Finding global maximizers and minimizers I want to  find if global maximum or minimum exists in $ f(x,y)=e-^{(x^2+y^2)}$    
I found that (0,0) is the only critical point.
In the Hessian matrix $H_{(f)}(0,0)$
was negative definite and so (0,0) is a local maximizer.
There are few things I need to clarify .
1)As $(0,0)$ is the only critical point and it is a local maximum does it indicate that it is a global maximum as well, because it is the only critical point.  
2)Using principal minor method in Hessian matrix $\partial^2f \over \partial^2x$=$2(2x^2-1) *e-^{(x^2+y^2)}$ which is the first minor. And this is $<0$ for (0,0) but for (2,0) it is $>0$.
For a global min/max shouldn't Hessian be positive/negative definite for all $x \in R^2$.
So in this case to have a global maximum  isn't it necessary that the first minor be negative for all $x \in R^2$  
3) $\lim\limits_{x,y \rightarrow +\infty}f(x,y) =0$. Also $\lim\limits_{x,y \rightarrow -\infty}f(x,y) =0$.  From this how to determine if the function has a global maximum or global minimum at (0,0)

While it is always nice to practice multi-variable extremal analysis, with partial derivatives, gradients, Hessians, etc, sometimes a much simpler argument will do. Since the function $e^{-(x^2+y^2)}$ depends only on the length of the vector $(x,y)$, we can look at the function $e^{-r^2}$. We can justify this formally by passing to polar coordinates: $x=r\cos\theta$ and $y=r\sin\theta$. Now it is evident that $e^{-r^2}$ has a global maximum at $r=0$ and only there, and that it has no minimum, not global nor local. So $f$ has a global maximum at the origin $x=y=0$ and only there, and has no minimum anywhere.
