# Question on stationary points

• Oct 29th 2010, 05:31 PM
goaway716
Question on stationary points
Find and classify the stationary points of :
f(x,y)=(x+y)*e^(-x²-y²)
• Oct 29th 2010, 05:33 PM
lilaziz1
I think it means find the critical points.
• Oct 29th 2010, 05:41 PM
goaway716
Thing is that i can do stationary points with 1 variable involved but this one has two,not used to it.
help plsss
• Oct 29th 2010, 06:46 PM
lilaziz1
Um then I'm assuming you don't know partial derivatives. Just in case you do, you should do the following:

Find $f_x$ and $f_y$, set them equal to 0, and solve for x and y.

I got this for $f_x$ and $f_y$:

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

$f_y=(x+y)e^{-x^2-y^2}(\frac{1}{x+y}-2y)$
• Oct 30th 2010, 01:59 AM
mr fantastic
Quote:

Originally Posted by goaway716
Thing is that i can do stationary points with 1 variable involved but this one has two,not used to it.
help plsss

Haven't you been taught this material? If not, then why are you attempting this question. If you have been taught it, then your class notes or textbook should provide enough help to at least get you started - what have you tried and where are you stuck?
• Oct 30th 2010, 01:45 PM
goaway716
yes i have done whatever i could have but i just want to make sure whether i am on the right track and need some guidance.
This is what i have tried :

Take the partial derivatives d/dx and d/dy. So:

(d/dx) f(x,y)=(d/dx)[(x+y)*e^(-x²-y²)] =e^(-x²-y²) +(e^(-x²-y²))(-2x)= e^(-x²-y²)[1-2x]=0 iff x=1/2.

The y-case is symmetrical, so (d/dy) f(x,y)=0 iff y=1/2.

They are then both equal to zero at the point

(1/2,1/2, z). Substituting , this is (1/2,1/2,e^(-1/2))

this is it..any mistake?