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...

$\begin{equation} \begin{aligned} \text{cost } &= 10 \left[ \left( \sqrt[3]{\frac{180}{40}} \right) \left( 2\sqrt[3]{\frac{180}{40}} \right) \right] \left[ 2 \left( \sqrt[3]{\frac{180}{40}} \right) \left( \frac{5}{\left(\frac{180}{40}\right)^{2/3}} \right) +2 \left( \frac{5}{\left(\frac{180}{40}\right)^{2/3}} \right) \left( 2 \left( \sqrt[3]{\frac{180}{40}} \right) \right) \right] \\ \\ \text{cost } &= \$163.54 \end{aligned} \end{equation}$