Answer
The dimensions of the box of maximum volume are $\left( {\frac{{2a}}{{\sqrt 3 }},\frac{{2b}}{{\sqrt 3 }},\frac{{2c}}{{\sqrt 3 }}} \right)$.
Work Step by Step
Referring to Figure 6, let $ \pm x$, $ \pm y$ and $ \pm z$ denote the vertices of the box. So, the volume of the box is given by $f\left( {x,y,z} \right) = \left( {2x} \right)\left( {2y} \right)\left( {2z} \right) = 8xyz$.
Our task is to maximize the function $f\left( {x,y,z} \right)$ subject to the constraint $g\left( {x,y} \right) = {\left( {\frac{x}{a}} \right)^2} + {\left( {\frac{y}{b}} \right)^2} + {\left( {\frac{z}{c}} \right)^2} - 1 = 0$.
Step 1. Write out the Lagrange equations
Using Theorem 1, the Lagrange condition $\nabla f = \lambda \nabla g$ yields
$\left( {8yz,8xz,8xy} \right) = \lambda \left( {\frac{{2x}}{{{a^2}}},\frac{{2y}}{{{b^2}}},\frac{{2z}}{{{c^2}}}} \right)$
So, the Lagrange equations are
(1) ${\ \ \ }$ $8yz = \frac{{2\lambda x}}{{{a^2}}}$, ${\ \ }$ $8xz = \frac{{2\lambda y}}{{{b^2}}}$, ${\ \ }$ $8xy = \frac{{2\lambda z}}{{{c^2}}}$
Step 2. Solve for $\lambda$ in terms of $x$ and $y$
Since $\left( {0,0,0} \right)$ does not satisfy the constraint, we have the following cases by examining equation (1):
Case 1. $x=0$, $y=0$, $z \ne 0$
Substituting $x=0$, $y=0$ in the constraint gives
$\frac{{{z^2}}}{{{c^2}}} = 1$
So, $z = \pm c$.
The critical points are $\left( {0,0,c} \right)$ and $\left( {0,0, - c} \right)$.
Case 2. $x=0$, $y \ne 0$, $z=0$
Substituting $x=0$, $z=0$ in the constraint gives
$\frac{{{y^2}}}{{{b^2}}} = 1$
So, $y = \pm b$.
The critical points are $\left( {0,b,0} \right)$ and $\left( {0, - b,0} \right)$.
Case 3. $x \ne 0$, $y=0$, $z=0$
Substituting $y=0$, $z=0$ in the constraint gives
$\frac{{{x^2}}}{{{a^2}}} = 1$
So, $x = \pm a$.
The critical points are $\left( {a,0,0} \right)$ and $\left( { - a,0,0} \right)$.
Case 4. $x \ne 0$, $y \ne 0$, $z \ne 0$
From equation (1), we obtain
(2) ${\ \ \ \ }$ $\lambda = \frac{{4yz{a^2}}}{x} = \frac{{4xz{b^2}}}{y} = \frac{{4xy{c^2}}}{z}$
Step 3. Solve for $x$ and $y$ using the constraint
From equation (2) we obtain
$\frac{{4yz{a^2}}}{x} = \frac{{4xz{b^2}}}{y}$, ${\ \ \ }$ $\frac{{4yz{a^2}}}{x} = \frac{{4xy{c^2}}}{z}$
${y^2} = \frac{{{b^2}}}{{{a^2}}}{x^2}$, ${\ \ \ \ }$ ${z^2} = \frac{{{c^2}}}{{{a^2}}}{x^2}$
So, $y = \pm \frac{b}{a}x$ and $z = \pm \frac{c}{a}x$.
Substituting these in the constraint gives
$\frac{{{x^2}}}{{{a^2}}} + \frac{{{x^2}}}{{{a^2}}} + \frac{{{x^2}}}{{{a^2}}} - 1 = 0$
$\frac{{3{x^2}}}{{{a^2}}} = 1$
So, $x = \pm \frac{a}{{\sqrt 3 }}$.
Using $y = \pm \frac{b}{a}x$ and $z = \pm \frac{c}{a}x$, we obtain the critical points: $\left( {\frac{a}{{\sqrt 3 }},\frac{b}{{\sqrt 3 }},\frac{c}{{\sqrt 3 }}} \right)$, $\left( {\frac{a}{{\sqrt 3 }}, - \frac{b}{{\sqrt 3 }},\frac{c}{{\sqrt 3 }}} \right)$, $\left( {\frac{a}{{\sqrt 3 }},\frac{b}{{\sqrt 3 }}, - \frac{c}{{\sqrt 3 }}} \right)$, $\left( {\frac{a}{{\sqrt 3 }}, - \frac{b}{{\sqrt 3 }}, - \frac{c}{{\sqrt 3 }}} \right)$, $\left( { - \frac{a}{{\sqrt 3 }},\frac{b}{{\sqrt 3 }},\frac{c}{{\sqrt 3 }}} \right)$, $\left( { - \frac{a}{{\sqrt 3 }}, - \frac{b}{{\sqrt 3 }},\frac{c}{{\sqrt 3 }}} \right)$, $\left( { - \frac{a}{{\sqrt 3 }},\frac{b}{{\sqrt 3 }}, - \frac{c}{{\sqrt 3 }}} \right)$, $\left( { - \frac{a}{{\sqrt 3 }}, - \frac{b}{{\sqrt 3 }}, - \frac{c}{{\sqrt 3 }}} \right)$.
Step 4. Calculate the critical values
We evaluate the extreme values and list them in the following table:
$\begin{array}{l}
\begin{array}{*{20}{c}}
{{\rm{Critical{\ }point}}}&{f\left( {x,y,z} \right)}\\
{\left( {0,0,c} \right)}&0\\
{\left( {0,0, - c} \right)}&0\\
{\left( {0,b,0} \right)}&0\\
{\left( {0, - b,0} \right)}&0\\
{\left( {a,0,0} \right)}&0\\
{\left( { - a,0,0} \right)}&0\\
{\left( {\frac{a}{{\sqrt 3 }},\frac{b}{{\sqrt 3 }},\frac{c}{{\sqrt 3 }}} \right)}&{\frac{{8abc}}{{3\sqrt 3 }}}\\
{\left( {\frac{a}{{\sqrt 3 }}, - \frac{b}{{\sqrt 3 }},\frac{c}{{\sqrt 3 }}} \right)}&{ - \frac{{8abc}}{{3\sqrt 3 }}}
\end{array}\\
\begin{array}{*{20}{c}}
{\left( {\frac{a}{{\sqrt 3 }},\frac{b}{{\sqrt 3 }}, - \frac{c}{{\sqrt 3 }}} \right)}&{ - \frac{{8abc}}{{3\sqrt 3 }}}\\
{\left( {\frac{a}{{\sqrt 3 }}, - \frac{b}{{\sqrt 3 }}, - \frac{c}{{\sqrt 3 }}} \right)}&{\frac{{8abc}}{{3\sqrt 3 }}}\\
{\left( { - \frac{a}{{\sqrt 3 }},\frac{b}{{\sqrt 3 }},\frac{c}{{\sqrt 3 }}} \right)}&{ - \frac{{8abc}}{{3\sqrt 3 }}}\\
{\left( { - \frac{a}{{\sqrt 3 }}, - \frac{b}{{\sqrt 3 }},\frac{c}{{\sqrt 3 }}} \right)}&{\frac{{8abc}}{{3\sqrt 3 }}}
\end{array}\\
\begin{array}{*{20}{c}}
{\left( { - \frac{a}{{\sqrt 3 }},\frac{b}{{\sqrt 3 }}, - \frac{c}{{\sqrt 3 }}} \right)}&{\frac{{8abc}}{{3\sqrt 3 }}}\\
{\left( { - \frac{a}{{\sqrt 3 }}, - \frac{b}{{\sqrt 3 }}, - \frac{c}{{\sqrt 3 }}} \right)}&{ - \frac{{8abc}}{{3\sqrt 3 }}}
\end{array}
\end{array}$
From the results in this table, we conclude that the maximum volume of a box inscribed in the ellipsoid is $\frac{{8abc}}{{3\sqrt 3 }}$, where the vertices of the box are $\left( { \pm x, \pm y, \pm z} \right) = \left( { \pm \frac{a}{{\sqrt 3 }}, \pm \frac{b}{{\sqrt 3 }}, \pm \frac{c}{{\sqrt 3 }}} \right)$. Hence, the dimensions of the box of maximum volume are $\left( {\frac{{2a}}{{\sqrt 3 }},\frac{{2b}}{{\sqrt 3 }},\frac{{2c}}{{\sqrt 3 }}} \right)$.
As an illustration, please see the figure attached.
