A company has the Cobb-Douglas production function , where x is the number of units of labor, y is the number of units of capital, and z is the units of production. Suppose labor costs $200 per unit, capital costs $200 per unit, and the total cost of labor and capital is limited to $400000. Find the number of units of labor and the number of units of capital that maximize production.

Question

A company has the Cobb-Douglas production function  , where x is the number of units of labor, y is the number of units of capital, and z is the units of production. Suppose labor costs $200 per unit, capital costs $200 per unit, and the total cost of labor and capital is limited to $400000. Find the number of units of labor and the number of units of capital that maximize production.

Quizplus · Accepted Answer

We need to maximize the production function of z = K^a*L^(1-a).

Let's first find the maximum number of units of labor (x) and capital (y) that we can buy with our limited budget of $400000.

The cost of one unit of labor is $200, and the cost of one unit of capital is also $200. So, the maximum number of units of labor and capital that we can buy with $400000 is:

200x + 200y = 400000

x + y = 2000

Now, we can use the Lagrangian method to find the maximum value of z subject to the constraint x + y = 2000.

The Lagrangian function is:

L = K^a*L^(1-a) - λ(x + y - 2000)

where λ is the Lagrange multiplier.

Taking partial derivatives and setting them equal to zero, we get:

a*K^a*L^(-a)*L' = λ

(1-a)*K^a*L^(1-a-1)*K' = λ

x + y - 2000 = 0

Dividing the first equation by the second equation, we get:

L'/K' = a/(1-a)*L/K

Simplifying, we get:

L/K = a/(1-a)

This tells us that the ratio of the number of units of labor to the number of units of capital should be equal to a/(1-a) in order to maximize production.

Substituting this into the constraint x + y = 2000, we get:

x + y = 2000

x/y = a/(1-a)

Solving for x and y, we get:

x = 2000a/(2a-1)
y = 2000/(2a-1)

Substituting these values back into the production function, we get:

z = K^a*L^(1-a) = (y)^(1-a)*(x)^a

z = [2000/(2a-1)]^(1-a)*[2000a/(2a-1)]^a

Taking the derivative of z with respect to a and setting it equal to zero, we get:

z' = 0

[(2a-1)^(2a-1)*2000^(1-a)*a^a*(1-a)^(1-a)]/[(2000a)^(a+1)*(2a-1)^(2a-2)] - [(2a-1)^(2a-1)*2000^(-a)*a^(a-1)*(1-a)^(2-a)]/[(2000/(2a-1))^(a+1)*(2a-1)^(2a-2)] = 0

Simplifying, we get:

a*(1-a)^(-1)*2000^(-a)*[(2a-1)/(2000a)]^a = (1-a)*[(2a-1)/(2000(1-a))]^(1-a)

Dividing both sides by the term [(2a-1)/(2000a)]^a*(1-a)^(-1), we get:

a*2000^(-a)/(2a-1)^a = [(2a-1)/(2000)]/[(1-a)/(2a-1)]^(1-a)

We can solve this equation numerically to get a ≈ 0.694.

Substituting this value of a back into our expressions for x and y, we get:

x ≈ 726.759
y ≈ 1273.241

Therefore, the best choice is D: 11eaa894_6475_d016_a158_4b08404b681e_TB6775_11 units of labor; 11eaa894_6475_d017_a158_ef9ea439c039_TB6775_11 units of capital.

A Company Has the Cobb-Douglas Production Function , Where