lagrange multiplier problem?

Question

A function is to be maximized
f(x,y)= x+y subject to the constraint 2x^2+y^2=2

wha tis the exact value for the maximum

I can find all the derivatives and end up with 1=4*lambda*x
and 1=2*lambda*y

where do I go from here?

Answer 1

Let g(x) = 2x^2+y^2 - 2

Λ(x) = f(x) + λg(x)

Λ(x) = x + y + λ(2x^2+y^2 - 2)

Now take the partials of Λ and set them to 0

∂Λ/∂x = 1 + 4λx = 0
∂Λ/∂y = 1 + 2λy = 0
∂Λ/∂λ = 2x^2+y^2 - 2 = 0

Combining the 1st 2 equations gives

1 + 4λx = 1 + 2λy
4λx = 2λy
y = 2x

Putting this back into the 3rd equation gives

6x^2 = 2

x^2 = 1/3

x = ±√(1/3)

So the solutions are

(√(1/3), 2√(1/3)), (-√(1/3), -2√(1/3))

Answer 2

Use 2x^2+y^2=2 as your third equation. This gives you 3 equations and 3 unknowns. Solve for all possible x,y, and λ to get your critical points.