Given the function:
$$f(x,y)=y\exp{(x(1-y^2))}$$
Determine the global maximum and global minimum for $f$ in the point set:
$$A=\{(x,y)\,|\,x\in[-1,1],y\in[-1,1]\}$$
I know how to find the stationary points. Those are $(\frac{1}{2},1)$ and $(\frac{1}{2},-1)$. Both are saddle points without the restriction though. How do I find the global extrema when given a point set?
Note that $A=[-1,1]\times [-1,1]$ is indeed a compact set, then your (continuous) function attains its minimum and maximum. First check for critical points on the interior of $A$, $\dot{A}$, which you can find by its gradient $$ \nabla f = \binom{y(1-y^2)\exp(x(1-y^2))}{\exp(x(1-y^2))-2y^2\exp(x(1-y^2))} = \binom{0}{0}. $$ I guess that the two point that you've mentioned are exactly these. Next, you should check for extreme points on the boundary of $A$, i.e., $\partial A$. Such that for $y=-1$ you have $f(x,-1)=-1$ and $f(x,1)=1$, you should check also what happens where $x= \pm 1$ (i.e., find critical points of the univariate function $f(\pm1,y)$ in $A$). And lastly, check out the vertices - that is easy in this case as for all the vertices $(\pm1, \pm1)$, $f(x,y)= \pm 1$. Finally, compare all the critical points that you've found.