# Homework Help: Finding Critical Points of a multivariable function

1. Jul 4, 2011

### arisachu

1. The problem statement, all variables and given/known data
I'm supposed to find (and classify, which I'm really not sure what that means) critical points of f(x,y) = x2y- x2 - 2y2 - 2x4 - 3y4

2. Relevant equations

3. The attempt at a solution
I found the partial derivatives,
fx(x,y) = 2xy - 2x - 8x3
and
fy(x,y) = x2 - 4y - 12y3

I gather that x = 0 and x = sqrt(y-1/4)
I can't figure out how to solve for y, if I'm even supposed to.
I was under the impression to set both equations to 0 and solve, but I don't know if I am correct.
I guess my [STRIKE]main[/STRIKE] only question is if I am supposed to take the y solution and use it in the x? I want to solve it 100% on my own, so I only want broad answers, if that makes any sense?

Thanks to anyone who can help me! :)

Last edited: Jul 4, 2011
2. Jul 4, 2011

### micromass

Hi arisachu!

You'll need to know when fx and fy simultaniously become 0. So you'll need to solve the system

$$\left\{\begin{array}{c} f_x(x,y)=0\\ f_y(x,y)=0\\ \end{array}\right.$$

Thus

$$\left\{\begin{array}{c} 2xy - 2x - 8x^3=0\\ x^2 - 4y - 12y^3=0\\ \end{array}\right.$$

From your first equation, you have correctly deduced that x=0 or

$$x=\sqrt{\frac{y-1}{4}}$$

Now, what does that give you when you plug that in the second equation?

3. Jul 4, 2011

### arisachu

Thanks for you reply and help! :D

Ohhh, so it's as simple as that? :O
And then I solve for [STRIKE]x[/STRIKE] y from that? :)

4. Jul 4, 2011

### micromass

Yes, solve for y in the second equation, and then you can solve for x.

5. Jul 4, 2011

### arisachu

Gosh, I had been thinking you solve for y in the second equation without the first equation's x, and then set those two solutions equal to each other and I couldn't for the life of me figure it out!
You're a life (and sanity haha) saver! Thank you so much!