economics

Solution to Maximization Problems October 30, 2020

1. [Revenue Maximization]
Maximize the following total revenue function (TR) and the profit π by finding out
(a) the critical values (the output values); (b) checking the second order conditions
and (c) the maximum values.
Note I: First Order Condition (F.O.C.) The critical values are the set of points in
the domain the derivative of the total revenue or profit function is equal to zero.
Total Revenue

T R = 32Q − Q2.

(a) F. O. C.
dT R
dQ

= 32 − 2Q = 0,

yields the critical point (the output Q where the F.O.C. is satisfied) as

Q∗ =
32
2

= 16.

(b) The maximum total revenue is

T R∗ = T R(Q∗) = 32 × 16 − 162 = 512 − 256 = 256.

Page 1 of 4

Solution to Maximization Problems October 30, 2020

2. [Profit Maximization] Total Profit Function (π)

π = −
Q3

3
− 5Q2 + 2000Q − 326.

(a)

F. O. C.
dπ
dQ

= −
3Q2

3
− 5 ×(2Q)+ 2000 = 0,

which simplifies to
−Q2 − 10Q + 2000 = 0,

and can be factored as

−Q2 − 50Q + 40Q + 2000 = −Q(Q + 50)+ 40(Q + 50) = (Q + 50)(−Q + 40) = 0,

which yields the critical point as

Q∗ = 40.

Observe that Q = −50 also satisfies the F. O. C. but is not meaningful and hence we
ignore it.
(b) The maximum profit is

π∗ = π(Q∗) = −
403

3
−5×402 +2000×40−326 = −

64000
3

−8000+80000−326 = 50340.67.

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Solution to Maximization Problems October 30, 2020

3. [Minimum Average Cost] Minimize the average cost for each of the following total
cost function (TC) by finding out (a) average cost function, (b) the critical values
(the output values) at which AC is minimized; and (c) the minimum average cost.
Total Cost Function

T C = Q3 − 21Q2 + 500Q.

(a) The average cost is

AC =
T C
Q

=
Q3 − 21Q2 + 500Q

Q
= Q2 − 21Q + 500.

F. O. C.
dAC
dQ

= 2Q − 21 = 0,

which yields the critical point as

Q∗ = 10.5.

(b) The minimum AC is

AC∗ = AC(Q∗) = (10.5)2 − 21 × 10.5 + 500 = 110.25 − 220.5 + 500 = 389.75.

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Solution to Maximization Problems October 30, 2020

4. [Profit Maximization – II]
Given the following total revenue function (TR) and the total cost function (TC),
maximize profit π by following steps (a) set up the profit function

π = T R − T C,

(b) the critical values (the output values) where the profit is at a relative extremum;
(c) checking the second order conditions and (d) the maximum profit value.

T R = 4350Q − 13Q2;

T C = Q3 − 5.5Q2 + 150Q + 675.

(a) The profit is

π = 4350Q − 13Q2 − Q3 + 5.5Q2 − 150Q − 675 = 4200Q − 7.5Q2 − Q3 − 675.

F. O. C.
dπ
dQ

= 4200 − 15Q − 3Q2 = 0,

which can be factored as

4200 − 120Q + 105Q − 3Q2 = 120(35 − Q)+ 3Q(35 − Q) = (35 − Q)(105 + 3Q) = 0.

This yields the critical point as
Q∗ = 35,

as the other solution Q = −40 is not economically meaningful.
(b) The maximum profit is

π∗ = π(Q∗) = 4200 × 35 − 7.5(35)2 −(35)3 − 675 = 94262.5.

Page 4 of 4

Instructor: Ram Sewak Dubey ECON 317 Linear Algebra

Linear algebra is the branch of mathematics dealing with (among many other things) matri-
ces and vectors. It’s intuitively easy to see why linear algebra is important for econometrics
and statistics. Economic data is arranged in matrix format (rows corresponding to observa-
tions, columns corresponding to variables), so the body of theory governing matrices should
help us analyze data. It is harder to see the connection between matrix theory and the
optimization that we do in micro theory, but there are some important links. We’ll cover
the basics and some of the necessary detail here, but more detailed coverage will be offered
in the core courses

.

1 Vectors
You may be familiar with vectors from physics courses, in which a vector is a pair giving
the magnitude and direction of a moving body. The vectors we use in economics are more
general, in that they can have any finite number of elements (rather than just 2), and the
meaning of each element can vary with the context (rather than always signifying magnitude
and direction). Formally speaking a vector can be defined as a member of a vector space

,

but we don’t need to deal with such a definition here. For our purposes:

Definition 1. A vector is an ordered array of elements with either one row or one column.

The elements are usually numbers. A vector is an n × k matrix for which either n = 1, k =

1

or both (see the definition of a matrix below). A general vector, for which the number of
elements is not specified but left as n, will sometimes be called an “n-vector”. We also refer
to these as “vectors in Rn”. A vector can be written in either row or column form:

Row Vector: x ∈ Rn

=

(
x1 x2 …

xn

)
; Column Vector: x ∈ Rn =




x1
x2
…

xn


.

Although you will sometimes be able to switch between thinking of a vector as a row or a
column without restriction, there are certain operations that require a vector to be oriented
in a certain way, so it is good to distinguish between row and column vectors whenever
possible. Most people use x to refer to the vector in column form and x′ to refer to it in
row form, but this is not universal. Also, we usually use lowercase letters for vectors

and

uppercase letters for matrices.

1.1 Special Vectors

Null vector 0n×1 =


0…

0




Sum vector un×1 =


1…

1



Page 1 of 10

Instructor: Ram Sewak Dubey ECON 317 Linear Algebra

Unit vector : In Rn there are n unit vectors.
The ith unit vector, called ei, has all elements 0 except for the i th, which is equal to 1. The
definition of a unit vector is specific to the vector space in which it sits. For example:

e2 ∈ R3 =


01

0


 (1)

and

e2 ∈ R4 =




0
1
0
0


 (2)

1.2 Vector Relations and Operations
Definition 2.

(a) Equality :

Vectors x ∈ Rn, y ∈ Rm are equal if n = m and xi = yi ∀ i.

(b) Inequalities : ∀x,y ∈ Rn:

x ≥ y if xi ≥ yi ∀ i = 1,··· ,n;
x > y if xi ≥ yi ∀ i = 1,··· ,n and xi > yi for at least one i;
x ≫ y if xi > yi ∀i = 1,··· ,n.

(c) Addition :
∀x,y ∈ Rn, x + y = z ∈ Rn where zi = xi + yi, ∀i.

(d) Scalar Multiplication :∀x ∈ Rn, and α ∈ R, we define the scalar product as

αx =




αx1
αx2

…
αxn


 (3)

(e) Vector Multiplication : This is essentially an inner product rule applied to Rn. See the
rules for matrix multiplication below, as they also apply for vectors.

Page 2 of 10

Instructor: Ram Sewak Dubey ECON 317 Linear Algebra

2 Matrices
Definition 3. A matrix is a rectangular array of elements (usually numbers, for our purposes).

A matrix is characterized as n × k when it has n rows and k columns. To represent the n × k
matrix A, we can write:

[A]
n×k

=
[
ai j

]
n×k =




a11 a12 … a1k
a21 a22 … a2k
…

…
…

…
an1 an2 … ank




The matrix An×k is a null matrix if ai j = 0 for i = 1,··· ,n, j = 1,··· ,k.
The matrix An×k is square if n = k. In this case we refer to it as an n × n matrix.
The square matrix An×n is symmetric if ai j = a ji ∀ i, j.
The symmetric matrix An×n is diagonal if ai j = 0 whenever i ̸= j.
The diagonal matrix An×n is an identity matrix if ai j = 1 whenever i = j.
The square matrix An×n is lower triangular if ai j = 0 ∀ i < j. The square matrix An×n is upper triangular if ai j = 0 ∀ i > j.
It’s worthwhile to check your understanding of each of the above definitions by writing out
a matrix that satisfies each. Then note this next definition carefully:
The k × n matrix B is called the transpose of An×k if bi j = a ji ∀ i, j.
We write the transpose of A as either AT or A′. If A is symmetric then A′ = A. This is an
obvious statement, but you could try proving it formally. It should only take a few lines.

2.1 Matrix Operations
2.1.1 Addition

Matrix addition is only defined for matrices of the same size. If A is n × k and B is n × k then

[A]
n×k

+ [B]
n×k

= [C]
n×k

(4)

where
ci j = ai j + bi j ∀ i = 1,··· ,n, j = 1,··· ,k. (5)

We say that matrix addition occurs “element wise” because we move through each element
of the matrix A, adding the corresponding element from B.

2.1.2 Scalar Multiplication

Scalar multiplication is also an element wise operation. That is,

∀ λ ∈ R, λ

· [A]
n×k

=



λa11 λa12 ··· λa1k
λa21 λa22 ··· λa2k

…
…
…
…

λan1 λan2 ··· λank


 (6)

Page 3 of 10

Instructor: Ram Sewak Dubey ECON 317 Linear Algebra

2.1.3 Matrix Multiplication

Matrix multiplication is defined for matrices [A]
m× j

and [B]
n×k

if j = n or m = k. That is, the

number of columns in one of the matrices must be equal to the number of rows in the other.
If matrices A and B satisfy this condition, so that A is m × j and B is j × k, their product
[C]
m×k

≡ [A]
m× j

· [B]
j×k

is given by ci j = Ai · B j, where Ai is the ith row of A and B j is the jth column

of B. For example, suppose

[A]
2×2

=

[

1 2
3 4

]
and [B]

2×

3

=

[
6 5 4
3 2 1

]
Multiplication between A and B is only defined if A is on the left and B is on the right. It
must always be the case that the number of columns in the left hand matrix is the same as
the number of rows in the right hand matrix. In this case, if we say AB = C, then element

c11 =
[
1 2

]
·
[

6
3

]
= 1 · 6 + 2 · 3 = 12

Likewise

c12 = 1 · 5 + 2 · 2 = 9
c13 = 1 · 4 + 2 · 1 = 6
c21 = 3 · 6 + 4 · 3 = 30
c22 = 3 · 5 + 4 · 2 = 23
c23 = 3 · 4 + 4 · 1 = 16

which gives
[A]
2×2

· [B]
2×3

= [C]
2×3

=

[
12 9 6
30 23 16

]
Note that matrix multiplication is not a symmetric operation. In general, AB ̸= BA, and
in fact it is often the case that the operation will only be defined in one direction. In our
example BA is not defined because the number of columns of B = (3) is not equal to the
number of rows of A = (2). For both AB and BA to be defined

[A]
n×k

· [B]
k×n

= [C]
n×n

,

and
[B]
k×n

· [A]
n×k

= [D]
k×k

.

2.2 Some Fun facts about matrix multiplication
(i) Even if n = k,

AB ̸= BA.

A =

[

1 2
3 4

]
, B =

[
0 −1
6 7

]
, AB =

[
12 13
24 25

]
, BA =

[
−3 −4
27 40

]
.

Page 4 of 10

Instructor: Ram Sewak Dubey ECON 317 Linear Algebra

(ii) AB may be null matrix even when A ̸= 0 and B ̸= 0.

A =
[

2 4
1 2

]
, B =

[
−2 4
1 −2

]
, AB =

[
0 0
0 0

]
.

(iii) CD = CE ; D = E even when C ̸= 0.

C =
[

2 3
6 9

]
, D =

[
1 1
1 2

]
, E =

[
−2 1
3 2

]
, CD = CE =

[
5 8

15 24

]
.

2.3 Rules for matrix operations
A + B = B + A

A +(B +C) = (A + B)+C

(AB)C = A(BC)

(A + B)C = AC + BC

A(B +C) = AB + AC

Check that you have a clear understanding of the restrictions needed on the number of rows
and columns of A,B and C in order for the above to work. More matrix rules, involving the
transpose:

(A′)′ = A (7)
(A + B)′ = A′ + B′ (8)

(AB)′ = B′A′ (9)
Note the reversal of the order of the matrices in the last operation.

2.4 Inverse of a square matrix:
Definition 4. A square matrix [A]

n×n
is invertible if there exist [B]

n×n
such that [A]

n×n
· [B]

n×n
= [B]

n×n
·

[A]
n×n

= [I]
n×n

. Then B is called inverse of A.

Rules for the inverse: (
A−1

)−1
= A (10)

(AB)−1 = B−1A−1 (11)(
A′
)−1

=
(
A−1

)′ (12)

Page 5 of 10

Instructor: Ram Sewak Dubey ECON 317 Linear Algebra

3 Determinant of a matrix
Determinant is defined only for square matrices. The determinant is a function depending
on n that associates a scalar, det(A), to an n × n square matrix A. The determinant of an
1-by-1 matrix A is the only entry of that matrix: det(A) = A11. The determinant of a 2 by
2 matrix

A =
[

a b
c d

]
is det(A) = ad − bc.

Definition 5. The cofactor Ai j of the element ai j is defined as (−1)i+ j times the determinant
of the sub matrix obtained from A after deleting row i and column j.

Example 1. Let

A =
[
1 2
3 4

]
A11 = (−1)1+1 · 4 = 4, A12 = (−1)1+2 · 3 = −3
A21 = (−1)2+1 · 2 = −2, A22 = (−1)2+2 · 1 = 1.

Definition 6. The determinant of an n × n matrix A is given by

det(A) =
n

∑
j=1

a1 jA1 j =
n

∑
i=1

ai1Ai1. (13)

Example 2. Let

A =


 a b cd e f

g h i


.

Then

det(A) = a(−1)1+1 det
[

e f
h i

]
+ b(−1)1+2 det

[
d f
g i

]
+ c(−1)1+3 det

[
d e
g h

]
= a(ei − f h)− b(di − f g)+ c(dh − eg).

3.1 Properties of Determinants
(a)

det(A) = det
(
A′
)

(14)

(b) Interchanging any two rows will alter the sign but not the numerical value of the deter-
minant.

(c) Multiplication of any one row by a scalar k will change the determinant k− fold.

(d) If one row is a multiple of another row, the determinant is zero.

Page 6 of 10

Instructor: Ram Sewak Dubey ECON 317 Linear Algebra

(e) The addition of a multiple of one row to another row will leave the determinant un-
changed.

(f) If A and B are n × n matrices, then

det(AB) = det(A)· det(B).

(g) Properties (b)−(e) are valid if we replace row by columns everywhere.

A =
[
1 2
3 4

]
, det(A) = −2; A′ =

[
1 3
2 4

]
det

(
A′
)
= −2

B =
[

3 4
1 2

]
, det(B)

= 2.

4 System of Linear Equations
The system of linear equation is

Ax = b (15)
where matrix A is of dimension n × k, x is a column vector k × 1 and b is column vector n × 1.
This is a system of n equations with k unknowns.

Example 3. The system of two linear equations,

5x + 3y = 1
6x + y = 2

can be written as [

5 3
6 1

][
x
y

]
=

[
1
2

]
When b = 0, the system is called a homogeneous system. When b ̸= 0, it is called a non-
homogeneous system.

Definition 7. Column vector x∗ is called a solution to the system if Ax∗ = b.

There are three important questions in this context.

(a) Does a solution exist?

(b) If there exists a solution, is it unique?

(c) If a solution exists, how do we compute it?

Claim 1. A homogeneous system Ax = 0 always has a solution (Trivial x = 0). But there
might be other solutions (solution may not be unique).

Claim 2. For a non-homogeneous system Ax = b, a solution may not exist.

Page 7 of 10

Instructor: Ram Sewak Dubey ECON 317 Linear Algebra

Example 4. Following system of two linear equations

2x + 4y = 5
x + 2y = 2

does not have a solution. Multiply second equation by 2. Then LHS of both equations
become same which leads to 5 = 4 which is a contradiction.
Example 5. Following system of two linear equations

2x + 4y = 2
x + 2y = 1

has many solution.
Given [A]

n×k
and {b}

k×1
, the n×(k + 1) matrix [Ab]

n×(k+1)
=
[
A1 A2 ··· Ak b

]
is called the augmented

matrix. Note Ai is the ith column of A.

Example 6. Let A =
[

5 3
6 1

]
, b =

[
1
2

]
⇒ Ab =

[
5 3 1
6 1 2

]
.

Theorem 1. The system of equations

[A]
n×k

· {x}
k×1

= {b}
n×1

has a unique solution if and only if the matrix A has an inverse (is invertible).
Consider the case of n equations in n unknowns. In this case, A is n × n. If Ax = b has a
solution and if det(A) ̸= 0 then the solution is characterized by

{x∗}
n×1

= [A−1]
n×n

· {b}
n×1

(16)

Example 7. The system of linear equations

2x + y = 0
2x + 2y = 0

gives us
A =

[
2 1
2 2

]
, b =

[
0
0

]
.

It is easy to verify that the matrix A is invertible. Hence solution exists and is unique.
Example 8. The system of linear equations

2x + y = 0
4x + 2y = 0

leads to
A =

[
2 1
4 2

]
, b =
[
0
0
]
.

It is again easy to verify that matrix A is not invertible. Hence even though solution exists,
there is no unique solution.

Page 8 of 10

Instructor: Ram Sewak Dubey ECON 317 Linear Algebra

5 Cramer’s Rule
Recall that we wanted to calculate the (unique) solution of a system of n equations in n
unknowns given by

Ax = c (17)
where A is an n × n matrix, and c is a vector in Rn.
To obtain a unique solution, we saw that we must have A non-singular, which now translates
to the condition “|A| ̸= 0”. The unique solution to (17) is then

x = A−1c =
adj A

|A|

c (18)

Let us evaluate x1, using (18). This can be done by finding the inner product of x with the
first unit vector, e1 = (1,0,··· ,0). Thus,

x1 = e1x =
e1 · adj A

|A|
c

=
[A11A21 An1]c

|A|

= [c1A11 + c2A21 +···+ cnAn1]/|A|

=

∣∣∣∣∣∣∣
c1 a12 ··· a1n
…

cn an2 ··· ann

∣∣∣∣∣∣∣ |A|
−1

This gives us an easy way to compute the solution of x1. In general, in order to calculate xi,
replace the ith column of A by the vector c and find the determinant of this matrix. Dividing
this number by the determinant of A yields the solution xi. This rule is known as Cramer’s
Rule.

Example 9. General Market Equilibrium with three goods

Consider a market for three goods. Demand and supply for each good are given by:

D1 =5 − 2P1 + P2 +

P3

S1 =− 4 + 3P1 + 2P2
D2 =6 + 2P1 − 3P2 + P3
S2 =3 + 2P2
D3 =20 + P1 + 2P2 − 4P3
S3 =3 + P2 + 3P3

Page 9 of 10

Instructor: Ram Sewak Dubey ECON 317 Linear Algebra

where Pi is the price of good i; i = 1; 2; 3. The equilibrium conditions are: Di = Si; i = 1; 2; 3,
that is

5P1 + P2 − P3 = 9
−2P1 + 5P2 − P3 = 3
−P1 − P2 + 7P3 =

17

This system of linear equations can be solved at least in two ways.

(a) Using Cramer’s rule:

A1 = det


 9 1 −13 5 −1

17 −1 7


 = 356.

A = det


 5 1 −1−2 5 −1

−1 −1 7


 = 178.

P∗1 =
A1
A

=
356

178

= 2.

Similarly P∗2 = 2 and P
∗
3 = 3. The vector of (P

∗
1 , P

∗
1 , P

∗
3 ) describes the general market

equilibrium.

(b) Using the inverse matrix rule. Let us denote

A =

 5 1 −1−2 5 −1
−1 −1 7


, P =


 P1P2

P3


, B =


 93

17




The matrix form of the system is AP = B, which implies P = A−1B.

A−1 =
1

det A


 34 −6 415 34 7

7 4 27




P =
1

178

 34 −6 415 34 7
7 4 27


 ·


 93
17


 =


 22

3



Again, P∗1 = 2, P
∗
1 = 2, and P

∗
3 = 3.

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