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\textbf{Applications of Partial Differentiation}

\textit{\textbf{Extremes}}
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\textbf{Question}

Given the ellipsoid $$\frac{x^2}{a^2} + \frac{y^2}{b^2}+
\frac{z^2}{c^2},$$ find the volume of the largest rectangular box that
can be placed inside. (The faces must be parallel to the coordinate
planes.)


\textbf{Answer}

Let $(x,y,z)$ be the coordinates of the corner of the box that is the
first octant of space.

Thus $x, \ y, \ z \ge 0$ and $$\frac{x^2}{a^2}+\frac{y^2}{b^2}
+\frac{z^2}{c^2} = 1.$$
The volume of the box is $$V =(2x)(2y)(2z)=8cxy\sqrt{1-\frac{x^2}{a^2}
- \frac{y^2}{b^2}}$$
for $x \ge 0$, $y \ge 0$ and $(x^2/a^2) + (y^2/b^2) = 1$, the maximum
value of $V^2$ and hence of $V$, will occur at a critical point of
$V^2$ where $x>0$ and $y>0$.

For CP:
\begin{eqnarray*}
0 = \frac{\partial V^2}{\partial x} & = & 64c^2 \left ( 2xy^2 -
\frac{4x^3y^2}{a^2} -\frac{2xy^4}{b^2} \right )\\
& = & 128c^2xy^2\left ( 1 - \frac{2x^2}{a^2} -\frac{y^2}{b^2} \right
)\\
0 = \frac{\partial V^2}{\partial y} & = & 128c^2x^2y \left ( 1 -
\frac{x^2}{a^2} -\frac{2y^2}{b^2} \right )
\end{eqnarray*}
Hence we must have
$$\frac{2c^2}{a^2} + \frac{y^2}{b^2} = 1 = \frac{x^2}{a^2}+
\frac{2y^2}{b^2}.$$
so that $x^2/a^2=y^2/b^2=1/3$ and $x=a/\sqrt{3}$, $y=b/\sqrt{3}$.

The largest box has volume
$$V=\frac{8abc}{3}\sqrt{1-\frac{1}{3}-
\frac{1}{3}}=\frac{8abc}{3\sqrt{3}} \ textrm{cubic units}.$$

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