I want to find global max and/or min of the function
$$h(x,y)=(x^2+y^2)e^{-xy}$$
on the domain $\frac{x}{2}\geq y\geq2x$, or prove that it does not exist.
I see by looking at the function that it is always bigger than or equal to $0$. I'm thinking that I could approximate the function to also be less than some function that goes to zero when x goes to infinity. Then I know that since the function $h$ also goes to $0$ (when $x^2+y^2$ goes to infinity)then the global maximum and minimum will be somewhere inside the domain $\frac{x}{2}\geq y\geq2x$.
In other words say if I find two stationary points than I know that I can find a value less than the greater value of the points, and also one that is bigger than the least value.
I would try to write $$0\leq(x^2+y^2)e^{-xy}\leq (some function)$$
Or maybe there is another way to solve this problem.
Best Answer
The exponential part of the function tends to $0$ when $x$ and $y$ both go to positive infinity or negative infinity, and tends to infinity when one of $x$ and $y$ goes to positive infinity and the other goes to negative infinity. Therefore there can be no global maximum, and the only global minimum would have to take a non-positive value.
Your function $h$ cannot take negative values, therefore we search for points such that $h=0$. The exponential part of the function is never zero, and the $(x^2 + y^2)$ part is only zero when $x=y=0$. Therefore, the only global minimum happens at $x=y=0$, where $h=0$.
I hope this helps!
EDIT: How about local extrema? A set of necessary conditions for local extrema are $\partial h/\partial x=0$ and $\partial h/\partial y=0$. We compute \begin{align} h(x,y) &= (x^2+y^2) e^{-xy} \\\implies \frac{\partial h}{\partial x} &= (2x-y(x^2+y^2)) e^{-xy}, \\\implies \frac{\partial h}{\partial y} &= (2y-x(x^2+y^2)) e^{-xy}. \end{align}
Solving $\partial h/\partial x=0$ implies $2x-y(x^2+y^2)=0$ since the exponential part can never be zero. One solution is $x=y=0$, which we have already considered. If we let one of $x$ or $y$ be zero without the other being zero, we cannot find a solution. Therefore, we assume both $x$ and $y$ are distinct from $0$.
Looking at $\partial h/\partial y=0$ we recover $2y-x(x^2+y^2)=0$ instead. We can add both equations to get \begin{align} 2x-y(x^2+y^2) + 2y-x(x^2+y^2) = (x+y)(2-(x^2+y^2))=0. \end{align} To solve $(x+y)(2-(x^2+y^2))=0$ we need either $(x+y)=0$ or $(2-(x^2+y^2))=0$
We only have to check the points $(x,y) = (1,1)$ and $(x,y) = (-1,-1)$. In fact, since your function is invariant under the transformation $x\mapsto -x$, $y\mapsto -y$, we have symmetry along the $y=-x$ diagonal, which means we only have to check $(x,y) = (1,1)$, the other point will have the same characteristic. Since $(x,y) = (1,1)$ is a maximum of $h$ along the $y=x$ line, it may be a maximum or a saddle of $h$, not a minimum.
Now, you could use the Hessian determinant to check whether this is a maximum, but we can also do the following: Instead of considering the point $(x,y) = (1,1)$, we can change to polar coordinates, $(r,\theta)$, with the change $x=r\cos(\theta)$, $y=r\sin(\theta)$. The function becomes $r^2 e^{-r^2\cos(\theta)\sin(\theta)}$, which at $x^2+y^2=r^2=2$ is equal to $2 e^{-2\cos(\theta)\sin(\theta)} = 2 e^{-\sin(2 \theta)}$. The point $(x,y)=(1,1)$ translates to $(r,\theta)=(\sqrt{2},\pi/4)$, and $\theta = \pi/4$ is a maximum of $\sin(2 \theta)$, therefore a minimum of $2 e^{-\sin(2 \theta)}$. Thus, the point $(x,y) = (1,1)$ acts as a maximum of $h$ along the $y=x$ line but as a minimum along the $x^2+y^2=2$ circle, which means it cannot be a maximum.
Therefore, the only extremum is the global minimum $x=y=0$.