Answer
The maximum volume is $V\left( {1,\frac{3}{2}} \right) = \frac{3}{4}$.
The size of the box when the volume is maximum:
$\left( {x,y,z} \right) = \left( {1,\frac{3}{2},\frac{1}{2}} \right)$.
Work Step by Step
Step 1. Determine the function to be maximized
For $x,y,z \ge 0$, let $P = \left( {x,y,z} \right)$ be the corner of the box touching the paraboloid (see the figure attached). Thus, the the box has volume $V = xyz$.
Using $z = 1 - \frac{{{x^2}}}{4} - \frac{{{y^2}}}{9}$ we obtain the volume as a function of $x$ and $y$:
$V\left( {x,y} \right) = xy\left( {1 - \frac{{{x^2}}}{4} - \frac{{{y^2}}}{9}} \right) = xy - \frac{{{x^3}y}}{4} - \frac{{x{y^3}}}{9}$
Let the domain $D$ be the region $OAB$ in the $xy$-plane. So,
$D = \left\{ {\left( {x,y} \right)|x,y \ge 0,\frac{{{x^2}}}{4} + \frac{{{y^2}}}{9} \le 1} \right\}$
Since $D$ is closed and bounded, by Theorem 3, the maximum value occurs either at the critical points in the interior of $D$ or at points on the boundary of $D$.
Step 2. Examine the critical points
We have $V\left( {x,y} \right) = xy - \frac{{{x^3}y}}{4} - \frac{{x{y^3}}}{9}$ on the domain $D = \left\{ {\left( {x,y} \right)|x,y \ge 0,\frac{{{x^2}}}{4} + \frac{{{y^2}}}{9} \le 1} \right\}$.
The partial derivatives are
${V_x} = y - \frac{{3{x^2}y}}{4} - \frac{{{y^3}}}{9}$
${V_y} = x - \frac{{{x^3}}}{4} - \frac{{x{y^2}}}{3}$
To find the critical points we solve the equations ${V_x} = 0$ and ${V_y} = 0$:
${V_x} = y - \frac{{3{x^2}y}}{4} - \frac{{{y^3}}}{9} = 0$
${V_y} = x - \frac{{{x^3}}}{4} - \frac{{x{y^2}}}{3} = 0$
We get
$y\left( {1 - \frac{{3{x^2}}}{4} - \frac{{{y^2}}}{9}} \right) = 0$
$x\left( {1 - \frac{{{x^2}}}{4} - \frac{{{y^2}}}{3}} \right) = 0$
From the first equation we get $y=0$. Substituting it in the second equation gives $x = \pm 2$.
From the second equation we get $x=0$. Substituting it in the first equation gives $y = \pm 3$.
From the set of equations: $1 - \frac{{3{x^2}}}{4} - \frac{{{y^2}}}{9} = 0$ and $1 - \frac{{{x^2}}}{4} - \frac{{{y^2}}}{3} = 0$ we obtain the solution $\left( {1,\frac{3}{2}} \right)$.
So, the critical points that are included in $D$ are $\left( {0,0} \right)$, $\left( {2,0} \right)$, $\left( {0,3} \right)$ and $\left( {1,\frac{3}{2}} \right)$. The corresponding extreme values are
$\begin{array}{*{20}{c}}
{{\rm{Critical{\ }points}}}&{V\left( {x,y} \right)}\\
{\left( {0,0} \right)}&0\\
{\left( {2,0} \right)}&0\\
{\left( {0,3} \right)}&0\\
{\left( {1,\frac{3}{2}} \right)}&{\frac{3}{4}}
\end{array}$
Step 2. Check the boundaries
We restrict the volume $V$ along the edges and find the maximum values. The results are given in the following table:
$\begin{array}{*{20}{c}}
{}&{{\rm{Restriction{\ }of}}}\\
{{\rm{Edge}}}&{V\left( {x,y} \right){\rm{to{\ }Edge}}}\\
{Bottom:0 \le x \le 2,y = 0}&{h\left( x \right) = V\left( {x,0} \right) = 0}\\
{Left:x = 0,0 \le y \le 3}&{m\left( y \right) = V\left( {0,y} \right) = 0}\\
{Right:0 \le x \le 2,y = 3\sqrt {1 - {x^2}/4} }&{n\left( x \right) = V\left( {x,3\sqrt {1 - {x^2}/4} } \right) = 0}
\end{array}$
So, we have $V=0$ for all points on the boundary of $D$.
Step 3. Compare the results
Comparing the results from Step 1 and Step 2 we obtain the maximum volume of $V\left( {1,\frac{3}{2}} \right) = \frac{3}{4}$.
Using $z = 1 - \frac{{{x^2}}}{4} - \frac{{{y^2}}}{9}$, we obtain the size of the box when the volume is maximum: $\left( {x,y,z} \right) = \left( {1,\frac{3}{2},\frac{1}{2}} \right)$.
