The Lagrangian Method of Constrained Optimization: Pracitce Problems and Solutions

Note: Here, I will present solve problems typical of those offered in a mathematical economics or advanced microeconomics course. The problems were authored by Dr. Charles N. Steele and are reprinted with his generous permission. The solutions to the problems are my own work and not necessarily the only way to solve the problems.

3. Find the maximum values of the objective function F subject to the given constraint for each of the following, using the Lagrangian method.

a. F(x, y) = xy, subject to 5x + 2y = 20

b. F(x, y) = 2x^1/2y^1/2 subject to x² + y² = 8

c. F(x, y, z) = xyz subject to x² + y² + z² = 12

d. F(x, y, z) = x + y + z subject to x² + y² + z² = 12

Solution 3a. Lagrangian: L(x, y, λ) = xy + λ[20 - 5x - 2y]

L_x = y - 5λ ≡ 0

L_y = x - 2λ ≡ 0

L_λ = 20 - 5x - 2y ≡ 0

Thus, 2λ = x and 5λ = y (from the transformed for L_x and L_y).

So 20 - 5x - 2y = 20 -5*2λ - 2*5λ = 20 - 20λ = 0, so 20 = 20λ and λ =1,

whereby x = 2 and y = 5.

Solution 3b. Lagrangian: L(x, y, λ) = 2x^1/2y^1/2 + λ[8 - x² - y²]

L_x = x^-1/2y^1/2 - 2λx ≡ 0

L_y = x^1/2y^-1/2 - 2λy ≡ 0

L_λ = 8 - x² - y² ≡ 0

x^-1/2y^1/2 - 2λx ≡ 0 implies 2λx = x^-1/2y^1/2 and 2λ = x^-3/2y^1/2

Thus, λ = (1/2)x^-3/2y^1/2

x^1/2y^-1/2 - 2λy ≡ 0 implies 2λy = x^1/2y^-1/2 and 2λ = x^1/2y^-3/2

x^1/2y^-3/2 = x^-3/2y^1/2 implies that x² = y² and thus 8 = 2x² and x = 2, y = 2.

λ = (1/2)x^-3/2y^1/2 = (1/2)(2)^-3/2(2)^1/2 = λ = ¼

Solution 3c. Lagrangian: L(x, y, z, λ) = xyz + λ[12 - x² - y² - z²]

L_x = yz - 2λx ≡ 0

L_y = xz - 2λy ≡ 0

L_z = xy - 2λz ≡ 0

L_λ = 12 - x² - y² - z² ≡ 0

yz - 2λx ≡ 0 implies 2λx = yz and λ = yz/2x

xz - 2λy ≡ 0 implies 2λy = xz and λ = xz/2y

xy - 2λz ≡ 0 implies 2λz = xy and λ = xy/2z

yz/2x = xz/2y = xy/2z implies

y²z/x = xz = xy²/z implies

y²z² = x²z² = x²y²

x²z² = x²y² implies z² = y²

y²z² = x²z² implies x² = y²

Thus, x² + y² + z² = 12 implies 3x² = 12 and x = 2, y = 2, z = 2

λ = xy/2z = (2*2)/(2*2) = λ = 1

Solution 3d. Lagrangian: x + y + z + λ[12 - x² - y² - z²]

L_x = 1 - 2λx ≡ 0

L_y = 1 - 2λy ≡ 0

L_z = 1 - 2λz ≡ 0

L_λ = 12 - x² - y² - z² ≡ 0

Rearranging the expressions for L_x, L_y, and L_z, we get x = y = z = 1/2λ

Thus, x² + y² + z² = 12 implies 3x² = 12 and x = 2, y = 2, z = 2

z = 1/2λ means that 2λ = 1/z and λ = 1/2z = λ = ¼

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