the production function is f(L, M)  4L^1/2 M^1/2, where L is the number of units of labor and M is the number of machines used.If the cost of labor is $9 per unit and the cost of machines is $81 per unit, then the total cost of producing 10 units of output will be

A) $270.
B) $90.
C) $135.
D) $450.
E) None of the above.

The total cost of production is the sum of the cost of labor and the cost of machines used:

Cost of labor: $9 per unit of labor * L units of labor
Cost of machines: $81 per unit of machines * M units of machines

We need to find the values of L and M that minimize the total cost of production, subject to the constraint that 10 units of output are produced. We can solve for L and M using the Lagrange multiplier method:

minimize C = 9L + 81M
subject to f(L, M) = 10

The Lagrangian function is:

L = 9L + 81M + λ(10 - f(L, M))

Taking partial derivatives of L with respect to L and M, and setting them equal to 0, we get:

∂L/∂L = 9 - λF1^1/2 = 0
∂L/∂M = 81 - λF1^1/2 = 0

We also need to take the partial derivative of L with respect to λ:

∂L/∂λ = 10 - f(L, M) = 0

Solving these equations simultaneously, we get:

F1^1/2 = 1/9
81/9 = F1^1/2
F1^1/2 = 9

Substituting F1^1/2 = 9 into the production function, we get:

f(L, M) = 10 = 4L^1/2 * 9M^1/2

Simplifying, we get:

L^(1/2)M^(1/2) = 10/36

We want to find the values of L and M that satisfy this equation and minimize the cost function. Substituting L^(1/2)M^(1/2) = 10/36 into the cost function, we get:

C = 9L + 81M = 9(36L^(1/2)M^(1/2)) + 81M = 324(LM)^(1/2) + 81M

Substituting LM = (10/36)^2, we get:

C = 324(10/36) + 81(10/36) = $90

Therefore, the total cost of producing 10 units of output is $90, and the best choice is (C).