# Extrema on compact Region

$f(x,y)={ e }^{ xy }\sin(y)\\ R=\left\{ (x,y):\left| x \right| \le 1;\left| y \right| \le 1 \right\}$

The partial Derivatives are: $f^{ x }\left( x,y \right) =y{ e }^{ xy }\sin(y)\\ f^{ y }\left( x,y \right) =(x\sin(y)+\cos(y)){ e }^{ xy }$

After setting both of them to zero, we get the requirement that $y=0$ or $y=k\pi$. But those Points dont statisfie the second equation,in result showing us there are no Critical Points in the inside of $R$. Since we know the set is compact Max/Min need to exist, so we need to evaluate the boundaries. Here I'm Stuck. Since the boundary is $\left| x \right| =1\quad \left| y \right| =1$ we would take those values and plug it into our function thus being only dependend on one Variable but here I get functions here i can't get a Max/min off. Show how would I procced from here?

• The boundary is a square. Choose say the east side where $x=1$, Substitute that value in $f(x,y)$. You end up with a function of a single variable whose max/min you know how to find. Repeat for the other three sides and compare. – NickD Sep 20 '17 at 12:37

If $x\in[-1,1]$, then $f(x,-1)=e^{-x}\sin(-1)=-\frac{\sin 1}{e^x}$. Therefore, in $[-1,1]\times\{-1\}$ the maximum is $-\frac{\sin1}e$ and the minimum is $-e\sin1$. Now, do the same thing in $[-1,1]\times\{1\}$, in $\{-1\}\times[-1,1]$ and in $\{1\}\times[-1,1]$.
• @johnka Every continuous function from a compact into $\mathbb R$ attains its maximum and its minimum somwhere. If it is not at the interior of $K$, then it is at its boundary. I have no idea why you decided to differentiatt the function $x\mapsto f(x,-1)$. Isn't it obvious where its maximim and its minimum are (when $x\in[-1,1]$)? – José Carlos Santos Sep 20 '17 at 12:56
the searched Minimum is $$-e\sin(1)$$ for $$x=-1,y=-1$$ and Maximum is $$e\sin(1)$$ for $$x=1,y=1$$