Question

A closed box has a square base with side length $\mathrm{L}\mathrm{feet}$ and height $\mathrm{h}\mathrm{feet}$.

Given that the volume of the box is $39\mathrm{cubic}\mathrm{feet}$, express the surface area of the box in terms of $\mathrm{L}$ only.

Answer 1

Step 1: Determine the height of the box:

The area of the square $A$ is calculated $A=s^2$ where, $s$ represents the side of the square.

Here, the side of the square is $\mathrm{L}$.

The area of square $A$ is ${\mathrm{L}}^2$.

The volume of the box is $39\mathrm{cubic}\mathrm{feet}$.

The volume of the box $V$ is calculated by $\mathrm{V}=\mathrm{A}·\mathrm{h}$.

$\begin{array}{rcl}39& =& {\mathrm{L}}^2·\mathrm{h}\\ & \Rightarrow & \mathrm{h}=\frac{39}{{\mathrm{L}}^2}\end{array}$

Step 2: Determine the surface area of the box in terms of $\mathbf{L}$:

The surface area of the box is calculated by $\mathrm{S}=2\left(\mathrm{lb}+\mathrm{bh}+\mathrm{lh}\right)$.

Where $\mathrm{l}$ represents the length of the box, $b$ represents the breadth of the box and $h$ represents the height of the box.

$\begin{array}{rcl}\mathrm{S}& =& 2\left(L\times L+L·\frac{39}{L^2}+L·\frac{39}{L^2}\right)\\ & =& 2\left(L^2+\frac{39}{L}+\frac{39}{L}\right)\\ & =& 2\left(L^2+\frac{78}{L}\right)\\ & =& \left(2L^2+\frac{156}{L}\right)f{t}^2\end{array}$

Hence, the surface area of the box in terms of $\mathbf{L}$ is $\mathbf{(}\mathbf{2}{\mathbf{L}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{156}}{\mathbf{L}}\mathbf{)}\mathit{f}{\mathit{t}}^{\mathbf{2}}$.