# Find all critical points, and identify them as minima, maxima, or

1. Aug 31, 2013

### Jormungandr

1. The problem statement, all variables and given/known data
Consider the function f(x,y) = (x2 + 4y2)e(1-x2-y2)

Find all critical points, and identify them as maxima, minima, or saddle points.

3. The attempt at a solution

I took the partial of x and the partial of y, and set them equal to 0. This is what I got:

fx(x,y) = 4y2 - x2 - 1 = 0

fy(x,y) = 4y - 2x2 + 4 = 0

But from here I'm kind of lost. I took Calc 3 a year ago, last fall, and now I've had to recall a lot of it for my PChem class. This is one of our homework questions, and I'm just having a lot of trouble remembering what the steps are for this. Help is greatly appreciated. Thanks!

2. Aug 31, 2013

### Jufro

Check your derivatives first. I came up with something similar but a little different.

Then, what you are left with is two ellipses. Therefore, the solution set which satisfies both equations would be the intersection points of those two ellipses gained from fx and fy.

After finding those points you can use the second partials test to categorize the extrema as being a local max, local min or a saddle point.

For reference, http://en.wikipedia.org/wiki/Second_partial_derivative_test.

Hope this helps

3. Aug 31, 2013

### Zondrina

The partial with respect to $x$ is:

$\frac{∂}{∂x} [e^{1-x^2-y^2} (x^2+4y^2)] = -2x e^{1-x^2-y^2} (x^2+4y^2-1)$

Set that equal to zero. Note that $e^u > 0, \forall u \in ℝ^2$. So either $x=0$ or $x^2+4y^2-1 = 0$.

That gives you one coordinate of potential point(s), $x=0$. You have to solve the other one in terms of x or y.

Then take your partial with respect to y and run through the cases.

4. Sep 2, 2013

### Jormungandr

So I've been slowly plodding through this problem, and so far this is what I've done:

Took partial of $x$, factored out the $e$ term and canceled it, because it can never equal $0$.

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

Same with the partial of $y$: $x^2+4y^2 = 4$

Now, these are ellipses, but they don't intersect. Ever. So from partial of $x$, I got $x = 0$ and from partial of $y$, I got $y = 0$. I plugged in $x = 0$ into $x^2+4y^2 = 4$ and vice versa for the other, and the critical points I got were:

$(0,0)$
$(0,1)$
$(0,-1)$
$(1,0)$
$(-1,0)$

I then took the second partials of $x$ and $y$, and the mixed partial $xy$. I substituted each point into the formula:

$D(x,y) = f_{xx}(x,y) f_{yy}(x,y) - [f_{xy}(x,y)]^2$ and got:

$(0,0)$: Saddle point
$(0,1)$: Maximum
$(0,-1)$: Saddle point
$(1,0)$: Saddle point
$(-1,0)$: Saddle point

And... That just seems off to me. The way I see it, either I got the wrong critical points, or I took the wrong partial derivatives. I got:

$f_{xx}(x,y) = (2-10x^2-8y^2+4x^4+16x^2y^2)e^{1-x^2-y^2}$

$f_{yy}(x,y) = (8 - 2x^2 - 24y^2 - 16y + 4x^2 y^2 + 16y^4)e^{1-x^2-y^2}$

$f_{xy}(x,y) = (-20xy+4x^{3}y+16xy^3)e^{1-x^2-y^2}$

And this is where I am a bit stumped. The next part has me calculating the total differential, which doesn't seem too difficult. But I am at my wit's end, and short of redoing the entire problem I am not sure what could have gone wrong. If someone could diagnose what I've done incorrectly, I'd greatly appreciate it.

5. Sep 2, 2013

### Zondrina

Your partial derivatives look alright I think. Also, you got these points :

$(0,0)$: Saddle point
$(0,1)$: Maximum
$(0,-1)$: Saddle point
$(1,0)$: Saddle point
$(-1,0)$: Saddle point

I'm getting (0,0) as a minimum and (0,-1) as a max ( not a saddle ). Otherwise everything else looks good.

Last edited: Sep 2, 2013
6. Sep 2, 2013

### Jufro

I agree with Zondrina, the values are correct but not their classification.

To help visualize the function I attached its image.

File size:
169.3 KB
Views:
45