find all the critical points of the function f(x,y)=(x^2 -3xy+y^2 )e^-(x^2+y^2) * i got for d/dx=(2x - 3y) e^-(x^2+y^2) +(x^2 - 3 x y + y^2 ) e^-(x^2+y^2) (-2x) * i got for d/dy=(2y - 3x) e^-(x^2+y^2) +(x^2 - 3 x y + y^2 ) e^-(x^2+y^2) (-2y) i can't figure out how to get the critical points for this....

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find all the critical points of the function f(x,y)=(x^2 -3xy+y^2 )e^-(x^2+y^2) * i got for d/dx=(2x - 3y) e^-(x^2+y^2) +(x^2 - 3 x y + y^2 ) e^-(x^2+y^2) (-2x) * i got for d/dy=(2y - 3x) e^-(x^2+y^2) +(x^2 - 3 x y + y^2 ) e^-(x^2+y^2) (-2y) i can't figure out how to get the critical points for this....

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i remember how to get critical numbers for y=f(x). you find where f' is 0 and where f' DNE in the domain of f. would we do the same thing here? let me look it up
here is an example http://www.math.wvu.edu/~hjlai/Teaching/Tip-Pdf/Tip3-30.pdf
I have to make d/dx = 0 to find the critical but i can not figure out what it will be.

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you might have to solve one of the equations for y or x and plug it into the other equation to find either x or y depending on what you solved for in the first one let me take a deeper look
the only solution i got is through quadratic but is that the only way?
This problem is ridiculous. I can't imagine taking derivatives and using the second partials test CANNOT be the way to do it. Have you done lagrange multipliers or polar coordinates?
I got this for homework and I am killing myself over it
how would i use lagranges for this?
Ha, you can't use Lagrange because there's only one function and no constraints. Was this problem all of your homework assignment? I am definitely thinking there are polar coordinates involved because of all the squares. I guess finding Fxx, Fxy, and Fyy would be the next step... nasty.

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