables, you
should use the graphical method to find the optimal solution.

A graphical method involves formulating a set of linear inequalities subject to the

constraints. Then the inequalities are plotted on a X-Y plane. Once we have plotted
all the inequalities on a graph the intersecting region gives us a feasible region. The
feasible region explains what all values our model can take. And it also gives us the
optimal solution.

Let’s understand this with the help of an example.

Example: A farmer has recently acquired an 110 hectares piece of land. He has
decided to grow Wheat and barley on that land. Due to the quality of the sun and the
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region’s excellent climate, the entire production of Wheat and Barley can be sold. He
wants to know how to plant each variety in the 110 hectares, given the costs, net
profits and labor requirements according to the data shown below:

Variety Cost (Price/Hec) Net Profit (Price/Hec) Man-days/Hec

Wheat 100 50 10

Barley 200 120 30

The farmer has a budget of US$10,000 and an availability of 1,200 man-days during
the planning horizon. Find the optimal solution and the optimal value.

Solution: To solve this problem, first we gonna formulate our linear program.

Formulation of Linear Problem

Step 1: Identify the decision variables

The total area for growing Wheat = X (in hectares)

The total area for growing Barley = Y (in hectares)

X and Y are my decision variables.

Step 2: Write the objective function

Since the production from the entire land can be sold in the market. The
farmer would want to maximize the profit for his total produce. We are given net
profit for both Wheat and Barley. The farmer earns a net profit of US$50 for each
hectare of Wheat and US$120 for each Barley.

Our objective function (given by Z) is, Max Z = 50X + 120Y

Step 3: Writing the constraints

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 Mathematical Economics UNIT -4

1. It is given that the farmer has a total budget of US$10,000. The cost of producing
Wheat and Barley per hectare is also given to us. We have an upper cap on the total
cost spent by the farmer. So our equation becomes:

100X + 200Y ≤ 10,000

2. The next constraint is, the upper cap on the availability on the total number of
man-days for planning horizon. The total number of man-days available are 1200.
As per the table, we are given the man-days per hectare for Wheat and Barley.

10X + 30Y ≤ 1200

3. The third constraint is the total area present for plantation. The total available area
is 110 hectares. So the equation becomes,

X + Y ≤ 110

Step 4: The non-negativity restriction

The values of X and Y will be greater than or equal to 0. This goes without saying.

X ≥ 0, Y ≥ 0

We have formulated our linear program. It’s time to solve it.

Solving a LP through Graphical method

Since we know that X, Y ≥ 0. We will consider only the first quadrant.

To plot for the graph for the above equations, first I will simplify all the equations.

100X + 200Y ≤ 10,000 can be simplified to X + 2Y ≤ 100 by dividing by 100.

10X + 30Y ≤ 1200 can be simplified to X + 3Y ≤ 120 by dividing by 10.

The third equation is in its simplified form, X + Y ≤ 110.

Plot the first 2 lines on a graph in first quadrant (like shown below)

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 Mathematical Economics UNIT -4

The optimal feasible solution is achieved at the point of intersection where the
budget & man-days constraints are active. This means the point at which the
equations X + 2Y ≤ 100 and X + 3Y ≤ 120 intersect gives us the optimal solution.

The values for X and Y which gives the optimal solution is at (60,20).

To maximize profit the farmer should produce Wheat and Barley in 60 hectares and
20 hectares of land respectively.

The maximum profit the company will gain is,

Max Z = 50 * (60) + 120 * (20)

= US$5400

3. Solve Linear Program Using R

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 Mathematical Economics UNIT -4

R is an open source tool which is very popular among the data scientists for
essential data science tasks. Performing linear programming is very easy and we
can attain an optimum solution in very few steps. Come let’s learn.

Example: A toy manufacturing organization manufactures two types of toys A and

B. Both the toys are sold at Rs.25 and Rs.20 respectively. There are 2000 resource
units available every day from which the toy A requires 20 units while toy B requires
12 units. Both of these toys require a production time of 5 minutes. Total working
hours are 9 hours a day. What should be the manufacturing quantity for each of the
pipes to maximize the profits?

Here:

The objective function is:

Max.Z=25x+20y

where x are the units of pipe A

y are the units of pipe B

Constraints:
20x+12y<=2000

5x+5y<=540

Difference and Differential equations

Differential equations describe continuous functions. When possible, you
solve for the function in closed form.

Closed form solutions usually require regular geometries and simple shapes.
They are impossible for real life geometries.

That’s where approximate, numerical solutions come in.

All techniques - finite differences, finite elements, and boundary elements -

start by discretizing the irregular shape as a mesh of connected points and
approximating the differential equation as matrices. Linear algebra is used to
solve for an approximate solution.

These solutions are obtained using computers. They can involve thousands or
millions of unknowns.
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 Mathematical Economics UNIT -4

In simple words, Differential equations are based on analog machinery (

Continuity of signal ) while Difference equations are based on digital
machinery ( Discrete signals ). Moreover, differential equations are generally
used in analog circuits and conversely, difference equations are used in digital
circuits (systems that take measurements with digital I/O boards or GPIB
instruments).

Z transform is used to solve difference equations while Laplace transform is

used for differential equations.

Differential equation (D.E.) is an equation which involves in it the derivatives

(dy/dx) of a function y = f(x) . For example, dy/dx + py = q , while a difference
equation (d.e.) involves differences of terms in a sequence and it can be
expressed in terms of shift operator E or forward difference operator delta
. For example;

U(n + 2) - U(n +1) - U(n) = 0 … .. . . (1) and it can be written as;

[E^(2) - E - 1]U(n) = 0, n= 0, 1, 2, … ..n …. .

Difference equations (d.e.)as one given in eq.(1) are very useful in describing
the discrete problems involving discrete signals while D. E. are used in anolog
system involving continuous signals. In fact a d. e.is a discrete version of a D.
E. ; for example the d.e. (1) is a discrete form of the D. E. dy/dx = y

A differential equation is a mathematical equation that relates some

function with its derivatives. In applications, the functions
usually represent physical quantities, the derivatives represent their rates of
change, and the differential equation defines a relationship between the two.

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PART-1

1) and Then is
(a) 0 (b) 1 (c) 2 (d) 3

2) where is a matrix. Then

(a) A must be a zero matrix (b) A is an identity matrix

(c) rank of A is 1 or 0 (d) A is diagonalizable

3) The number of linearly independent eigen vectors of is

(a) 1 (b) 2 (c) 3 (d) 4

4) The minimal polynomial of is

(a) (b) (c) (d)

5) A is a unitary matrix. Then eigen value of A are

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(a) 1, -1 (b) 1, -i (c) i, -i (d) -1, i

6) is an operator on The invariant subspaces of the operator

are
(a) and the subspace with base {(0,1)} (b) and the zero subspace

(c) the zero subspace and the subspace with base {(1,1)} (d) only

7) Rank of the matrix is

(a) 2 (b) 3 (c) 4 (d) 5

8) The dimension of the subspace of spanned by

and is
(a) 1 (b) 2 (c) 3 (d) 4

9) U and V are subspace of such that

U = span [(1,2,3,4), (5,7,2,1), (3,1,4,-3)]

V=span [(2,1,2,3), (3,0,1,2), (1,1,5,3)].

Then the dimension of is

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(a) 1 (b) 2 (c) 3 (d) 4

10) Let be the set of all n-square symmetric matrices and the
characteristics polynomial of each is of the form
Then the dimension of over R is

(a) (b) (c) (d)

11) A is a matrix with Then is

(a) 6 (b) 4 (c) 9 (d) 100

12) A is matrix, all of whose entries are 1, then

(a) A is not diagonalizable (b) A is idempotent (c) A is nilpotent

(d) The minimal polynomial and the characteristics polynomial of A are not
equal.

13) A is an upper triangular with all diagonal entries zero, then I+A is
(a) invertible (b) idempotent (c) singular (d) nilpotent

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14) Number of linearly independent eigen vectors of is
(a) 1 (b) 2 (c) 3 (d) 4

15) A is a matrix over then

(a) is a minimal polynomial (b) is a characteristics polynomial

(c) both (a) and (b) are true (d) none of (a) and (b) is true

16) M is a 2-square matrix of rank 1, then M is

(a) diagonalizable and non singular (b) diagonalizable and nilpotent

(c) neither diagonalizable nor nilpotent (d) either diagonalizable or

nilpotent

17) A be a n-square matrix with integer entries and Then

(a) B is idempotent (b) exist (c) B is nilpotent (d) B-I is
idempotent

18) Let then is

(a) a minimal polynomial of A (b) a characteristics polynomial of A

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(c) both (a) and (b) are true (d) none of (a) and (b) is true

19) A is a 4-square matrix and Then

(a) (b) (c) (d)

20) and Then

(a) A and B are similar (b) A and B are not similar

(c) A and B are nilpotent (d) A and AB are similar

21) Let be subset of Then

(a) S is linearly independent (b) S is linearly dependent

(c) (2,-1,3), (0,1,1), (1,0,-2) are linearly dependent (d) S is a basis

of

22) such that Then rank of T is

(a) 1 (b) 2 (c) 3 (d) 4

23) such that and Then

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(a) is 1-1, is not (b) is 1-1, is not

(c) is onto and is 1-1 (d) and both are 1-1

24) such that then

(a) T is 1-1 (b) T is onto (c) T is both 1-1 and onto (d) T is neither 1-1
nor onto

25) W=L{(1,0,0,0), (0,1,0,0)}, then

(a)

(b)

(c)

(d) is a basis of

26) then
(a) A has zero image (b) all the eigen value of A are zero

(c) A is idempotent (d) A is nilpotent

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27) defined by Then
(a) T is nilpotent (b) T has at least one non-zero eigen value

(c) index of nilpotent is three (d) T is not nilpotent

28) where then

(a) A is not diagonalizable (b) A is idempotent

(c) A is nilpotent (d) minimal polynomial ≠ characteristics polynomial

29) and rank of A is 1, then

(a) A is diagonalizable (b) A is nilpotent

(c) both (a) and (b) are true (d) none of (a) and (b) is true

30) A is a 3-square matrix and the eigen values of A are -1, 0, 1 with
respect to the eigen vectors then 6A is

(a)

(b)

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(c)

(d)

31) The sum of eigen values of is

(a) -3 (b) -1 (c) 3 (d) 1

32) The matrix where has

(a) three real, non-zero eigen values (b) complex eigen values

(c) two non-zero eigen values (d) only one non-zero eigen value

33) is
(a) diagonalizable (b) nilpotent (c) idempotent (d) not diagonalizable

34) If a square matrix of order 10 has exactly 5 distinct eigen values, then
the degree of the minimal polynomial is
(a) at least 5 (b) at most 5 (c) always 5 (d) exactly 10
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35) defined by T(A)=BA, where Then rank of
T is
(a) 1 (b) 2 (c) 3 (d) 4

36) Then
(a) both and are diagonalizable (b) is diagonalizable but not

(c) and have the same minimal polynomial (d) is diagonalizable

but not

37) Rank of is 5 and that of is 3, then rank of AB is

(a) 1 (b) 2 (c) 3 (d) 4

38) A and B are n-square positive definite matrices. Then which of the
following are positive definite.
(a) A+B (b) ABA (c) AB (d)

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39) and then which of the following are
subspaces of
(a) (b)

(c) (d)

40) Let T be a linear operator on the vector space V and T be invariant

under the subspace W of V. Then
(a) (b) (c) (d) None of these

41) where Then the dimension of kernel of

A is
(a) 1 (b) 2 (c) 3 (d) 4

42) where Then the dimension of image of A

is
(a) 1 (b) 2 (c) 3 (d) 4

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43) Let u, v, w be three non-zero vectors which are linearly independent,
then
(a) u is linear combination of v and w (b) v is linear combination of u and
w

(c) w is linear combination of u and v (d) none of these

44) Let U and W be subspaces of a vector space V and is also a

subspace of V, then
(a) either or (b) (c) U=W (d) None of these

45) Let I be the identity transformation of the finite dimensional vector

space V, then the nullity of I is
(a) dimV (b) 0 (c) 1 (d) dimV – 1

46) such that for Then is a zero

of the polynomial:
(a) (b) (c) (d) none of above

47) The sum of the eigen values of the matrix is

(a) 4 (b) 23 (c) 11 (d) 12
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48) Let A and B are square matrices such that AB=I, then zero is an eigen
value of
(a) A but not of B (b) B but not of A (c) both A and B (d) neither A nor
B

49) The eigen values of a skew-symmetric matrix are

(a) negative (b) real (c) absolute value of 1 (d) purely imaginary or
zero

50) The characteristics equation of a matrix A is then

(a) does not exist (b) exit but cannot be determined from the
data

(c) (d)

ANSWER-

1) c , 2) d , 3) c , 4) c , 5) c, 6) b, 7) c, 8) b, 9) b, 10) c , 11) b, 12) d , 13)

d , 14) d , 15) d, 16) d , 17) b, 18) d, 19) c , 20) a & c, 21) a , 22) c , 23) a
, 24) c , 25) b, c & d, 26) b & d, 27) a, b & c , 28) d, 29) b, 30) b, 31) b,
32) d, 33) d, 34) a, 35) b, 36) a, 37) c, 38) a, 39) a, b, c & d, 40) d, 41) b,
42) b, 43) d, 44) a, 45) b, 46) d, 47) c, 48) d, 49) d, 50) d

PART - 2

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12
1. Decision Science approach is

a. Multi-disciplinary

b. Scientific

c. Intuitive

d. All of the above

2. For analyzing a problem, decision-makers should study

a. Its qualitative aspects

b. Its quantitative aspects

c. Both a & b

d. Neither a nor b

3. Decision variables are

a. Controllable

b. Uncontrollable

c. Parameters

d. None of the above

4. A model is

a. An essence of reality

b. An approximation

c. An idealization

d. All of the above

5. Managerial decisions are based on

a. An evaluation of quantitative data

b. The use of qualitative factors

c. Results generated by formal models

d. All of the above

6. The use of decision models

a. Is possible when the variables value is known

b. Reduces the scope of judgement & intuition known with certainty in decision-making
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c. Require the use of computer software

d. None of the above

7. Every mathematical model

a. Must be deterministic

b. Requires computer aid for its solution

c. Represents data in numerical form

d. All of the above

8. A physical model is example of

a. An iconic model

b. An analogue model

c. A verbal model

d. A mathematical model

9. An optimization model

a. Provides the best decision

b. Provides decision within its limited context

c. Helps in evaluating various alternatives

d. All of the above

10. The quantitative approach to decision analysis is a

a. Logical approach

b. Rational approach

c. Scientific approach

d. All of the above

11. The qualitative approach to decision analysis relies on

a. Experience

b. Judgement

c. Intuition

d. All of the above

12. The mathematical model of an LP problem is important because

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a. It helps in converting the verbal description & numerical data into mathematical expression

b. Decision-makers prefer to work with formal models

c. It captures the relevant relationship among decision factors

d. It enables the use of algebraic technique

13. Linear programming is a

a. Constrained optimization technique

b. Technique for economic allocation of limited resources

c. Mathematical technique

d. All of the above

14. A constraint in an LP model restricts

a. Value of objective function

b. Value of a decision variable

c. Use of the available resources

d. All of the above

15. The distinguishing feature of an LP model is

a. Relationship among all variables is linear

b. It has single objective function & constraints

c. Value of decision variables is non-negative

d. All of the above

16. Constraints in an LP model represents

a. Limitations

b. Requirements

c. Balancing limitations & requirements

d. All of the above

17. Non-negativity condition is an important component of LP model because

a. Variables value should remain under the control of the decision-maker

b. Value of variables make sense & correspond to real-world problems

c. Variables are interrelated in terms of limited resources

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d. None of the above

18. Before formulating a formal LP model, it is better to

a. Express each constrain in words

b. Express the objective function in words

c. Verbally identify decision variables

d. All of the above

19. Maximization of objective function in an LP model means

a. Value occurs at allowable set of decisions

b. Highest value is chosen among allowable decisions

c. Neither of above

d. Both a & b

20. Which of the following is not a characteristic of the LP model

a. Alternative courses of action

b. An objective function of maximization type

c. Limited amount of resources

d. Non-negativity condition on the value of decision variables.

21. The best use of linear programming technique is to find an optimal use of

a. Money

b. Manpower

c. Machine

d. All of the above

22. Which of the following is not a characteristic of the LP

a. Resources must be limited

b. Only one objective function

c. Parameters value remains constant during the planning period

d. The problem must be of minimization type

23. Non-negativity condition in an LP model implies

a. A positive coefficient of variables in objective function

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b. A positive coefficient of variables in any constraint

c. Non-negative value of resources

d. None of the above

24. Which of the following is an assumption of an LP model

a. Divisibility

b. Proportionality

c. Additivity

d. All of the above

25. Which of the following is a limitation associated with an LP model

a. The relationship among decision variables in linear

b. No guarantee to get integer valued solutions

c. No consideration of effect of time & uncertainty on LP model

d. All of the above

26. The graphical method of LP problem uses

a. Objective function equation

b. Constraint equations

c. Linear equations

d. All of the above

27. A feasible solution to an LP problem

a. Must satisfy all of the problem’s constraints simultaneously

b. Need not satisfy all of the constraints, only some of them

c. Must be a corner point of the feasible region

d. Must optimize the value of the objective function

28. Which of the following statements is true with respect to the optimal solution of an LP problem

a. Every LP problem has an optimal solution

b. Optimal solution of an LP problem always occurs at an extreme point

c. At optimal solution all resources are completely used

d. If an optimal solution exists, there will always be at least one at a corner

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29. An iso-profit line represents

a. An infinite number of solutions all of which yield the same profit

b. An infinite number of solution all of which yield the same cost

c. An infinite number of optimal solutions

d. A boundary of the feasible region

30. If an iso-profit line yielding the optimal solution coincides with a constaint line, then

a. The solution is unbounded

b. The solution is infeasible

c. The constraint which coincides is redundant

d. None of the above

31. While plotting constraints on a graph paper, terminal points on both the axes are connected by

a straight line because

a. The resources are limited in supply

b. The objective function as a linear function

c. The constraints are linear equations or inequalities

d. All of the above

32. A constraint in an LP model becomes redundant because

a. Two iso-profit line may be parallel to each other

b. The solution is unbounded

c. This constraint is not satisfied by the solution values

d. None of the above

33. If two constraints do not intersect in the positive quadrant of the graph, then

a. The problem is infeasible

b. The solution is unbounded

c. One of the constraints is redundant

d. None of the above

34. Constraints in LP problem are called active if they

a. Represent optimal solution

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b. At optimality do not consume all the available resources

c. Both a & b

d. None of the above

35. The solution space (region) of an LP problem is unbounded due to

a. An incorrect formulation of the LP model

b. Objective function is unbounded

c. Neither a nor b

d. Both a & b

36. While solving a LP model graphically, the area bounded by the constraints is called

a. Feasible region

b. Infeasible region

c. Unbounded solution

d. None of the above

37. Alternative solutions exist of an LP model when

a. One of the constraints is redundant

b. Objective function equation is parallel to one of the constraints

c. Two constraints are parallel

d. All of the above

38. While solving a LP problem, infeasibility may be removed by

a. Adding another constraint

b. Adding another variable

c. Removing a constraint

d. Removing a variable

39. If a non-redundant constraint is removed from an LP problem then

a. Feasible region will become larger

b. Feasible region will become smaller

c. Solution will become infeasible

d. None of the above

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40. If one of the constraint of an equation in an LP problem has an unbounded solution, then

a. Solution to such LP problem must be degenerate

b. Feasible region should have a line segment

c. Alternative solutions exist

d. None of the above

41. The initial solution of a transportation problem can be obtained by applying any known method.

However, the only condition is that

a. The solution be optimal

b. The rim conditions are satisfied

c. The solution not be degenerate

d. All of the above

42. The dummy source or destination in a transportation problem is added to

a. Satisfy rim conditions

b. Prevent solution from becoming degenerate

c. Ensure that total cost does not exceed a limit

d. None of the above

43. The occurrence of degeneracy while solving a transportation problem means that

a. Total supply equals total demand

b. The solution so obtained is not feasible

c. The few allocations become negative

d. None of the above

44. An alternative optimal solution to a minimization transportation problem exists whenever

opportunity cost corresponding to unused route of transportation is:

a. Positive & greater than zero

b. Positive with at least one equal to zero

c. Negative with at least one equal to zero

d. None of the above

45. One disadvantage of using North-West Corner rule to find initial solution to the transportation
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problem is that

a. It is complicated to use

b. It does not take into account cost of transportation

c. It leads to a degenerate initial solution

d. All of the above

46. The solution to a transportation problem with ‘m’ rows (supplies) & ‘n’ columns (destination) is

feasible if number of positive allocations are

a. m+n

b. m*n

c. m+n-1

d. m+n+1

47. If an opportunity cost value is used for an unused cell to test optimality, it should be

a. Equal to zero

b. Most negative number

c. Most positive number

d. Any value

48. During an iteration while moving from one solution to the next, degeneracy may occur when

a. The closed path indicates a diagonal move

b. Two or more occupied cells are on the closed path but neither of them represents a corner

of the path.

c. Two or more occupied cells on the closed path with minus sign are tied for lowest circled

value

d. Either of the above

49. The large negative opportunity cost value in an unused cell in a transportation table is chosen to

improve the current solution because

a. It represents per unit cost reduction

b. It represents per unit cost improvement

c. It ensure no rim requirement violation

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d. None of the above

50. The smallest quantity is chosen at the corners of the closed path with negative sign to be

assigned at unused cell because

a. It improve the total cost

b. It does not disturb rim conditions

c. It ensure feasible solution

d. All of the above

51. When total supply is equal to total demand in a transportation problem, the problem is said to

be

a. Balanced

b. Unbalanced

c. Degenerate

d. None of the above

52. Which of the following methods is used to verify the optimality of the current solution of the

transportation problem

a. Least cost method

b. Vogel’s approximation method

c. Modified distribution method

d. All of the above

53. The degeneracy in the transportation problem indicates that

a. Dummy allocation(s) needs to be added

b. The problem has no feasible solution

c. The multiple optimal solution exist

d. a & b but not c

54. An assignment problem is considered as a particular case of a transportation problem because

a. The number of rows equals columns

b. All xij = 0 or 1

c. All rim conditions are 1

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d. All of the above

55. An optimal assignment requires that the maximum number of lines that can be drawn through

squares with zero opportunity cost be equal to the number of

a. Rows or columns

b. Rows & columns

c. Rows + columns – 1

d. None of the above

56. While solving an assignment problem, an activity is assigned to a resource through a square with

zero opportunity cost because the objective is to

a. Minimize total cost of assignment

b. Reduce the cost of assignment to zero

c. Reduce the cost of that particular assignment to zero

d. All of the above

57. The method used for solving an assignment problem is called

a. Reduced matrix method

b. MODI method

c. Hungarian method

d. None of the above

58. The purpose of a dummy row or column in an assignment problem is to

a. Obtain balance between total activities & total resources

b. Prevent a solution from becoming degenerate

c. Provide a means of representing a dummy problem

d. None of the above

59. Maximization assignment problem is transformed into a minimization problem by

a. Adding each entry in a column from the maximization value in that column

b. Subtracting each entry in a column from the maximum value in that column

c. Subtracting each entry in the table from the maximum value in that table

d. Any one of the above

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60. If there were n workers & n jobs there would be

a. n! solutions

b. (n-1)! solutions

c. (n!)n

solutions

d. n solutions

61. An assignment problem can be solved by

a. Simplex method

b. Transportation method

c. Both a & b

d. None of the above

62. For a salesman who has to visit n cities which of the following are the ways of his tour plan

a. n!

b. (n+1)!

c. (n-1)!

d. n

63. The assignment problem

a. Requires that only one activity be assigned to each resource

b. Is a special case of transportation problem

c. Can be used to maximize resources

d. All of the above

64. An assignment problem is a special case of transportation problem, where

a. Number of rows equals number of columns

b. All rim conditions are 1

c. Values of each decision variable is either 0 or 1

d. All of the above

65. Every basic feasible solution of a general assignment problem, having a square pay-off matrix of

order, n should have assignments equal to

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a. 2n+1

b. 2n-1

c. m+n-1

d. m+n

66. To proceed with the MODI algorithm for solving an assignment problem, the number of dummy

allocations need to be added are

a. n

b. 2n

c. n-1

d. 2n-1

67. The Hungarian method for solving an assignment problem can also be used to solve

a. A transportation problem

b. A travelling salesman problem

c. A LP problem

d. Both a & b

68. An optimal solution of an assignment problem can be obtained only if

a. Each row & column has only one zero element

b. Each row & column has at least one zero element

c. The data is arrangement in a square matrix

d. None of the above

69. Customer behavior in which the customer moves from one queue to another in a multiple

channel situation is

a. Balking

b. Reneging

c. Jockeying

d. Altering

70. Which of the following characteristics apply to queuing system

a. Customer population
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b. Arrival process

c. Both a & b

d. Neither a nor b

71. Which of the following is not a key operating characteristics apply to queuing system

a. Utilization factor

b. Percent idle time

c. Average time spent waiting in the system & queue

d. None of the above

72. Priority queue discipline may be classified as

a. Finite or infinite

b. Limited & unlimited

c. Pre-emptive or non-pre-emptive

d. All of the above

73. The calling population is assumed to be infinite when

a. Arrivals are independent of each other

b. Capacity of the system is infinite

c. Service rate is faster than arrival rate

d. All of the above

74. Which of the cost estimates & performance measures are not used for economic analysis of a

queuing system

a. Cost per server per unit of time

b. Cost per unit of time for a customer waiting in the system

c. Average number of customers in the system

d. Average waiting time of customers in the system

75. A calling population is considered to be infinite when

a. All customers arrive at once

b. Arrivals are independent of each other

c. Arrivals are dependent upon each other

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d. All of the above

76. The cost of providing service in a queuing system decreases with

a. Decreased average waiting time in the queue

b. Decreased arrival rate

c. Increased arrival rate

d. None of the above

77. Service mechanism in a queuing system is characterized by

a. Server’s behavior

b. Customer’s behavior

c. Customers in the system

d. All of the above

78. Probabilities of occurrence of any state are

a. Collectively exhaustive

b. Mutually exclusive

c. Representing one of the finite numbers of states of nature in the system

d. All of the above

79. In a matrix of transition probability, the probability values should add up to one in each

a. Row

b. Column

c. Diagonal

d. All of the above

80. In a matrix of transition probability, the element aij where i=j is a

a. Gain

b. Loss

c. Retention

d. None of the above

81. In Markov analysis, state probabilities must

a. Sum to one
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b. Be less than one

c. Be greater than one

d. None of the above

82. State transition probabilities in the Markov chain should

a. Sum to 1

b. Be less than 1

c. Be greater than 1

d. None of the above

83. If a matrix of transition probability is of the order n*n, then the number of equilibrium

equations would be

a. n

b. n-1

c. n+1

d. None of the above

84. In the long run, the state probabilities become 0 & 1

a. In no case

b. In same cases

c. In all cases

d. Cannot say

85. While calculating equilibrium probabilities for a Markov process, it is assumed that

a. There is a single absorbing state

b. Transition probabilities do not change

c. There is a single non-absorbing state

d. None of the above

86. The first-order Markov chain is generally used when

a. Transition probabilities are fairly stable

b. Change in transition probabilities is random

c. No sufficient data are available

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d. All of the above

87. A problem is classified as Markov chain provided

a. There are finite number of possible states

b. States are collectively exhaustive & mutually exclusive

c. Long-run probabilities of being in a particular state will be constant over time

d. All of the above

88. The transition matrix elements remain positive from one point to the next. This property is

known as:

a. Steady-state property

b. Equilibrium property

c. Regular property

d. All of the above

89. Markov analysis is useful for:

a. Predicting the state of the system at some future time

b. Calculating transition probabilities at some future time

c. All of the above

d. None of the above

90. Which of the following is not one of the assumptions of Markov analysis:

a. There are a limited number of possible states

b. A future state can be predicted from the preceding one

c. There are limited number of future periods

d. All of the above

91. An advantage of simulation as opposed to optimization is that

a. Several options of measure of performance can be examined

b. Complex real-life problems can be studied

c. It is applicable in cases where there is an element of randomness in a system

d. All of the above

92. The purpose of using simulation technique is to

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a. Imitate a real-world situation

b. Understand properties & operating characteristics of complex real-life problems

c. Reduce the cost of experiment on a model of real situation

d. All of the above

93. Which of the following is not the special purpose simulation language

a. BASIC

b. GPSS

c. GASP

d. SIMSCRIPT

94. As simulation is not an analytical model, therefore the result of simulation must be viewed as

a. Unrealistic

b. Exact

c. Approximation

d. Simplified

95. While assigning random numbers in Monte Carlo simulation, it is

a. Not necessary to assign the exact range of random number interval as the probability

b. Necessary to develop a cumulative probability distribution

c. Necessary to assign the particular appropriate random numbers

d. All of the above

96. Analytical results are taken into consideration before a simulation study so as to

a. Identify suitable values of the system parameters

b. Determine the optimal decision

c. Identify suitable values of decision variables for the specific choices of system parameters

d. All of the above

97. Biased random sampling is made from among alternatives which have

a. Equal probability

b. Unequal probability

c. Probability which do not sum to 1

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d. None of the above

98. Large complicated simulation models are appreciated because

a. Their average costs are not well-defined

b. It is difficult to create the appropriate events

c. They may be expensive to write and use as an experimental device

d. All of the above

99. Simulation should not be applied in all cases because it

a. Requires considerable talent for model building & extensive computer programming efforts

b. Consumes much computer time

c. Provides at best approximate solution to problem

d. All of the above

100. Simulation is defined as

a. A technique that uses computers

b. An approach for reproducing the processes by which events by chance & changes are created

in a computer

c. A procedure for testing & experimenting on models to answer what if ___, then so & so ___

types of questions

d. All of the above

101. The general purpose system simulation language

a. Requires programme writing

b. Does not require programme writing

c. Requires predefined coding forms

d. Needs a set of equations to describe a system

102. Special simulation languages are useful because they

a. Reduce programme preparation time & cost

b. Have the capability to generate random variables

c. Require no prior programming knowledge

d. All of the above

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103. Few causes of simulation analysis failure are

a. Inadequate level of user participation

b. Inappropriate levels of detail

c. Incomplete mix of essential skills

d. All of the above

104. To make simulation more popular, we need to avoid

a. Large cost over runs

b. Prolonged delays

c. User dissatisfaction with simulation results

d. All of the above

105. The important step required for simulation approach in solving a problem is to

a.Test & validate the model

b. Design the experiment

c. Conduct the experiment

d. All of the above

ANSWER –

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PART – 3

MULTIPLE CHOICE QUESTIONS

CHAPTER – 9 DIFFERENTIAL EQUATIONS
 d 2 y   dy 2 sin  dy 1 0 is:
3

Q1. The degree of the differential equation  2    dx 

dx  dx 
     


(a) 3 (b) 2 (c) 1 (d) not defined

2 d2y dy
Q2. The order of the differential equation 2x 3  y  0 is:
dx2 dx
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 (a) 2 (b) 1 (c) 0 (d) not defined

Q3. The solution of the differential equation  e x 1 y dy   y 1 e x dx is:

(b) e y  c  e x  y 1
 
(a) e y  c  e x 1  y 1 (c) y   e x 1  y 1 (d) none of these

dy
1 x  y  xy is:
Q4. The solution of the differential equation
dx
1 y x2
y x2
x2
(a) e x c (b) log 1 y  x  c (c) e  x  c (d) none of these
2
2 2

dy
Q5. The solution of the differential equation   0 is:
1 y2
dx
1 x2

(a) (b) sin1 y sin1 x  c (c) sin1 y sin1 x  c (d) none of these
y sin1 y sin1 x  c

1 x2
Q6. The solution of the differential equation x
1 y dx  y
2 dy  0 is:

1 x2 1 y2
(a) sin1 x sin1 y  c (b)  c (c) tan1 x  tan1 y 
(d) none of these
c

dy y2 x 2
Q7. The solution of the differential equation
 is:
dx 2xy

(a)
x2  y2  cx (b) x2  y2  cy (c) x2  y2  cx (d) none of these

dy
Q8. The solution of the differential equation  y tan x sec x is:
dx

(a) (b) y sin x  c cos x (c)  y sin xsin x  c

y sin x  c cos x (d) none of these

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34
 dy
Q9. The integrating factor of the differential equation  x log x  y  2 log x is:
dx

(a) log log x (b) ex (c) log x (d) x

dy y
Q10. The solution of the differential equation  sin x is:
dx x

x  y cos x  sin x  c (b) x  y cos x sin x  c

(a) (c) x  y cos x  cos x  c (d) none of these

dy
Q11. The solution of the differential equation 2x  y  3 represents:
dx

(a) circles (b) straight lines (c) ellipses (d) parabolas

dy y
Q12. The solution of the differential equation x  y  x tan is:
dx x

x x y
(a) sin  x  c y
(b) sin  cx (c) sin  cy (d) sin  cy
y y x
x

dy
Q13. The solution of the differential equation x  y  4 is:
2 2

dx

(a) (b) x2  y 2 3x  c `(c) (d) x3  y3 12x c

x2  y 2 12x  c x3  y3 3x  c

Q14. The solution of the differential equation

xdx  ydy  x2 ydy  y2 xdx is:

  
(a) x2 1 c 1 y2  (b) x2 1 c 1 y2  (c) x3 1 c 1 y3  (d) x3 1 c 1 y3 

Q15. The solution of the differential equation  x 2 1   y 2 1  0 is:

dy
dx


1 x 1 x
(a) y  2  x 2
(b) y  (c) y  x  x 1 (d) y 
1 x 1 x
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 dy x2  xy  y2
Q16. The solution of the differential equation  is:
dx x2

1 x  1 y  x 

(a) tan log y  c (b) tan log x  c (c) tan1 log x  c (d) none of these
     y 
 y   x   

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 ANSWERS

1. d 2. a 3. a 4. b 5. c 6. b 7. c 8. b 9. c 10. a 11. d 12. b 13. d 14. a

15.
d 16. d

PART - 4

MULTIPLE CHOICE QUESTIONS CHAPTER – 6

APPLICATIONS OF DERIVATIVES

1
Q1. For the function f  x  x  , x1, 3, the value of c for the mean value theorem is:
x

(c) 2 (d) none of these

(a) 1 (b) 3

Q2. The value of c in Rolle ’s Theorem when f  x  2x3 5x 2  4x  3 , x1 3,3 is:

(a) 2 (b) 1 3 (c) 2 (d) 2 3

Q3. For the function f  x  x  x  2  , x1, 2, the value of c for the mean value theorem is:

(a) 1 (b) 1 2 (c) 2 3 (d) 3 2

Q4. The value of c in Rolle ’s Theorem when f  x  ex sin x , x0,  is:

(a)  6 (b)  4 (c)  2 (d) 3 4

Q5. The cost function of a firm is C 3x 2  2x 3. Then the marginal cost, when x  3 is:

(a) 10 (b) 20 (c) 5 (d) 25

Q6. The function f  x  tan x  x is:

(a) Always increasing (b) always decreasing

(c) not always decreasing (d) sometimes increasing and sometimes decreasing
Q7. The function f  x  x  6x 15x 12 is:
3 2

DIWAKAR EDUCAION HUB Page 1

 (a) Strictly decreasing on R (b) Strictly increasing on R
(b) Increasing on  , 2and decreasing in  2,  (d) none of these
Q8. The function f  x  4 3x  3x2  x3 is:

(a) Decreasing on R (b) Increasing on R

(c) Strictly increasing on R (d) Strictly increasing on R

f  x   x
Q9. The function is:
sin x
(a) Increasing in 0,1 (b) Decreasing in 0,1
 1 1 

(c) Increasing in 0, and decreasing in ,1   (d) none of these
 2  2 
 

   

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 Q10. The function f  x  xx is decreasing in the interval:

(a)  0, e   1
(b) 0, 
 (c) 0,1  (d) none of these
 e 


  
Q11. The function
f  x    x  x 3 is increasing in:
2

(a)  0,  (b)  ,0 (c) 1, 3  3

(d) 0, 3, 


 2 


  


x 
Q12. The function f  x 
x 1 is increasing in:
2

(a) 1,1 (b) 1,  (c) , 11,  (d) none of these

Q13. The function f  x cot1 x  x increases in the interval:

(a) 1,  (b) 1,  (c) ,  (d) 0, 


Q14. If the function f  x  kx3 9x 2  9x  3is monotonically increasing in every interval, then:

(a) k  3 (b) k  3 (c) k  3 (d) k  30.

Q15. Side of an equilateral triangle expands at rate of 2 cm / sec. The rate of increase of its area when
each side is 10 cm is (in cm2 /sec ):
(a) 10 2 (b) 10 3 (c) 10 (d) 5

Q16. The radius of a sphere is changing t the rate of 0.1 cm/sec. The rate of change of its surface area
when the radius is 200 cm (in cm2 /sec ) is:

(a) 8 (b) 12 (c) 160 (d) 200

Q17. A cone whose height is always equal to its diameter is increasing in volume at the rate of
40cm3 / sec. At what rate is the radius increasing when its circular base area is1m2 ?

(a) 1 mm/sec (b) 0.001 cm/sec (c) 2 mm/sec (d) 0.002 cm/sec

Q18. A cylindrical vessel of radius 0.5 m is filled with oil at the rate of 0.25 m3 / min. The rate at
which the surface of the oil is rising (in m / min ) is:

(a) 1 (b) 2 (c) 5 (d) 1.25

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Q19. The coordinates of the point on the ellipse 16x2  9y2  400 where the ordinate decrease at the
same rate at which the abscissa increases, is:
 16   16   1
(a)  3, 16 
 
(b) 3, 

(c) 3,  

(d) 3, 
 3   3   3   3 
   

       

DIWAKAR EDUCAION HUB Page 4

 Q20. If the rate of change of volume of a sphere is equal to the rate of change of its radius, then its
radius is equal to:

1 1 
(a) 1 unit (b) 2 unit (c) unit (d) unit
2 2 

Q21. A man of height 6 ft walks at a uniform speed of 9 ft/sec from a lamp post fixed at 15 ft height.
The length of his shadow is increasing at the rate of:
(a) 15 ft/sec (b) 9 ft/sec (c) 6 ft/sec (d) none of these

Q22. If there is an error of a% in measuring the edge of a cube, then percentage error in its surface is:
a
(a) 2a% (b) % (c) 3a% (d) none of these
2

Q23. In an error of k% is made in measuring radius of a sphere, then percentage error in its volume is:
k
(a) k% (b) 3k% (c) 2k%
(d)
3

Q24. A sphere of radius 100 mm shrinks to radius 98 mm, then approximate decrease in its volume is:
(a)
12000 (b) 800 mm3 (c) 8000 mm3 (d) 120 mm3
mm3

Q25. The approximate value of  33 is:

15

(a) 2.0125 (b) 2.1 (c) 2.01 (d) none of these

Q26. The circumference of a circle is measured as 28 cm with an error of 0.01 cm. The percentage
error in its area is:
1 1
(a) (b) 0.01 (c)
14 (d) none of these
7

Q27. The equation of normal to the curve x  2 is:

y  x sin x cos x at

(a) x  2 (b) x  (c) x    0 (d) 2x 


Q28. The point on the curve
y  x2 3x  2 where tangent is perpendicular to y  x is:

(a) 1 2,1 4 (b) 1 4,1 2 (c) 4, 2 (d) 1,1

DIWAKAR EDUCAION HUB Page 5

Q29. The point on the curve where tangent makes
y2  x 450 angle with x-axis is:

(a)  0, 0 (b) 2,16 (c) 3, 9 (d) none of these

Q30. The angle between the curves

y2  x and x2  y at 1,1 is:

(a) tan1 4 3 (b) tan1 3 4 (c) 900 (d) 450

DIWAKAR EDUCAION HUB Page 6

 Q31. The equation of tangent at those points where the curve y  x2  3x  2 meets x-axis are:

(a) (b) x  y 1  0 , x  y  2  0
x  y  2  0 , x  y 1  0

(c) x  y 1  0 , x  y  0 (d) x  y  0 , x  y  0

Q32. At what point the slope of the tangent to the curve

x2  y 2  2x 3 is zero?

(a) 3, 0 , 1, 0 (b)  3, 0  , 1, 2 (c) 1, 0,1, 2 (d) 1, 2,1,  2


Q33. If the curve ay  x2  7 and x3  y cut each other at 900 at 1,1 , then value of a is:

(a) 1 (b) 6 (c) 6 (d) 0

Q34. The equation of normal 4 is:
x  a cos3  , y  a sin3  at the point  

(a) x  0 (b) y  0 (c) x  y (d) x  y  a

Q35. The angle of intersection of the parabolas

y2  4ax and x2  4ay at the origin is:

(a)  6 (b)  3 (c)  2 (d)  4

Q36. The line y  x y2  4x at the point:
1touches

(a) 1, 2 (b) 2,1 (c) 1,  2 (d) 1, 2


Q37. The tangent to the curve x axis at:

y  e2 x at the point 0,1
meets

(a) 0,1  1 
(b)  ,0  (d) 0, 2
 (c) 2, 0
 2 


  

Q38. The curves y  4x2  2x 8 and y  x3  x 13touch each other at the point:

(a) 3, 23 x (c) 34,3

(b) 23,  3

 f x 
log x
is:
Q39. The maximum value of
DIWAKAR EDUCAION HUB Page 7
 (d) 3,34

1 2
(a) (b)
e (c) e (d) 1
e
250
x2  is:
Q40. The minimum value of
x

(a) 0 (b) 25 (c) 50 (d) 75

DIWAKAR EDUCAION HUB Page 8

Q41. The maximum value of
f  x    x  2  x  3 is:
2

7
(a) 4
3 (b) 3 (c) (d) 0
27
Q42. The least value of f  x  ex  e x is:

(a) 2 (b) 0 (c) 2 (d) can’t be determine

1 x  x2
Q43. For all real values of x , the minimum value of y  is:
1 x  x2

1
(a) 0 (b) 1 (c) 3 (d)
3

Q44. The maximum value of

y sin x.cos x is:
1
(a)
4 (c) (d) 2 2
1 2
(b)
2
f  x   x  ax2  bx 1is maximum at x  0 and minimum at
3

Q45. If the function x 1, then:

3 3
(a) 2
a  ,b  0 (b) a   , b  0 (c) a  0,b  (d) none of these

3 2 2

ax  b
y has a turning point at P  2, 1 . The value of
Q46. If a and b so that y is maximum
 x 1 x  4
at P is:

(a) (b) a  1 , b  0 (c) a  1 , b  2 (d) a  2 , b  1

a0,b1

Q47. The smallest value of polynomial 3x4 8x3 12x2  48x 1in 1, 4 is:

(a) 49 (b) 59 (c) 59 (d) 257

Q48. The function f  x  2x3 3x2 12x  4 , has:

(a) Two points of local maximum (b) Two points of local minimum
(c) one maxima and one minima (d) no maxima or minima
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Q49. The sum of two non-zero numbers is 8, the minimum value of the sum of their reciprocals is:

(a) 1 4 (b) 1 2 (c) 1 8 (d) none of these

Q50. The point on the curve
x2  2 y which is nearest to the point 0, 5 is:


(a) 2 2 , 4  
(b) 2 2 , 0 
(c) 0, 0 (d) 2, 2

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 UNIT -4 ECONOMICS

ANSWERS

1. b 2. a 3. d 4. d 5. b 6. a 7. b 8. a 9. a 10. b 11. d 12. a 13. c 14. c

15. b 16. c 17. d 18. a 19. a 20. d 21. c 22. a 23. b 24. c 25. a 26. a 27. d 28. b

29. b 30. b 31. b 32. d 33. c 34. c 35. c 36. a 37. b 38. d 39. a 40. d 41. c 42. c

43. d 44. b 45. b 46. b 47. c 48. c 49. b 50. a

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