# Multivariable Maximum and Minimum Problem

Let $$f(x, y) = 2xye^{-x^2-y^2}.$$ Find critical points. Use second derivative to to determine whether a critical point corresponds to local min, max or saddle.

Hi so for this question I am able to get my critical points by taking all the partial derivatives but when I get to inputting my critical points into the equation $$f_{xx}f_{yy}-(f_{xy})^2$$. I don't get the right answer. For instance one of the critical points is $$(0, 0)$$ but when I input it into the equation I get $$0$$ which is not correct as it is supposed to be $$-4$$. Can someone show me how to use the second derivative for this question.

EDIT: I took the partial derivatives to get Fxx= 4(2x^2-3)xye^(-x^2-y^2) Fyy = 4(2y^2-3)xye^(-x^2-y^2) Fxy = 2(2y^2-1)xe^(-x^2-y^2)

D(x, y) = fxx*fyy - (fxy)^2

Therefore, whenever I put the required equations in and set them to 0 I do not get the answer expected which is -4.

• Please edit your post to include your work. Otherwise we have to be mind readers to figure out why you don't get the right answer. – Ted Shifrin Jan 20 '20 at 3:11
• The question has been EDITED I included the partial derivatives and the formula used to check what D(x, y) is equal to. – OGK Jan 20 '20 at 6:02

Your value for $$f_{xy}$$ cannot be correct, as $$x$$ and $$y$$ do not appear symmetrically in it. (Note that $$f(x,y)=f(y,x)$$, so everything should be symmetric.) Indeed, $$f_{xy} = 2(2x^2-1)(2y^2-1)e^{-(x^2+y^2)},$$ so $$f_{xy}(0,0) = 2$$.