# A max/min problem.

1. Apr 3, 2005

### ArnfinnS

hi...
i need to find the mex and min point of the function f(x,y) = x^2*y*e^(-x^2 - 2y^2) for (x,y) in R^2

here is what i tried : i found the partial derivatives :

f_x = 2x*y*e^8-x^2 - 2y^2) + x^2*y*e^(-x^2 -2y^2)*(-2x)

and f_y = x^2*y*(-4y)*e^(-x^2 - 2y^2)

i see that those partials equals 0 in the point (0,0). is this the only stationary point here?

what is max / and what is minimum?
can anyone help me?

2. Apr 3, 2005

### Zurtex

I assume you mean the points where the plane tangent to the surface is parallel to the x-y plane, would this be where Fxy = 0? I've not quite got that far on my calc course.

Edit: oh by the way my graph below is a picture of what the general equation looks like,

#### Attached Files:

• ###### Clipboard02.jpg
File size:
62.8 KB
Views:
128
Last edited: Apr 3, 2005
3. Apr 3, 2005

### SpaceTiger

Staff Emeritus
Alright, this'll be a bit easier to read in Latex:

$$f(x,y)=x^2ye^{-x^2-2y^2}$$

Alright, here's a first problem. You have:

$$f_x = 2xye^{-x^2 - 2y^2} - 2x^3ye^{-x^2 -2y^2}$$
$$f_y = -4x^2y^2e^{-x^2 - 2y^2}$$

The partial with respect to x looks fine, but you missed a term in the y partial. It should be:

$$f_y = x^2e^{-x^2 - 2y^2}-4x^2y^2e^{-x^2 - 2y^2}$$

To find the critical points, you just set these to zero. This will eliminate those pesky exponentials.

$$-x^2+1=0$$
$$-4y^2+1=0$$

This will give you your critical points. I'll leave that part to you. Now, if you want to classify them, you have to calculate the second derivatives at your critical points and see if they're positive or negative in each direction. Do you know how to do that?

Last edited: Apr 3, 2005