Lagrange Calculus Problem!

Question

Use Lagrange multipliers to find the maximum and minimum values of f(x,y) = 2x−5y subject to the constraint x^2+2y^2 = 66, if such values exist.

maximum =

minimum =

Answer 1

Objective function: f(x,y) = 2x-5y

Constraint function: g(x,y) = x²+2y²-66

Lagrangian function: L(x,y,λ) = f(x,y) - λg(x,y) = 2x-5y - λ(x²+2y²-66) = 2x - 5y - x²λ - 2y²λ + 66λ

Find partial x and y and set both equal to 0 to solve for x and y in terms of λ

∂L/∂x = 2 - 2xλ = 0 --> 2 = 2xλ --> x = 1/λ

∂L/∂y = -5 - 4yλ = 0 --> -5 = 4yλ --> y = -5/(4λ)

Plug the x and y values from before into the constraint function

x² + 2y² - 66 = 0

(1/λ)² + 2(-5/(4λ))² = 66

1/λ² + 2(25/(16λ²)) = 66

1/λ² + 25/(8λ²) = 66

8+25 = 528λ²

33 = 528λ²

1/16 = λ²

±1/4 = λ

Plug possible values of λ back into x and y

x = 1/λ = 1/(±1/4) = ±4

y = -5/(4λ) = -5/(4(±1/4)) = ±5

Check all possible ordered pairs for the objective function to see which ones are the maximum and minimum

f(4,5) = 2(4)-5(5) = 8-25 = -17

f(4,-5) = 2(4)-5(-5) = 8+25 = 33 <-- Maximum

f(-4,5) = 2(-4)-5(5) = -8-25 = -33 <-- Minimum

f(-4,-5) = 2(-4)-5(-5) = -8+25 = 17

Thus, f(x,y) has a maximum of 33 at (4,-5) and a minimum of -33 at (-4,5)

Hope this helped!