# Help findng solutions to equations

1. Mar 20, 2013

### nateja

1. The problem is to find the critical points and what they are (saddle point, max, min) for the equation: f(x,y) = (x^2+y^2)*e^(y^2-x^2).

2. You will need to take partial derivatives of the function and set them equal to zero in order to solve for the critical points.

3. I did a lot of messy algerbra and got the partial derivatives. I found a minimum at (0,0). However, that is not my problem with the question. I already spoke to my professor about it and he said I was missing two critical points (-1, 0) and (1,0) - everything else was correct.

When setting fx(x,y) = 0, you get to a point where the equation factors out as:
(e^(y^2-x^2))(-2x)(x^2+y^2-1) = 0
I know (e^(y^2-x^2)) will never equal zero, so there is no critical point from that. -2x gives us critical value x = 0. And where I'm having trouble, is that x^2+y^2-1 = 0 gives x = -1 and x = 1.

I can see if you set y = 0, you will get x^2 = 1 which gives x = ±1. But then you could argue that you can set x = 0 and get y = ±1. But (0,±1) is not a critical point on the graph.

How does this work out? The function is a circle, so if x = 0, y = ±1 and if y = 0, x = ±1. So how do I convey that only (±1,0) are the critical points?

2. Mar 20, 2013

### SammyS

Staff Emeritus

I agree that fx(x,y) = 0 on the entire unit circle, x2 + y2 = 1 .

Now, look at where fy(x,y) = 0 .

Last edited: Mar 20, 2013
3. Mar 21, 2013

### nateja

Ok, so fy(x,y) is only equal to 0 when y = 0 since (1 + x^2 +y^2) does not ever equal zero, except for imaginary numbers

so the critical points are (0,0) and (±1,0)! Thanks for the help!