# The production function

Second Midterm Exam

Questions

The production function for the Cosmos firm can be written as: Q = 500L + 90L2 – .

1. Graph the total product function.

Q                                                                         Q = 500L + 90L2

L

Questions

The production function for the Fringle firm can be written as: Q = 5K1/2L1/2

1. Graph the isoquant for Q = 350.

K

Q = 350

Isoquant

L

1. Assume the price per unit of K = \$500 and the price per unit of L = \$1200. Calculate the least cost way of making Q = 350.  How much K and L will you use?  What are your costs?

Min. C=1200L +  500K   s.t   Q=5K1/2L1/2

ℓ=1200L+ 500K+l[70- K1/2L1/2]]

Foc w.r.t  = K =  500-0.5lK-1/2L1/2=0

F.o.c w.r.t = L=  1200- lK1/2L-1/2=0

F.o.c w.r.t = l = 70- K1/2L1/2=0

2400^2/500^2=K/L=23.04

K=23.04L

K=336 units

L=15 units

Cost

C=1200L+ 500K

C=1200*15+500*336=\$186,000

1. Now assume that the price per unit of L drops to \$900 and the price per unit of K stays the same. Calculate the least cost way of making Q = 350.  How much K and L will you use?  What is the cost?

1800^2/500^2=K/L=12.96

K=12.96L

Given that 70- K1/2L1/2=0

Then, 70=[12.96L]^.5*L^.5

L=19.4 =20 units

K= 252

C=900*20+500*252=\$144,000

1. Now assume that you have been given \$100,000 to make as much output as possible. The price per unit of K is \$500 and the price per unit of L is \$1000.  What is the maximum amount of Q you can make?  How much K and L will you use?

In this case we maximize Q subject to costs

Max Q=5K1/2L1/2

Subject to

C=500K+1000L

L=5K1/2L1/2+l(100000-500K-1000L)

Foc r w.r.t = K= 2.5K^-.5L^.5-500 =0

Foc r w.r.t = L = 2.5 K^.5L^-.5-1000=0

Foc r w.r.t = l = 100000-500K-1000L=0

Therefore K=2L

100000=500*2L+1000L

L=50

K=2L=100

Q=5K1/2L1/2

Q=353 units

1. Now assume that the price per unit of L falls to \$600 and the price per unit of K stays the same. How much Q can you now make for \$100,000?  How much K and L will you use?

Max Q=5K1/2L1/2 s.t

C=500K+600L

L=5K1/2L1/2+l(100000-500K-600L)

First order conditions

K

2.5K^-.5L^.5-500 =0

L

2.5 K^.5L^-.5-600=0

l

100000-500K-600L=0

Therefore K=1.2L

100000=500*1.2L+600L

L=83.33 = 84 units

K=1.2L=100

Q=5K1/2L1/2

Q=458.25 units

1. Now assume that you have been given K = 6400 free of charge and each unit of L costs \$2000 and each unit of Q can be sold for \$100. How much K and L will you use to maximize profits?  What are your maximum profits?

1. What is the marginal product of L at the profit maximizing amount of L?

Questions

Sara has the following production function: Q = 10K1/2L1/2.  She faces the following demand function: Q = 11,000 – 5P.  The price per unit of K is \$2,000 and the price per unit of L is \$3,000.

1. Calculate the amount of K and L that Felicia will use to maximize profits.

Min C, st. Q

ℓ = 2000K + 3000L + ?(Q – 10PK1/2L1/2)

Foc w.r.t K = 2000 – 5?PK-1/2L1/2 =0———————————- (1)

Foc w.r.t L = 3000 – 5?PK1/2L-1/2 =0————————————-(2)

Foc w.r.t ? = Q – 10PK1/2L1/2

Therefore, Devide eq (i) with (ii), you get:

2/3 = L/K, hence L = 2/3K, replace L in eq (i)

2000 = 5P(2/3)

P = 600

Hence

1. Calculate the profit maximizing amount of output and the profit maximizing price.

1. Calculate the maximum profits of the firm.

1. Calculate the marginal revenue at the profit maximizing amount of Q.

1. Calculate the marginal product of labor at the K and L in 15.

1. At what price and quantity would total revenue be a maximum?  What is the maximum revenue?