Question

How do I solve this problem?

3. Skips provides repair services (R) using two inputs: labor (L), and tools (T). Skips' production function is R-L'4T1/4. Skips sells repair services a competitive market at a price of $100. Labor costs $1 per unit, and tools cost $5 per unit. a. Write down Skips' profit function. (2 points) a. b. suppose in the short run, tools are fixed at T-16. Find the first order condition for the short- run profit maximizing amount of labor. (2 points) Solve for the amount of labor that Skips uses in the short run to maximize profit. (3 points) In the long run, both inputs are variable. Find the first order conditions for the long-run c. d. profit maximizing amounts of both inputs. (3 points) Solve for the amount of labor and tools that Skips uses to maximize profit in the long run. (4 e. points)

Answer 1

Total Revenue=TR=P*R=100L^1/4T^1/4

Total Cost=TC=wL+rT=1*L+5T=L+5T

Profit=TR-TC=100L^1/4T^1/4-L-5T

a)

We have derived that

Profit=100L^1/4T^1/4-L-5T

b)

R=L^1/4T^1/4

Put T=16

R=L^1/416^1/4=2L^1/4

Marginal Product of labor=MPL=dR/dL=2(1/4)L^-3/4=0.50L^-3/4

Marginal Revenue Product of labor=MRPL=P*MPL=100*0.50L^-3/4

Set MRPL=w for profit maximization

100*0.50L^-3/4=1

50L^-3/4=1

c)

We have derived that

50L^-3/4=1

L^3/4=50

L=184.20

d)

Profit=TR-TC=100L^1/4T^1/4-L-5T

d(Profit)/dL=100*(1/4)L^-3/4T^1/4-1

d(Profit)/dT=100*(1/4)L^1/4T^-3/4-5

Set d(Profit)/dL=0 and d(Profit)/dT=0

100*(1/4)L^-3/4T^1/4-1=0

25L^-3/4T^1/4=1--------------------(1)

100*(1/4)L^1/4T^-3/4-5=0

25L^1/4T^-3/4=5 ---------------------(2)

e)

Divide equation 1 by equation 2, we get

(25L^-3/4T^1/4)/(25L^1/4T^-3/4)=1/5

(T/L)=(1/5)

L=5T

From equation 2, we get

25L^1/4T^-3/4=5

i.e.

L^1/4T^-3/4=5/25=0.20

Put L=5T

(5T)1/4T-3/4=0.20

5^1/4T^1/4T^-3/4=0.20

0.20T^1/2=5^1/4

T^1/2=5*51/4

T=25*5^1/2=55.9017

L=5T=5*55.9017=279.5085