Local max/min points, partial derivatives I'm having a lot of problems with figuring out how to properly do max/min with partial derivatives.
To my knowledge, we have:
$$D(x, y) = f_{xx}(x, y)f_{yy}(x, y) - (f_{xy}(x, y))^{2}$$
With the following conditions:
-If $D > 0$ and $f_{xx} < 0$, the critical point is a local max
-If $D > 0$ and $f_{xx} > 0$ then the critical point is a local min
-If $D < 0$ then the point is a saddle point
-If $D = 0$ then the test is inconclusive
I have the following function:
$$f(x,y) = 2xye^{-x^{2}-y^{2}}$$
I solved the first derivatives (not entirely sure that these are right):
$$f_{x}(x, y) =2ye^{-x^{2}-y^{2}} \cdot (-2x) = -4xye^{-x^{2}-y^{2}}$$
$$f_{y}(x, y) =2xe^{-x^{2}-y^{2}} \cdot (-2y) = -4xye^{-x^{2}-y^{2}}$$
But now I have no idea how to solve $f_{x} = 0$ and $f_{y} = 0$ to find my critical points. I suspect I have to do the natural logarithm of something?
Any help is appreciated, thanks!
 A: We have:
$$f(x,y) = 2xye^{-x^{2}-y^{2}}$$
A plot of this shows:

A different view shows:

A contour plot shows:

The partials are:


*

*$f_{x}(x, y) = 2 y e^{-x^2-y^2}-4 x^2 y e^{-x^2-y^2}$

*$f_{y}(x, y) = 2 x e^{-x^2-y^2}-4 x y^2 e^{-x^2-y^2}$

*$f_{xx}(x, y) = 8 x^3 y e^{-x^2-y^2}-12 x y e^{-x^2-y^2}$

*$f_{yy}(x, y) = 8 x y^3 e^{-x^2-y^2}-12 x y e^{-x^2-y^2}$

*$f_{xy}(x, y) = f_{yx}(x, y) = -4 x^2 e^{-x^2-y^2}+8 x^2 y^2 e^{-x^2-y^2}+2 e^{-x^2-y^2}-4 y^2 e^{-x^2-y^2}$


You should have five critical points as:
$$(x, y) = (0,0), \left(-\dfrac{1}{\sqrt{2}}, -\dfrac{1}{\sqrt{2}}\right), \left(-\dfrac{1}{\sqrt{2}}, \dfrac{1}{\sqrt{2}}\right), \left(\dfrac{1}{\sqrt{2}}, -\dfrac{1}{\sqrt{2}}\right), \left(\dfrac{1}{\sqrt{2}}, \dfrac{1}{\sqrt{2}}\right)$$
To find these we have $f_x = f_y = 0$, which leads to the following:
$$2 y - 4 x^2 y = 0, 2 x - 4 x y^2 = 0 \implies 2x^2 = 1, 2y^2 = 1, ~\mbox{and}~x=y=0$$
This gives a total of five roots (critical points) shown above.
Can you now proceed?
A: If $f(x,y)=2xye^{-x^2-y^2}$, then $$\frac{\partial f}{\partial x}=2ye^{-x^2-y^2}-4x^2ye^{-x^2-y^2}$$ Setting this equal to 0 gives $$y(1-2x^2)=0$$ and setting $f_y=0$ gives $$x(1-2y^2)=0$$ The possible points are $(0,0),\hspace{2mm}\left(\pm{1\over\sqrt{2}},\pm{1\over\sqrt{2}}\right)$ (the second point is really four different points). 
Now you would have to compute the second derivatives, and evaluate the discriminant $D$ at each of these points to determine what type of extrema they are.
