Answer
The maximum volume is $V\left( {\frac{1}{3},\frac{2}{3}} \right) = \frac{2}{9}$.
The size of the box when the volume is maximum:
$\left( {x,y,z} \right) = \left( {\frac{1}{3},\frac{2}{3},1} \right)$.
Work Step by Step
Step 1. Determine the function to be maximized
Let $P = \left( {x,y,z} \right)$ be the corner of the box touching the front face of the tetrahedron (see the figure attached). Thus, the the box has volume $V = xyz$.
Using $x + \frac{1}{2}y + \frac{1}{3}z = 1$ we obtain the volume as a function of $x$ and $y$:
$V\left( {x,y} \right) = xy\left( {3 - 3x - \frac{3}{2}y} \right) = - 3{x^2}y + 3xy - \frac{3}{2}x{y^2}$
Let the domain $D$ be the triangle $\vartriangle OAB$ in the $xy$-plane. So,
$D = \left\{ {\left( {x,y} \right)|x,y \ge 0,x + \frac{1}{2}y \le 1} \right\}$
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 $V\left( {x,y} \right) = - 3{x^2}y + 3xy - \frac{3}{2}x{y^2}$ on the domain $D = \left\{ {\left( {x,y} \right)|x,y \ge 0,x + \frac{1}{2}y \le 1} \right\}$.
The partial derivatives are
${V_x} = - 6xy + 3y - \frac{3}{2}{y^2}$
${V_y} = - 3{x^2} + 3x - 3xy$
To find the critical points we solve the equations ${V_x} = 0$ and ${V_y} = 0$:
${V_x} = - 6xy + 3y - \frac{3}{2}{y^2} = 0$
${V_y} = - 3{x^2} + 3x - 3xy = 0$
We get
$y\left( { - 6x + 3 - \frac{3}{2}y} \right) = 0$
$x\left( { - 3x + 3 - 3y} \right) = 0$
From the first equation we get $y=0$. Substituting it in the second equation gives $x=1$.
From the second equation we get $x=0$. Substituting it in the first equation gives $y=2$.
From the set of equations: $ - 6x + 3 - \frac{3}{2}y = 0$ and $ - 3x + 3 - 3y = 0$ we obtain $x = \frac{1}{3}$ and $y = \frac{2}{3}$.
So, the critical points are $\left( {0,0} \right)$, $\left( {1,0} \right)$, $\left( {0,2} \right)$ and $\left( {\frac{1}{3},\frac{2}{3}} \right)$. The corresponding extreme values are
$\begin{array}{*{20}{c}}
{{\rm{Critical{\ }points}}}&{V\left( {x,y} \right)}\\
{\left( {0,0} \right)}&0\\
{\left( {1,0} \right)}&0\\
{\left( {0,2} \right)}&0\\
{\left( {\frac{1}{3},\frac{2}{3}} \right)}&{\frac{2}{9}}
\end{array}$
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:
$\begin{array}{*{20}{c}}
{}&{{\rm{Restriction{\ }of}}}\\
{{\rm{Edge}}}&{V\left( {x,y} \right){\rm{to{\ }Edge}}}\\
{Bottom:0 \le x \le 1,y = 0}&{h\left( x \right) = V\left( {x,0} \right) = 0}\\
{Left:x = 0,0 \le y \le 2}&{m\left( y \right) = V\left( {0,y} \right) = 0}\\
{Right:0 \le x \le 1,y = 2 - 2x}&{n\left( x \right) = V\left( {x,2 - 2x} \right) = 0}
\end{array}$
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 $V\left( {\frac{1}{3},\frac{2}{3}} \right) = \frac{2}{9}$.
Using $x + \frac{1}{2}y + \frac{1}{3}z = 1$, we obtain the size of the box when the volume is maximum: $\left( {x,y,z} \right) = \left( {\frac{1}{3},\frac{2}{3},1} \right)$.
