Inside an elliptical paraboloid with an equation $z=\frac{x^2}{a^2}+ \frac{y^2}{b^2}$ bounded by $z=h$ draw an right-angle parallelepiped.. Inside an elliptical paraboloid with an equation $z=\frac{x^2}{a^2}+ \frac{y^2}{b^2}$ bounded by $z=h$ draw an right-angle parallelepiped.. with the largest possible volume. What confuses my most is this elliptical paraboloid. 
I know what it looks like in theory but I have no clue what $a,b$ determine in it's equation, or if the curve is facing $x,y$ plane? I am pretty sure this has to be done using Lagrange coefficients, conditional extremes, or global extremes at the very least. 
The problem is I don't know how to relate all that is given here to make a quantifying expression,function who's extreme I am to find. I also studied the volume of an n-dimensional parallelepiped using the determinant of vectors formula, but I don't think that this will be needed here, but just in case.. Also, I don't know how to find the volume of this elliptical paraboloid because of my lack of defectiveness of it's exact appearance. Help is very much appreciated.
 A: An elliptic paraboloid is a curve whose intersection with a plane parallel to the $z$ plane is an ellipse and whose intersection with a plane parallel to the $x$ or $y$ plane is a parabola.
The meanings of $a$ and $b$ are related to the ellipse. If you intersect the paraboloid with the plane $z=1$ you get an ellipse whose semimajor and semiminor axes are $a$ and $b$.
It seems obvious that the desired box (a.k.a. right-angle parallelepiped or rectangular prism) will have one face on the plane $z=h$ and the opposite face with all four vertices on the paraboloid. (If a vertex is not on the paraboloid you can extend the box until the vertex is on the surface, with the new box larger than the previous one.) The paraboloid is very symmetric, so if one vertex of that face is $(x,y,h)$ then so are $(-x,y,h)$, $(x,-y,h)$, and $(-x,-y,h)$. That means that the dimensions of the face are from $-x$ to $x$, and $-y$ to $y$. The third, $z$, dimension goes from the paraboloid to the plane $z=h$, i.e. from $\frac{x^2}{a^2}+ \frac{y^2}{b^2}$ to $h$. Therefore the volume of the box is
$$V=2x\cdot 2y\cdot\left(h-\left[\frac{x^2}{a^2}+ \frac{y^2}{b^2}\right]\right)$$
Next you should find the limits on $x$ and on $y$, which is pretty easy. Now find the maximum of that expression for $V$ on the two variables $x$ and $y$ given those limits. That is a two-variable optimization, which you can probably do.

To avoid fractions, let's do the linear change of variables $r=\frac xa$ and $s=\frac yb$. We then have the restriction
$$h\ge r^2+s^2$$
so our limits on the independent variables are then $0\le r\le \sqrt h$, $0\le s\le \sqrt h$.  We are maximizing
$$\begin{align}
V &= 2ar\cdot 2bs\cdot(h-r^2-s^2) \\
  &= (4ab)(hrs-r^3s-rs^3)
\end{align}$$
To simplify more, let's define $W=\frac V{4ab}$, so we are maximizing
$$W=hrs-r^3s-rs^3$$
We can easily see that at any of the limits for $r$ and $s$ we get $W=0$, so the maximum will be at a stable point of $W$. We that that stable point in the usual way:
$$\begin{align}
\frac{\partial W}{\partial r} &= hs-3r^2s-s^3 \\
  &= s(h-3r^2-s^2) \\
\frac{\partial W}{\partial s} &= hr-r^3-3rs^2 \\
  &= r(h-r^2-3s^2) \\
\end{align}$$
Setting those equal to zero and removing the non-zero $r$ and $s$ gives us the simultaneous equations
$$3r^2+s^2=h, \qquad r^2+3s^2=h$$
Subtracting three times the second equation from the first leads gives
$$8r^2=2h$$
which means
$$r=\frac{\sqrt h}2$$
Subtracting three times the first equation from the second leads gives
$$8s^2=2h$$
which means
$$s=\frac{\sqrt h}2$$
Changing back to the original variables gives the maximum volume at

$$x=\frac{a\sqrt h}2, \qquad y=\frac{b\sqrt h}2$$

which gives the maximum volume of

$$V=\frac{ab}2h^2$$

Comparing the limits on $r$ and $s$ with their limits (or $x$ and $y$ with their limits) show us that the base of the maximum-volume box has sides that equal the semimajor and semiminor axes of the ellipse formed by intersecting the elliptic paraboloid with the plane $z=h$. Calculating the third dimension $z$ shows that it is half the height of the paraboloid section, $h/2$. That makes it easy to draw, which is the point of the given question.
