Question

This extreme value problem has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the extreme values of the function subject to the given constraint.
f(x, y) = xy; 4x2 + y2 = 8

Answer 1

Answer: minimum value of f = -2  and maximum value  of f = 2

Step-by-step explanation:

Given that;

f(x, y) = xy; 4x² + y² = 8

S(x,y) = xy + λ(4x² + y² - 8)

dS/dλ = 4x² + y² - 8 = 0

dS/dx = y + 8λx = 0 -> y = -8λx  

dS/dy = x + 2yλ = 0 -> x + 2λ(-8λx) = x(1 - 16λ²) = x(1-4λ)(1+4λ) = 0

-> x = 0 , λ = 1/4 , λ = -1/4

x = 0 -> y = 0 -> it doesn't satisfy 4x² + y² = 8

λ = 1/4 -> y = -8λx = -2x -> 4x² + y² = 8x² = 8 -> x = -1 , 1 -> y = 2 , -2

λ = -1/4 -> y = -8λx = 2x -> 4x² + y² = 8x² = 8 -> x = -1 , 1 -> y = -2 , 2

we have 4 critical points:

(-1, 2) -> f = -2

(1, 2) -> f = 2

(-1, -2) -> f = 2

(1 , -2) -> f = -2  

Therefore; minimum value of f = -2  and maximum value  of f = 2