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: 1 2 2 1 |
|
L x1 4x2 16 0
x2
x1 4
x1 4x2 16
4 4 16
* 2; x1* 4 2 8; x2 * 2; u 8, 4 84 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
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
1 362
1
2 1002
3x1 5x2 60
1 1
36 100
25
1
1 1
u 2 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 ) xx 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 1xp1 0 1 2
p1
L2 1 p2 0 L p1x1 p2x2 Y 0
1 xx2 p2
p1x1 p2x2 Y
1 xx2
p1 p2
x2 1 x1
p1 p2
1 p1x1
2
p2
p1x1 p2 1 p1x1 Y
x1* Y , x2 1 p1 1 Y
p1 p2 p2
5. By solving the Lagrangean conditions, identify six stationary points ofthe function
f (x1 , x2 ) xx2 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 4x1 0 (eq 1)
L x 2x2 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
Ifx2 2:
Substitute x2 into (eq 2):
x 2(2) 0
x 42 0
x 42
x1 2
1 2
2 42 22 3
122 3
2 1
2022-05-10