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\textbf{Applications of Partial Differentiation}

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

Find the the dimensions of an open top rectangular box. It is given
that the box has volume $V$ and the five faces give the box the least
possible surface area.


\textbf{Answer}

Let the length, width and height of the box be $x$, $y$ and $z$
respectively. Then $V=xyz$. The total surface area of the bottom and
sides is
\begin{eqnarray*}
S & = & xy+2xz+2yz = xy+2(x+y)\frac{V}{xy}\\
& = & xy + \frac{2V}{x}+ \frac{2V}{y}
\end{eqnarray*}
where $x>0$ and $y>0$. Since $S\to\infty$ as $x\to 0+$ or $y\to 0+$ or
$x^2+y^2\ to \infty$, $S$ must have a minimum value at a critical
point in the first quadrant.

For CP:
\begin{eqnarray*}
0 = \frac{\partial S}{\partial x} & = & y-\frac{2V}{x^2}\\
0 = \frac{\partial S}{\partial y} & = & x - \frac{2V}{y^2}
\end{eqnarray*}
Thus 
$x^2y=2V=xy^2$, so that $x=y=(2V)^{1/3}$ and z =$V/(2V)^{2/3} =
2^{-2/3}V^{1/3}$.


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