# Multivariable Calculus: finding relative extrema/saddle points

1. Feb 22, 2010

### Spatulatr0n

1. The problem statement, all variables and given/known data

I was given an assignment to find all relative extrema and saddle points of the equation

f(x,y) = 1/3x^4 + 1/2y^4 - 4xy^2 +2x^2 + 2y^2 + 3

I derived the first partial with respect to x and the first partial with respect to y, but when I tried to find where they both equal 0, the problem became really complicated. I don't know what I am suppose to do. Rather stuck.

2. Relevant equations

partial x: 4/3x^3 - 4y^2 + 4x

0 = 1/3x^3 - y^2 + x

partial y: 2y^3 - 8xy + 4y

0 = 2y(y^2 - 4x +2)

3. The attempt at a solution

I have tried using substition after setting the equations equal to each other, by using the partial derivatives to find

x = (y^2 + 2)/4

and

y = + or - sqrt(4x - 2)

y = + or - sqrt(1/3x^3 + x)

I have tried setting these two equal to each other to solve for x, but I get a cubic that I don't know what to do with. Is there something like "completing the cube" I could use? Haha.

(1/3x^3 - 3x + 2 = 0)

2. Feb 23, 2010

### Tinyboss

For starters, (0,0) is clearly a critical point.

3. Feb 23, 2010

### Spatulatr0n

I found that one, at the very least. :)

The assignment was due today, posting this was a last hurrah, I suppose. Now, however, I'm merely curious as to how to solve it.