# Finding local extremes from stationary points

1. Feb 27, 2008

### goatsebear

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

Find the stationary points and local extreme values of f(x,y) = xy^2e^-($$\frac{x^2 + y^2)}{2}$$

2. Relevant equations
You need to find the gradient for the function and then set it equal to 0. So then df/dy equals df/dx and you can solve for a solution set.

3. The attempt at a solution
df/dx = y^2(-x^2 + 1)*e^-($$\frac{x^2 + y^2)}{2}$$

df/dy = y*x(-y^2 + 2)*e^-($$\frac{x^2 + y^2)}{2}$$

So since e^-($$\frac{x^2 + y^2)}{2}$$ cannot equal zero, y^2(-x^2 + 1) = 0
and y*x(-y^2 + 2) = 0

So then, y^2(-x^2 + 1) goes to y^2 - x^2y^2 = 0 and then $$\pm$$1 = x

Plugging this into the dy/dx equation gives me the following x, y critical points
(0,0) (1,-$$\sqrt{2}$$) (1, -$$\sqrt{2}$$) (-1, $$\sqrt{2}$$) (-1, $$\sqrt{-2}$$)

So then to find out if they are max or mins or saddle points, I solve for the second partials of fx, fy, fxy

fxx=A = x(-1 + 2y^2 - x^2y^2) * e^-($$\frac{x^2 + y^2)}{2}$$

fyy=C = y(x - 2y + 3yx) * e^-($$\frac{x^2 + y^2)}{2}$$

fxy=B = y(2- 2x^2 + y^2 = Y2x^2) * e^-($$\frac{x^2 + y^2)}{2}$$

Now I use the equation D = AC - B^2

Can someone check my math out? It seems that something is wrong somewhere. I know this is a huge problem so thanks for any help.

Last edited: Feb 27, 2008
2. Feb 27, 2008

### EnumaElish

The critical points are:
(x, 0) for any x
(-1, -sqrt 2)
(-1, sqrt 2)
(1, -sqrt 2)
(1, sqrt 2)

In each 2nd order derivative, the polynomial term should simplify to:
x(-3 + x^2)y^2 ... for "xx"
x(2 - 5 y^2 + y^4) ... for "yy"
y(-1 + x^2)(-2 + y^2) ... for "xy"

Last edited: Feb 27, 2008
3. Feb 27, 2008

### goatsebear

So I was mostly right other than that being (x,0). Now I just plug those points in for x and y in that D = AC - B^2 right? How would I do that with the (x,0) point?

4. Feb 27, 2008

### EnumaElish

Plug in y = 0, leave x as x.