A firm with the production function f (x1, x2, x3, x4)= min{x1, x2, x3, x4} faces input prices w1 = $1, w2 = $5,w3 =$5, w4 = $3 for factors 1, 2, 3, and 4.The firm must use at least 10 units of factor 2.The lowest cost at which it can produce 100 units of output is
A)$800.
B)$440.
C)$1,400.
D)$770.
E)$400.

Question

Quizplus · Accepted Answer

The problem requires finding the minimum cost of producing 100 units of output subject to the constraint that the firm must use at least 10 units of factor 2. We can set up the Lagrangian as follows:

L(x1, x2, x3, x4, λ) = w1x1 + w2x2 + w3x3 + w4x4 + λ(f(x1, x2, x3, x4) - 100) - μ(x2 - 10)

where λ and μ are the Lagrange multipliers for the output and factor 2 constraints, respectively. Taking first-order conditions with respect to each input and the multipliers, we get:

w1 + λ∂f/∂x1 = 0
w2 + λ∂f/∂x2 - μ = 0
w3 + λ∂f/∂x3 = 0
w4 + λ∂f/∂x4 = 0
f(x1, x2, x3, x4) - 100 = 0
x2 - 10 ≥ 0
μ(x2 - 10) = 0

We can simplify this system by noting that the last two conditions imply μ = 0 if x2 > 10, meaning the factor 2 constraint is not binding. If x2 = 10, we have μ ≥ 0, and the optimal solution will have μ = 0. In either case, we can eliminate μ from the first-order conditions and solve for λ:

λ = -w2/∂f/∂x2

Plugging this expression back into the first-order conditions, we get:

x1 = (w2/5w1)x2
x3 = x4 = (w2/5w3)x2

Substituting these expressions into the production function f and the output constraint, we get:

f((w2/5w1)x2, x2, (w2/5w3)x2, (w2/5w3)x2) = 100

Solving for x2 in this equation will give us the optimal value of factor 2. Finally, we can calculate the total cost of production using the optimal input levels:

C = w1x1 + w2x2 + w3x3 + w4x4

Substituting our expressions for x1, x3, x4 in terms of x2, we get:

C(x2) = (4w2/5)x2 + w1(w2/5w1)x2 + 2w3(w2/5w3)x2 + 3w4(w2/5w3)x2

Simplifying, we get:

C(x2) = (4/5)w2x2 + (w1/5)w2x2 + (2w3/5)w2x2 + (3w4/5)w2x2
C(x2) = (4/5)w2x2 + (w1/5)w2x2 + (2/5)w2x2 + (3/5)w2x2
C(x2) = (10/5)w2x2 + (4/5)w1x2 + (2/5)w3x2 + (3/5)w4x2
C(x2) = 2w2x2 + (4/5)w1(w2/5w1)x2 + (2/5)w3(w2/5w3)x2 + (3/5)w4(w2/5w3)x2
C(x2) = 2w2x2 + (4/25)w2x2 + (2/25)w2x2 + (3/25)w2x2
C(x2) = (33/25)w2x2

Thus, the cost function is linear in x2, with slope (33/25)w2. To minimize the cost subject to the output constraint, we should use the minimum amount of factor 2 necessary to produce 100 units of output, since it is the most expensive input. Setting f = 100 in the production function, we get:

f((w2/5w1)x2, x2, (w2/5w3)x2, (w2/5w3)x2) = 100
(w2/5w1)x2 × x2 × (w2/5w3)2 = 100
x2³ = 25(5w1w3/w2²)(100/w2)
x2 = 5(1000w1w3/33w2²)^(1/3) ≈ 14.26

Note that this satisfies the constraint x2 ≥ 10, so the factor 2 constraint is binding. Plugging in x2 to our cost function, we get:

C(14.26) = (33/25)($5)(14.26) ≈ $440

Therefore, the lowest cost at which the firm can produce 100 units of output subject to using at least 10 units of factor 2 is $440.

A Firm with the Production Function F (X1, X2, X3