Additional Practice Problems Involving the Kuhn-Tucker Conditions

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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.

Hence, our function-maximizing values are (x, y) = (39.64285714, 184.1428571), whereby

F(x, y) = 106,578,510.2

Problem KTC2. Use the Kuhn-Tucker conditions to find maximum for the following:

F(x, y) = xy, subject to constraints (vii) x² + y² ≤ 3481 and (viii) 2x + 4y ≤ 120.

Assume that a and b are constants and that x and y must each be greater than or equal to 0.

Solution KTC2.

Lagrangian: L(x, y, λ, μ) = ax³y+ bxy³ + λ[3481 - x² - y²] + μ[120 - 2x - 4y].

FOC: L_x = y- 2λx - 2μ ≡ 0

(x*L_x = 0, but if x is zero, then F(x, y) = 0, so L_x must be 0).

L_y = x - 2λy- 4μ ≡ 0

(y*L_y = 0, but if y is zero, then F(x, y) = 0, so L_y must be 0).

L_λ = 3481 - x² - y² ≥ 0, λ ≥ 0 with complementary slackness.

L_μ = 120 - 2x - 4y ≥ 0, μ ≥ 0 with complementary slackness.

Case I: We first find the intersection of x² + y² = 3481 and 2x + 4y = 120.

4y = 120 - 2x, so y = 30 - 0.5x and x² + (30 - 0.5x)² = 3481. This gives a positive solution for x as x = 58.9979, implying that y = 30 - 0.5*58.9979 = y = 0.50105. When

(x, y) = (58.9979, 0.50105), F(x, y) = 29.5608978

Case II: If λ = 0 and μ > 0, then y- 2μ ≡ 0 and 2μ = y

Likewise, x - 4μ ≡ 0, so 4μ = x and x = 2y. Since μ > 0, L_μ = 0, so 120 - 2x - 4y = 0 and

120 - 4x = 0, so x = 30 and y = 15. (x, y) = (30, 15) meets constraint (vii), as 30² + 15² ≤ 3481 and meets constraint (viii), as 2*30 + 4*15 = 120. F(x, y) = 30*15 = 450

Case III: If μ = 0 and λ > 0, then x - 2λy = 0 and x = 2λy, so 2λ = x/y.

Likewise, y- 2λx = 0 and 2λ = y/x. Thus, x/y = y/x, so x = y and 3481 - x² - y² = 0

(Since λ > 0, L_λ = 0). Thus, 3481 = 2x², and x = y = 41.71930009. But this fails to meet constraint (viii), since 2*41.71930009 + 4*41.71930009 = 250.3158005 > 120.

Thus, the maximum for F(x, y), given the inequality constraints, is (x, y) = (30, 15) and F(x, y) = 450.

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