Question

A rectangular tank with a square base, an open top, and a volume of $4000{\mathrm{ft}}^3$ is to be constructed of sheet steel.

Find the dimensions of the tank that has the minimum surface area.

The dimensions of the tank with minimum surface area are ____?? ft.

Answer 1

Find the dimensions of the tank with the minimum surface area:

Let $r$ be the length, $r$ be the width, and $h$ be the height of the rectangular tank with a square base.

The formula for the volume is,

$\begin{array}{rcl}V& =& \mathrm{length}\times \mathrm{width}\times \mathrm{height}\\ & \Rightarrow & V=r\times r\times h\\ \therefore V& =& r^{2}h{\mathrm{ft}}^3\end{array}$

Since the volume of the box is $V=4000{\mathrm{ft}}^3$. So,

$\begin{array}{rcl}{r}^{2}h& =& 4000\\ & \Rightarrow & h=\frac{4000}{r^2}\end{array}$

The formula for the surface area of the open-top box:

$\begin{array}{rcl}A& =& \left(\mathrm{length}\times \mathrm{width}\right)+2\left(\mathrm{length}\times \mathrm{height}+\mathrm{width}\times \mathrm{height}\right)\\ & \Rightarrow & A\mathit{=}\mathit{(}\mathit{r}\mathit{\times }\mathit{r}\mathit{)}\mathit{+}\mathit{2}\mathit{(}\mathit{r}\mathit{\times }\mathit{h}\mathit{+}\mathit{r}\mathit{\times }\mathit{h}\mathit{)}\\ & \mathit{\Rightarrow }& \mathit{}\mathit{}\mathit{}A\mathit{=}{r}^{\mathit{2}}\mathit{+}\mathit{2}\mathit{\left(}\mathit{2}rh\mathit{\right)}\\ \mathit{\therefore }\mathit{}\mathit{}\mathit{}A& \mathit{=}& r^{\mathit{2}}\mathit{+}\mathit{4}rh\end{array}$

Substitute $\begin{array}{rcl}h& =& \frac{4000}{r^2}\end{array}$in the surface area:

$\begin{array}{rcl}A& =& r^2+4rh\\ & \Rightarrow & A=r^2+4r\left(\frac{4000}{r^2}\right)\\ & \Rightarrow & A=r^2+4\left(\frac{4000}{r}\right)\\ \therefore A& =& r^2+\frac{16000}{r}\end{array}$

Step-1: Find the length of the tank.

Differentiate the surface area with respect to $r$:

$\begin{array}{rcl}\frac{dA}{dr}& =& \frac{d}{dr}\left(r^2+\frac{16000}{r}\right)\\ & \Rightarrow & \frac{dA}{dr}=\frac{d}{dr}\left(r^2\right)+\frac{d}{dr}\left(\frac{16000}{r}\right)\\ & \Rightarrow & \frac{dA}{dr}=\left(2r\right)+\left(16000\right)\left(-r^{-1-1}\right)\\ \therefore \frac{dA}{dr}& =& 2r-\frac{16000}{r^2}\end{array}$

To minimize the surface area, so $\frac{dA}{dr}=0$

$\begin{array}{rcl}2r-\frac{16000}{r^2}& =& 0\\ & \Rightarrow & 2r=\frac{16000}{r^2}\\ & \Rightarrow & 2r^3=16000\\ & \Rightarrow & r^3=\frac{16000}{2}\\ & \Rightarrow & r^3=8000\\ & \Rightarrow & r=\sqrt[3]{8000}\\ \therefore r& =& 20\mathrm{ft}\end{array}$

Step-2: Find the width of the tank.

Again differentiate the surface area:

$\begin{array}{rcl}\frac{d^{2}A}{d{r}^2}& =& \frac{d}{dr}\left(2r-\frac{16000}{r^2}\right)\\ & \Rightarrow & \frac{d^{2}A}{d{r}^2}=2-16000\left(-\frac{2}{r^3}\right)\\ \therefore \frac{d^{2}A}{d{r}^2}& =& 2+\frac{32000}{r^3}\end{array}$

At $r=20\mathrm{ft}$,

$\begin{array}{rcl}\frac{d^{2}A}{d{r}^2}& =& 2+\frac{32000}{{20}^3}\\ & \Rightarrow & \frac{d^{2}A}{d{r}^2}=2+\frac{32000}{8000}\\ & \Rightarrow & \frac{d^{2}A}{d{r}^2}=2+4\\ & \Rightarrow & \frac{d^{2}A}{d{r}^2}=6>0\end{array}$

It follows that, $r=20\mathrm{ft}$ the surface area is minimum.

Step-3: Find the height of the tank.

Substitute $r=20\mathrm{ft}$ in $\begin{array}{rcl}h& =& \frac{4000}{r^2}\end{array}$:

$\begin{array}{rcl}h& =& \frac{4000}{{20}^2}\\ & \Rightarrow & h=\frac{4000}{400}\\ \therefore h& =& 10\mathrm{ft}\end{array}$

Therefore, the dimensions of the rectangular tank will be $\mathbf{20}\mathbf{}\mathbf{ft}\mathbf{\times }\mathbf{20}\mathbf{}\mathbf{ft}\mathbf{\times }\mathbf{10}\mathbf{}\mathbf{ft}$.