(f) What is the consumer’s (shortrun) indirect utility function?
19. This question is a minor variation of Question 19 in Shannon (1995,
p.508).
Consider a firm that that is a pricetaker in input markets.
Suppose that this firm has a one output and two input production
technology that can be represented by a production function of the
form
Q
=
f
(
L, K
) =
AK
α
L
β
.
This is known as a generalised CobbDouglas production function. The
input price for labour is
w
and the input price for capital is
r
. Suppose
that this firm wants to minimise the cost of producing a particular
output level,
Q
. Its cost minimisation problem is
min
wL
+
rK
:
AK
α
L
β
=
Q
.
6
(a) Find the Lagrangian for this constrained minimisation problem.
(Use
λ
to denote the Lagrange multiplier.)
(b) Obtain the values of
K
,
L
and
λ
that satisfy the firstorder con
ditions for this problem.
(c) What are the firm’s (minimum) cost function and outputconditional
input demand functions?
(d) What is the partial derivative of the firm’s (minimum) cost func
tion with respect to the chosen output level,
Q
?
(e) Compare your answer to part (d) of this question with the value
for
λ
that you found in part (b) of this question.
What is the
interpretation of the value for
λ
that you found in part (b) of this
question?
1
Partial Derivatives Answers
See separate document (which is an extract from Shannon and McDonald
1995).
2
Unconstrained Optimisation Answers
2.1
Question 11
2.1.1
The objective function
π
(
q
1
, q
2
) =

90 + 20
q
2
1
+ 5
q
2
2

8
q
1

5
q
1
q
2
.
2.1.2
The firstorder partial derivatives
∂f
∂q
1
= 40
q
1

8

5
q
2
.
∂f
∂q
2
= 10
q
2

5
q
1
.
2.1.3
The critical points
The firstorder conditions for an unconstrained optimisation problem are:
∂f
∂q
1
= 40
q
1

8

5
q
2
= 0,
7
and
∂f
∂q
2
= 10
q
2

5
q
1
= 0.
These FOCs can be rearranged to obtain the following system of two
equations in two unknown endogenous variables:
40
q
1

5
q
2
=
8

5
q
1
+ 10
q
2
=
0
.
(In this example, both equations are linear. This will not always be the
case.)
Upon solving this linear equation system, we obtain the following
unique critical point for the objective function:
(
q
*
1
, q
*
2
) =
16
75
,
8
75
.
2.1.4
The secondorder partial derivatives
∂
2
f
∂q
2
1
= 40.
∂
2
f
∂q
2
2
= 10.
∂
2
f
∂q
2
∂q
1
=
∂
2
f
∂q
1
∂q
2
=

5.
2.1.5
The Hessian matrix
The Hessian matrix for the objective function is
H
=
40

5

5
10
.
2.1.6
The secondorder conditions
Note that
det(
H
1
) = 40
>
0
and
det(
H
2
= det(
H
) = (40(10)

(

5)(

5) = 400

25 = 375
>
0
.
Thus we can conclude that
H
is a positive definite matrix for all values of
(
q
1
, q
2
). This means that the critical point will yield a minimum value of the
objective function.
8
2.1.7
The optimal value of the objective function
The minimum value of the objective function is
π
(
q
*
1
, q
*
2
) =
π
16
75
,
8
75
=

90 + 20(
q
*
1
)
2
+ 5(
q
*
2
)
2

8
q
*
1

5
q
*
1
q
*
2
=

90 + 20
16
75
2
+ 5
8
75
2

8
16
75

5
16
75
8
75
=
506
,
250 + 5
,
120 + 320

9
,
600

640
5
,
625
=
511
,
690

10
,
240
5
,
625
=
501
,
450
5
,
625
=
102
,
338
1
,
125
.
2.2
Question 12
2.2.1
The objective function
π
(
L, K
) = 500
L
0
.
3
K
0
.
7

2
L

5
K
.
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 Optimization, utility maximisation problem