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!


We have:

$$f(x,y) = 2xye^{-x^{2}-y^{2}}$$

A plot of this shows:

enter image description here

A different view shows:

enter image description here

A contour plot shows:

enter image description here

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?

  • $\begingroup$ It completely passed over my head to use product rule. That's what sleep deprivation gets you when you try to do calculus with no sleep! This helps a lot. Thank you for going into such detail :) $\endgroup$ – KingDan Apr 7 '14 at 0:10
  • $\begingroup$ Cool graphs! 8-) $\endgroup$ – amWhy Apr 7 '14 at 11:45

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.


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