Global Max and Min Problem I'm working on a problem which asks me to find local and global extrema of the following function.
$$f(x,y) = x^2y^2e^{(-x^2 - 2y^2)}$$
I went through and found all of the relevant partial derivatives.
\begin{align*}
f_x &= (2xy^2)(e^{(-x^2 - 2y^2)}) + (x^2y^2)(e^{(-x^2 - 2y^2)})(-2x)\\
f_x &= (e^{(-x^2-2y^2)})(2xy^2 -2x^3y^2)\\
\\
f_y & = (2x^2y)(e^{(-x^2-2y^2)}) + (x^2y^2)(e^{(-x^2-2y^2)})(-4y)\\
f_y &= (e^{(-x^2-2y^2)})(2x^2y-4x^2y^3)\\
\\
f_{xx} &= (e^{(-x^2-2y^2)})(-2x)(2xy^2 -2x^3y^2) + (e^{(-x^2-2y^2)})(2y^2 -6x^2y^2)\\
f_{xx} &= (e^{(-x^2-2y^2)})(-10x^2y^2 + 4x^4y^2 + 2y^2)\\
\\
f_{yy} &= (e^{(-x^2-2y^2)})(-4y)(2x^2y-4x^2y^3) + (e^{(-x^2-2y^2)})(2x^2 - 12x^2y^2)\\
f_{yy} &= (e^{(-x^2-2y^2)})(-20x^2y^2 + 16x^2y^4 + 2x^2)\\
\\
f_{xy} &= (e^{(-x^2-2y^2)})(-4y)(2xy^2 -2x^3y^2) + (e^{(-x^2-2y^2)})(4xy-4x^3y)\\
f_{xy} &= (e^{(-x^2-2y^2)})(-8xy^3 + 8x^3y^3 +4xy - 4x^3y)\\
\end{align*}
However, I'm not sure what to do after this. I thought I was  supposed to set $f_x$ and $f_y$ equal to 0 but I don't know how to solve the equations that I get. Can someone please help me? Did I make a mistake while I was determining my partial derivatives?
EDIT: I made a mistake calculating the partial derivatives and I edited that
 A: Well the global minimum is obviously $0$. As for the other part, yes you have to find the values of $x$ and $y$, for which $f_x$ and $f_y$ are bout equal to $0$. From then on (this is quite hard for me to formulate, because English is not my native language), you have to plug those values into a matrix $$\begin{pmatrix}
f_{xx} & f_{xy} \\ 
f_{yx} & f_{yy}
\end{pmatrix}$$ and if $D(a,b)>0$ and $f_{xx}(a,b)>0$ then the point $(a,b)$ is a local minimum of f.
If $D(a,b)>0$ and $f_{xx}(a,b)<0$ then the point $(a,b)$ is a local maximum of f.
If $D(a,b)<0$ then the point $(a,b)$ is a saddle point of f.
Hope this helps $\ddot\smile$.
A: Let $g(x,y) = \ln f(x,y)$. Note that, $g(x,y) = 2\ln x + 2\ln y - x^2 - 2y^2$ and
$$
\begin{cases}
\dfrac{\partial g}{\partial x} = \dfrac{2}{x} - 2x\\
\dfrac{\partial g}{\partial y} = \dfrac{2}{y} - 4y\
\end{cases}
$$
On the order hand,
$$
\begin{cases}
\dfrac{\partial g}{\partial x} = \dfrac{\partial}{\partial x}\ln f(x,y) = \dfrac{1}{f(x,y)}\dfrac{\partial f}{\partial x}\\
\dfrac{\partial g}{\partial y} = \dfrac{\partial}{\partial y}\ln f(x,y) = \dfrac{1}{f(x,y)}\dfrac{\partial f}{\partial y}
\end{cases}
$$
Thus, 
$$
\begin{cases}
\dfrac{\partial g}{\partial x} = x^2y^2e^{-x^2 -2y^2}\biggl(\dfrac{2}{x} - 2x\biggr) = 0\\
\dfrac{\partial g}{\partial y} = x^2y^2e^{-x^2 -2y^2}\biggl(\dfrac{2}{y} - 4y\biggr) = 0
\end{cases}
$$
This system, the critical points are $(0,0)$, $(-1,-\frac{1}{\sqrt{2}})$,  $(-1,\frac{1}{\sqrt{2}})$,  $(1,\frac{1}{\sqrt{2}})$,  $(1,-\frac{1}{\sqrt{2}})$.
A: Set $x^2=u,y^2=v$ then
$$f(x,y) = g(u,v)=uv e^{-u - 2v}$$
$$\frac{\partial g} {\partial u}=0 \implies (u-1)v=0$$
$$\frac{\partial g} {\partial v}=0 \implies (2v-1)u=0$$
The solutions are $u=v=0$ and $u=1,v=1/2$. the first is the global min. The second is a global max.
A: You have:
$$0= (e^{(-x^2-2y^2)})(2xy^2 -2x^3y^2)$$
$$0=(e^{(-x^2-2y^2)})(2x^2y-4x^2y^3)$$
Because $0 \neq e^{(-x^2-2y^2)}$:
$$0= 2xy^2 -2x^3y^2$$
$$0=2x^2y-4x^2y^3$$
You have two cases:
1)$x \neq 0$, then from second equation $0=y-2y^2$, so $y=0$ or $y=\frac{1}{2}$, so $(z,0)$ for all $z$ is potential extremum, and for $y=\frac{1}{2}$ from first $0=x-x^3$, so $x=1$ or $x=-1$.
2)$x=0$, then $(0,z)$ is potencial extermum for all $z$.
A: Another route is to change coordinates. If the exponent were $x^2+y^2$ then polar coordinates would be obvious; in lieu of this, we use the parameterization $$(x,y)=\left(r\cos\theta,\frac{1}{\sqrt{2}} r\sin\theta\right)$$ and so obtain $$f(r,\theta) = \frac{1}{2}r^4 e^{-r^2} \cos^2\theta\sin^2\theta=\frac{1}{8}r^4 e^{-r^2}\sin^22\theta.$$ This separates $f(r,\theta)$ into angular and radial parts which are easy  to maximize/minimize.
