Question

A cylindrical can is to hold $1$ cubic inches of orange juice.

The cost per square inch of constructing the metal top and bottom is four times the cost per square inch of constructing the cardboard side.

What are the dimensions of the least expensive can?

Answer 1

To find the dimensions:

The capacity of the cylindrical can is $1$ cubic inches.

Let $x$ be the radius and $y$ be the height of the cylindrical can.

Using the volume(capacity) formula of cylinder in below :

$\begin{array}{rcl}V& =& \mathrm{\pi }{x}^{2}y\left[\text{Volume of cylinder}\right]\\ & \Rightarrow & 1=\mathrm{\pi }{x}^{2}y\left[\text{Volume}\left(\text{capacity}\right)=1{\text{inches}}^3\right]\\ \therefore y& =& \frac{1}{\mathrm{\pi }{x}^2}...\left(1\right)\left[\text{Solve for}y\right]\end{array}$

Now, using the formula of total surface area of the cylinder :

$\begin{array}{rcl}\text{S}& =& {\mathrm{\pi x}}^2+{\mathrm{\pi x}}^2+2\mathrm{\pi xy}\left[\mathrm{Total}\mathrm{surface}\mathrm{area}\right]\end{array}$

Since the cost per square inch of constructing the metal top and bottom is four times the cost per square inch of constructing the cardboard side, so the total cost is :

$\begin{array}{rcl}\text{Cost(C)}& =& {\mathrm{\pi x}}^2+{\mathrm{\pi x}}^2+4\left(2\mathrm{\pi xy}\right)\left[\mathrm{Total}\text{cost}\right]\\ & \Rightarrow & \text{C}=2{\mathrm{\pi x}}^2+8\mathrm{\pi x}\left(\frac{1}{{\mathrm{\pi x}}^2}\right)\left[\mathrm{Substitute}\mathrm{y}=\frac{1}{{\mathrm{\pi x}}^2}\right]\\ \therefore \text{C}& =& 2{\mathrm{\pi x}}^2+\frac{8}{\mathrm{x}}...\left(2\right)\left[\mathrm{Simplifying}\right]\end{array}$

For the least cost, take derivative and set it equal to zero and solve for $x$ :

$\begin{array}{rcl}\frac{d\text{C}}{dx}& =& \frac{d}{dx}\left(2{\mathrm{\pi x}}^2+\frac{8}{\mathrm{x}}\right)\left[\text{Take derivative}\right]\\ & \Rightarrow & 0=\frac{d}{dx}\left(2{\mathrm{\pi x}}^2\right)+\frac{d}{dx}\left(\frac{8}{\mathrm{x}}\right)\left[\text{Substitute}\frac{d\text{C}}{dx}=0\right]\\ & \Rightarrow & 0=4\mathrm{\pi x}+\left(-\frac{8}{{\mathrm{x}}^2}\right)\left[\text{Simplifying}\right]\\ & \Rightarrow & 0=4\mathrm{\pi x}-\frac{8}{{\mathrm{x}}^2}\\ & \Rightarrow & \frac{8}{{\mathrm{x}}^2}=4\mathrm{\pi x}\\ & \Rightarrow & x^3=\frac{2}{\mathrm{\pi }}\\ \therefore x& \approx & 0.86\end{array}$

Now, substitute $x=0.86$ in equation $\left(1\right)$ :

$\begin{array}{rcl}y& =& \frac{1}{\mathrm{\pi }{\left(0.86\right)}^2}\\ \therefore y& \approx & 0.43\end{array}$

Therefore, the radius and the height of the can is about $0.86\text{and}0.43$, respectively.

Hence, the dimensions are $\mathbf{0}\mathbf{.}\mathbf{86}\text{and}\mathbf{0}\mathbf{.}\mathbf{43}$.