Define $f : \Bbb R^2 \to \Bbb R$ by $f(x,y)= y(1-x^2-y^2)$
Let $D:=${$(x,y)|x^2+y^2 \leq 1$}. Does $f$ take a maximum and a minimum on $D$? If so in which points?
So I calculated the critical points, which are $(1,0), (-1,0), (0, \frac{1}{\sqrt3}), (0, -\frac{1}{\sqrt3})$. I also calculated the values for these points, $f(1,0)=0, f(-1,0)=0, f(0, \frac{1}{\sqrt3})=\frac{2}{3\sqrt3}$ and$ f(0, -\frac{1}{\sqrt3})=-\frac{2}{3\sqrt3} $
I need to use the extreme value theorem, but I'm having difficulties, because when I use $x^2+y^2=1$ to find $x^2=1-y^2$, and plug that back into $f(x,y)$ I get $0$.
$f$ is a continuous function on the compact set $D$. By Weierstrass' theorem, it takes maximum and minimum on $D$.
Since $f=0$ on $\partial D$, while $f>0$ on the open set $A_1 := D^\circ \cap \{y>0\}$ and $f < 0$ on the open set $A_2 := D^\circ \cap \{y<0\}$, we have that this points are internal points of $D$. In particular, the maximum point must lie in $A_1$ and the minimum point in $A_2$.
Since $f\in C^1(D)$, at this points the gradient of $f$ must vanish, hence we are left with the four candidates you have already computed.