Single Variable Calculus, Chapter 4, 4.7, Section 4.7, Problem 14

Suppose that a rectangular storage container with an open top is to have a volume of $10m^3$. The length of its base is twice the width. material for the base costs \$10 per square meter, and \$6 per square meter for the sides. Find the cost of material for the cheapest such container.



Let $x,y$ and $z$ be the dimensions of the box.
Volume = $xyz = 10$
Recall that the Surface Area is equal to the sum of the area faces of the box.
Surface Area = $xy + xy + xz + yz + yz$
Surface Area = $2xy + 2yz + xz$

If we want to minimize the over all, we multiply the price per square meter of each faces of the box...
cost = $10(xz)+6(2xy+2yz)$ = minimum
It is stated in the problem that $z = 2x$
Substitute this value to the equation of volume to obtain...

$
\begin{equation}
\begin{aligned}
xyz &= 10\\
xy(2x) &= 10\\
2x^2y &= 10 ; \quad y = \frac{5}{x^2}
\end{aligned}
\end{equation}
$


Again, if we substitute $z =2x$ and $\displaystyle y = \frac{5}{x^2}$ to the cost function, we get...

$
\begin{equation}
\begin{aligned}
\text{cost } = 10 (x(2x)) + 6 \left( 2x \left( \frac{5}{x^2} \right) \right) + 2 \left( 2 \left( \frac{5}{x^2} \right)(2x) \right)
\end{aligned}
\end{equation}
$

If we trace the derivative of the cost function and equate it to zero,

$
\begin{equation}
\begin{aligned}
0 &= 40x - \frac{180}{x^2}\\
\\
\frac{180}{x^2} &= 40x \\
\\
x^3 &= \frac{180}{x^2}\\
\\
x &= \sqrt[3]{\frac{180}{40}}m, \text{ then}
\end{aligned}
\end{equation}
$


So if $\displaystyle x = \sqrt[3]{\frac{180}{40}}$
$\displaystyle z = 2x = 2 \left( \sqrt[3]{\frac{180}{40}} \right) = 2 \sqrt[3]{\frac{180}{40}}m \quad $ and $ \quad \displaystyle y = \frac{5}{x^2} = \frac{5}{\left( \sqrt[3]{\frac{180}{40}} \right)^2} = \frac{5}{\left( \frac{180}{40} \right)^{\frac{2}{3}}} m$
Therefore, the cheapest cost for the material would be...

$
\begin{equation}
\begin{aligned}
\text{cost } &= 10
\left[

\left( \sqrt[3]{\frac{180}{40}} \right)


\left( 2\sqrt[3]{\frac{180}{40}} \right)

\right]
\left[
2 \left( \sqrt[3]{\frac{180}{40}} \right)

\left( \frac{5}{\left(\frac{180}{40}\right)^{2/3}} \right)

+2 \left( \frac{5}{\left(\frac{180}{40}\right)^{2/3}} \right)

\left( 2 \left( \sqrt[3]{\frac{180}{40}} \right) \right)

\right]
\\
\\
\text{cost } &= \$163.54
\end{aligned}
\end{equation}
$