Econ 100A Problem Set #3: Constrained Optimization Spring 2022
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Econ 100A
Problem Set #3: Constrained Optimization
Spring 2022
1. Find the commodity bundle (x1*, x2*) that maximizes the utility function u(x1 , x2 )
x1x2
subject to the budget constraint x1
4x2
16 . Even though it’s fairly meaningless, find the maximum value of u and the value of the Lagrange multiplier, λ .
L
x1x2
![]()
x1
4x2
16
|
FOC:
|
|
L
x1
4x2
16
0
x2
x1
4
x1
4x2
16
4![]()
4 ![]()
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16
*
2; x1*
4
2
8; x2 *
2; u
8, 4
8
4
32
2. Find the commodity bundle (x*, m*) that maximizes the utility function
1
u(x, m)
12x2
m subject to the budget constraint 2x
m
24 . Even though it’s fairly meaningless, find the maximum value of u and the value of the Lagrange multiplier, λ .
L
12x2
m
![]()
2x
m
24
FOC:
1
L
6x 2
2![]()
0
x ![]()
Lm
1
![]()
0
*
1
L
2x
m
24
0
x*
9; m*
24
2
9
6;
2x
m
24
u
9, 6
12
9![]()
6
42
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3. Find the commodity bundle (x1*, x2*) that maximizes the utility function
1 1
u(x1 , x2 )
x
x
subject to the budget constraint 3x1
5x2
60 . Even though it’s fairly meaningless, find the maximum value of u and the value ofthe Lagrange multiplier, λ .
L
x
x
![]()
3x1
5x2
60
FOC:
1
1 1
1
L
3x1
5x2
60
0
1
1 36
2
1
2 100
2
3x1
5x2
60
![]()
1 1
36 100
25
![]()
1
1 1
u
![]()
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2
![]()
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2
4. Find the general expression (in terms of all the parameters) for the commodity bundle which maximizes the Cobb-Douglas utility function u(x1 , x2 )
x
x![]()
![]()
subject to the budget constraint p1x1
p2x2
Y . (Hint: you may find it easier to work with a transformation of u(x1, x2).)
I recommend working this with the transformation of v(u) = ln(u). Just to demonstrate different algebra, I’ll work it without the transformation. The optimal commodity bundle is the same.
L
![]()
p1x1
p2x2
Y
FOC:
L1
x![]()
1x![]()
p1
0
![]()
1 2
p1
L2
1
![]()
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p2
0 L
p1x1
p2x2
Y
0
![]()
1
![]()
x
x
2
p2
p1x1
p2x2
Y
![]()
1
![]()
x
x
2
p1 p2
x2
1
![]()
x1
p1 p2
1
![]()
p1x1
2
p2
p1x1
p2 ![]()
1
![]()
p1x1 ![]()
Y
x1*
Y , x2
1
![]()
p1 ![]()
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1
![]()
Y
p1
p2 p2
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5. By solving the Lagrangean conditions, identify six stationary points ofthe function
f (x1 , x2 )
x
x2 subject to the constraint 2x
x
3 . Which of these gives the highest
value off(x1, x2), and which gives the lowest value? (x1 and x2 can take on negative values. This problem is a lot longer than the previous ones.)
x1 ,x2 1 2 1 2
L
x x
2
![]()
2x
x
3
L
2x1x2
4
x1
0 (eq 1)
L
x
2
x2
0 (eq 2)
2x![]()
x
3 (eq 3)
We'll solve this using cases.
First simplify (eq 1):
1 2
1 2
1
Substitute x1 into (eq 3):
2(0)2
x
3
x2
3
x
3
Ifx2
3: Substitute x1 and x2 into (eq 2):
(0)2
2
3
0
![]()
0
Our first solution: x1
0,x2
3,![]()
0
f
0, 3
(0)2
3
0
Ifx2
3: Substitute x1 and x2 into (eq 2):
(0)2
2![]()
3
0
![]()
0
Our second solution: x1
0, x2
3,![]()
0
f
0,
3
(0)2
3
0
![]()
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Ifx2
2
:
Substitute x2 into (eq 2):
x
2
(2
)
0
x
4
2
0
x
4
2
x1
2
1 2
2
4
2
2![]()
2
3
12
2
3
2 1
2022-05-10