Additional Practice Problems Involving the Kuhn-Tucker Conditions

Mathematical Economics Problems and Solutions - Section 6

See Mr. Stolyarov's complete list of Mathematical Economics Problems and Solutions.

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Problem KTC1. Use the Kuhn-Tucker conditions to find maximum for the following:

F(x, y) = 2x²y², subject to constraints (v) 18x + 3y ≤ 1266 and (vi) 2x + 5y ≤ 1000.

Assume that x and y must each be greater than or equal to 0.

Solution KTC1.

Lagrangian: L(x, y, λ, μ) = 2x²y² + λ[1266 - 18x - 3y] + μ[1000 - 2x - 5y]

FOC: L_x = 4xy² - 18λ - 2μ ≡ 0. (x*L_x = 0, but if x is zero, then F(x, y) = 0, so L_x must be 0).

L_y = 4x²y - 3λ - 5μ ≡ 0. (y*L_y = 0, but if y is zero, then F(x, y) = 0, so L_y must be 0).

L_λ = 1266 - 18x - 3y ≥ 0, λ ≥ 0 with complementary slackness.

L_μ = 1000 - 2x - 5y ≥ 0, μ ≥ 0 with complementary slackness.

Case I: We first find the intersection of 18x + 3y = 1266 and

2x + 5y = 1000, which occurs when 3y = 1266 - 18x and y = 422 - 6x. Thus,

2x + 5(422 - 6x) = 1000 and x = 555/14 = x = 39.64285714, so y = 422 - 6x =

y = 184.1428571. Here, F(39.64285714, 184.1428571) = 106,578,510.2

Case II: If λ = 0 and μ > 0, then

4x²y - 5μ = 0 and 4xy² - 2μ = 0, so 4x²y = 5μ and 4xy² = 2μ, so (y/x) = 2/5 and so 5y = 2x Since μ > 0, L_μ = 0, so 1000 - 2x - 5y = 0 = 1000 - 4x, so 4x = 1000 and x = 250, while y = (2/5)x = y = 100. But (x, y) = (250, 100) is outside of our feasible space, since it violates the constraint 18x + 3y ≤ 1266, as 18*250 + 3*100 = 4800 > 1266.

Case III: If μ = 0 and λ > 0, then 4xy² - 18λ = 0 and 4xy² = 18λ.

Likewise, 4x²y - 3λ = 0, so 4x²y = 3λ. Thus, (y/x) = 6 and y = 6x, implying that 3y = 18x.

Since λ > 0, L_λ = 0, so 1266 - 18x - 3y = 0 and 1266 - 36x = 0, whereby x = 35.166666667

And y = 6x = 211. This is within the constraint 18x + 3y ≤ 1266, but violates the constraint

2x + 5y ≤ 1000, as 2*35.166666667 + 5*211 = 1125.3333333 > 1000. Thus, this solution is outside our feasible space.