Find maximum and minimum of $f(x, y) = xy$ on $D = \left\{ (x,y) \in \mathbb{R}^2: x^2+2y^2 \leq 1 \right\}$ I'm kinda stuck on this one : 

Find the minimum and maximum of the given function $f$ on $D$, where $$f(x, y) = xy$$ and $$D = \left\{(x,y) \in \mathbb{R}^2 : x^2+2y^2 \leq 1 \right\}$$

I don't know what to do with this domain $D$. I counted the first derivatives and got only point $(0,0)$ as a possible maximum/minimum inside $D$ but what about the boundary of $D$? Should I parametrize this ellipse or how should I approach this? Thanks for your tips!
 A: Obviously the function $f$ has no relative extreme inside the desired region $D$. In favt using the routine method we will find $(0,0)$ a saddle point in which $f_{xx}f_{yy}-(f_{xy}^2)<0$. Now consider $D$ and that $$x=\pm\sqrt{1-y^2}$$ Putting each parts $x=+$ and then $x=-$ separately we will find two one-variable functions $$f(y)=+y\sqrt{1-y^2}, ~~~f(y)=-y\sqrt{1-y^2}$$ I think you can find the relative extremes of these functions.... You have $4$ points as I plotted below:

A: As you said above the critical points at the rim can be found by parameterization. To parameterize the ellipse, use the following: $x=\cos(t)$ and $y=2\sin(t)$ As for the rest of the interval, first take the gradient of the function:
$$\nabla{f(x,y)}=\langle{y,x}\rangle$$
It equals $\vec{0}$ when $x=0$ and $y=0$, making $(0,0)$ the only critical point in the function (it also lies within the interval). Compare it to the critical point(s) you found on the boundary to find the minimum and maximum.
A: This can be solved rather easily by elementary methods without using calculus.
By AM-GM inequality, $$x^2+2y^2\geq 2(x^2\cdot 2y^2)^{1/2}=2\sqrt{2} xy$$ Hence we have $2\sqrt{2} xy\leq 1$, that is $xy\leq1/2\sqrt{2}$. Also note that the equality is achieved at $x=1/\sqrt{2}$ and $y=1/2$.
For the other inequality note that $$1\geq x^2+2y^2=(x+\sqrt{2}y)^2-2\sqrt{2}xy\geq -2\sqrt{2} xy$$ Hence we have $xy\geq -1/2\sqrt{2}$ and the equality is a achieved at $x=1/\sqrt{2}$ and $y=-1/2.$
