Question

Suppose that the production function of the firm is:

Q = 100L^1/2.K^1/2

 K= 100, P = $1, w =$50. and r = $40. Determine the quantity of labor that the firm should hire in order to maximize the profits. What is the maximum profit of this firm?

Answer 1

Given:

production function

$Q=100L^{\frac{1}{2}}K^{\frac{1}{2}}$

K=100

P=$1

W=$50

r=$40

To find the quantity of labor that maximizes the profit, let us equate the value of the marginal product of labor (MPL) and the wage.

Marginal product of labor(VMPL) $=\frac{\delta Q}{\delta L}$

$MPL=100\times \frac{1}{2}\times L^{\frac{-1}{2}}\times K^{\frac{1}{2}}$

Substitute value of K=100

$MPL=\frac{500}{L^{\frac{1}{2}}}$

To find the value of MPL, multiply the price of output (P) with MPL.

$VMPL=1\times\frac{500}{L^{\frac{1}{2}}}$

At equilibrium

VMPL=W

$\frac{500}{L^{\frac{1}{2}}}=50$

$L^{\frac{1}{2}}=\frac{500}{50}$

$L^{\frac{1}{2}}=10$

$L=10^2=100$

The firm will hire 100 labors to maximize profit.

 

Profit: profit refers to the difference between the total revenue and total cost.

Total revenue:

Substitute the value of L=100 and K=100 in the production function

$Q=100\times(100)^{\frac{1}{2}}\times(100)^{\frac{1}{2}}$

$Q=100\times10\times10$

$Q=10,000$

Total Revenue $=Q\times P=10000\times 1=\$10,000$

Now,

Total cost $=wL+rK=(50\times100)+(40\times100)=\$9,000$

Profit=Total Revenue-Total Cost

$Profit=10,000-9,000=\$1,000$

The maximum profit of the firm at L=100 is $1000.