a. optimal capital/labor ratio
MPL=5K
MPK=5L
equating MPL=MPK
K=L
b. We use the Lagrangian to maximize Q subject to cost constraint.
Max Q=5LK
subject to
64,000=20L+80K
Form the Lagrangian function
G=5LK+λ(64,000−20L−80K)
G=5LK+λ64,000−20λL−80λK
First order conditions are:
- 5K−20λ=0
- 5L−80λ=0
- 64,000−20L−80K=0
Solve (1) and (2) to eliminate λ.
5K−20λ=0
5L−80λ=0
20K−80λ=0
5L−80λ=0
20K−5L=0
Solve
64,000−20L−80K=0
−5L+20K=0
64,000−20L−80K=0
−20L+80K=0
64,000−40L=0
64,000=40L
L=1600
−5L+20K=0
−5(1600)+20K=0
−8000+20K=0
20K=8000
K=400
L=1600,K=400
Q=0.5(1600)(400)
Q=3,200,000
c.
For duality to be shown, we are supposed to show that the cost minimization approach leads to the same answer as the maximizing quantity.
MinimizeC=20L+80K
subjectto
3,200,000=5LK
Form Lagrangian function:
G=20L+80K+λ(3,200,000−5LK)
G=20L+80K+λ3,200,000−5λLK
First order conditions are:
- 20−5λK=0
- 80−5λL=0
- 3,200,000−5LK=0
Solve 1 and 2
20−5λK=0
80−5λK=0
K20−5λ=0
L80−5λ=0
K20−L80=0
Combine with equation (3) to solve for L and K
K20−L80=0
3,200,000−5LK=0
Multiply top equation by L2K.
20L 2−80LK=0
3,200,000−5LK=0
20L2−80LK=0
51,200,000−80LK=0
20L2−51,200,000=0
L2=2,560,000
L=1600
3,200,000−5(1600)K=0
−8000K=−3,200,000
K=400
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