I have an question: What is the max volume box which can be limit for $x \geq 0$, $y \geq 0$, and under the elliptic paraboloid $z = 50 - x^2 - 2y^2$?
In the textbook there is an example with a surface. Let x, y, z be the length, width and height. I started by letting Volume $V(x,y) = xy(50-x^2-2y^2)$. And then I tried to take the partial derivatives...
I got $V_x(x, y) = 50y - 3x^2y - 2y^3$ and $V_y(x, y) = 50x - x^3 - 6y^2x$. Solve for the critical points? I have a feeling I should not have multiplied the $xy$ in the volume function? Or is the partial derivative solvable for $0$ solution?
The system can be solved
$$ xV_x=0 \Rightarrow 50xy=3x^3y+2xy^3$$
$$ yV_x=y \Rightarrow 50xy=x^3y+6xy^3$$
subtracting these equations ...
$$2xy(x^2-2y^2)=0$$
this has 4 solutions: $x=0$ and $y=0$ are both clearly minima $x=-\sqrt 2 y$ is not in the domain
so go with the fourth solution $x=\sqrt 2 y$ and use this to solve $V_x = 0$.