# The production function

Second Midterm Exam

Questions

The production function for the Cosmos firm can be written as: Q = 500L + 90L^{2} – .

- Graph the total product function.

Q Q = 500L + 90L^{2}

L

Questions

The production function for the Fringle firm can be written as: Q = 5K^{1/2}L^{1/2}

^{ }

^{ }

- Graph the isoquant for Q = 350.

K

Q = 350

Isoquant

L

- 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=5K^{1/2}L^{1/2}

ℓ=1200L+ 500K+l[70- K^{1/2}L^{1/2]}]

Foc w.r.t = K = 500-0.5lK^{-1/2}L^{1/2}=0

** **F.o.c w.r.t = L= 1200- lK^{1/2}L^{-1/2}=0

F.o.c w.r.t = l = 70- K^{1/2}L^{1/2}=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

- 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- K^{1/2}L^{1/2}=0

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

L=19.4 =20 units

K= 252

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

- 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=5K^{1/2}L^{1/2}

Subject to

C=500K+1000L

L=5K^{1/2}L^{1/2}+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=5K^{1/2}L^{1/2}

Q=353 units

- 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=5K^{1/2}L^{1/2} s.t

C=500K+600L

L=5K^{1/2}L^{1/2}+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=5K^{1/2}L^{1/2}

Q=458.25 units

- 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?

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

Questions

Sara has the following production function: Q = 10K^{1/2}L^{1/2}. 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.

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

Min C, st. Q

ℓ = 2000K + 3000L + ?(Q – 10PK^{1/2}L^{1/2})

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

Foc w.r.t L = 3000 – 5?PK^{1/2}L^{-1/2} =0————————————-(2)

Foc w.r.t ? = Q – 10PK^{1/2}L^{1/2}

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

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

- Calculate the maximum profits of the firm.

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

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

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