Find and classify the stationary points of :

f(x,y)=(x+y)*e^(-x²-y²)

Printable View

- Oct 29th 2010, 04:31 PMgoaway716Question on stationary points
Find and classify the stationary points of :

f(x,y)=(x+y)*e^(-x²-y²) - Oct 29th 2010, 04:33 PMlilaziz1
I think it means find the critical points.

- Oct 29th 2010, 04:41 PMgoaway716
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, 05:46 PMlilaziz1
Um then I'm assuming you don't know partial derivatives. Just in case you do, you should do the following:

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

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

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

$\displaystyle f_y=(x+y)e^{-x^2-y^2}(\frac{1}{x+y}-2y)$ - Oct 30th 2010, 12:59 AMmr fantastic
- Oct 30th 2010, 12:45 PMgoaway716
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?