Use Lagrange multipliers to find the maximum and minimum values of f(x,y) = xy on the ellipse 8x^2+y^2=2?

Let C(x, y) = the constraint function (the equation of the ellipse written as if it had been set equal to zero) = 8x² + y² - 2

The Lagrange function is:

L(x, y, λ) = f(x, y) + λC(x, y)

L(x, y, λ) = xy + λ(8x² + y² - 2)

Take the partial derivatives:

∂L(x, y, λ)/∂x = y + 16λx

∂L(x, y, λ)/∂y = x + 2λy

∂L(x, y, λ)/∂λ = 8x² + y² - 2

Set the partial derivatives equal to zero and solve the system of equations:

0 = y + 16λx

0 = x + 2λy

0 = 8x² + y² - 2

16λx = -y

x = -2λy

0 = 8x² + y² - 2

-x/y = 1/16λ

-x/y = 2λ

2λ = 1/16λ

λ² = 1/32

λ = √2/8 and λ = -√2/8

The system of 3 equations just became 2 systems of two equations. It does not matter which of the first two original equations you use as your first equation but the second equation must be the ellipse.

Let's use:

0 = y + 16λx

0 = 8x² + y² - 2

y = 2√2x

0 = 8x² + 8x² - 2

x² = 1/8

x = √2/4 and x = -√2/4

y = 1 and y = -1

This will repeat for λ = -√2/8

There are two maximum points, (√2/4, 1) and (-√2/4, -1)

There are two minimum points, (-√2/4, 1) and (√2/4, -1)