Question

A rectangular tank with a square​ base, an open-top, and a volume of $864$ ft cubed is to be constructed of sheet steel. Find the dimensions of the tank that has the minimum surface area. Let $\mathit{\text{s}}$ be the length of one of the sides of the square base and let $\mathit{\text{A}}$ be the surface area of the tank. Write the objective function.

Answer 1

Find the objective function.

Let, $s$ denotes the side of the square base.

$h$ denotes the height of the tank.

$A$ denotes the surface area of the tank.

$V$ denotes the volume of the tank.

Step 1: The calculation for the value of $\mathit{h}$.

$\begin{array}{rcl}V& =& s^2\times h\\ & \Rightarrow & \frac{V}{s^2}=h\\ & \Rightarrow & h=\frac{V}{s^2}\end{array}$

Step 2: The surface area of the tank.

$\begin{array}{rcl}A& =& \text{Base Area}+4\left(\text{Lateral Area}\right)\\ & \Rightarrow & A=s^2+4\left(s\times h\right)\\ & \Rightarrow & A=s^2+4\left(s\times \frac{V}{s^2}\right)\\ & \Rightarrow & A=s^2+4\left(\frac{V}{s}\right)\\ & \Rightarrow & A=s^2+\frac{4V}{s}\end{array}$

The first derivative of the surface area.

$\begin{array}{rcl}{A}^{\text{'}}& =& 2s-\frac{4V}{s^2}\end{array}$

Substitute the value of $V$:

$\begin{array}{rcl}& \Rightarrow & A^{\text{'}}=2s-\frac{4\left(864\right)}{s^2}\\ & \Rightarrow & A^{\text{'}}=2s-\frac{3456}{s^2}\end{array}$

For minimum surface area, equate the first derivative to zero:

$\begin{array}{rcl}& \Rightarrow & 0=2s-\frac{3456}{s^2}\\ & \Rightarrow & \frac{3456}{s^2}=2s\\ & \Rightarrow & 3456=2s^3\\ & \Rightarrow & \frac{3456}{2}=s^3\\ & \Rightarrow & 1728=s^3\\ & \Rightarrow & s^3=1728\\ & \Rightarrow & s=\sqrt[3]{1728}\\ & \Rightarrow & s=12\end{array}$

Step 3: The calculation for the value $\mathit{h}$.

$\begin{array}{rcl}h& =& \frac{V}{s^2}\\ & \Rightarrow & h=\frac{864}{{12}^2}\\ & \Rightarrow & h=\frac{864}{144}\\ & \Rightarrow & h=6\end{array}$

Hence, the side of the square base is $\mathbf{12}$ ft and the height of the tank is $\mathbf{6}$ ft. The objective function is $\begin{array}{rcl}\mathit{A}& \mathbf{=}& {\mathit{s}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{4}\mathbf{V}}{\mathbf{s}}\end{array}$.