College Algebra, Exercise P, Exercise P.1, Section Exercise P.1, Problem 24

If a fruit crate has square ends and is twice as long as it is wide.

a.) Find the volume of the crate if its width is 20 inches.
Recall that the volume of a box is $V = L \times W \times H$. Since the box has a square ends, its width and height are equal. So,

$\begin{equation} \begin{aligned} V &= L \times W \times H \text{ ,where } W = H \text{ so}\\ \\ V &= L \times W \times W\\ \\ V &= L \times W^2 \text{ ,also from the condition } L = 2W \text{ so}\\ \\ V &= 2W \times W^2\\ \\ V &= 2W^3 && \text{model}\\ \\ V &= 2(20 \text{in})^3 && \text{Substitute } W = 20\text{in}\\ \\ V &= 16,000 \text{in}^3 && \text{Volume of the crate} \end{aligned} \end{equation}$


b.) Find a formula for the volume $V$ of the crate in terms of its width $x$.
From the model in part(a), we solve $V= 2W^3$ for $W = x$, so

$\begin{equation} \begin{aligned} \frac{V}{2} &= \frac{\cancel{2}x^3}{\cancel{2}} && \text{Divide both sides by 2}\\ \\ \sqrt[3]{\frac{V}{2}} &= \sqrt[3]{x^3} && \text{Take the cube root of both sides}\\ \\ \sqrt[3]{\frac{V}{2}} &= x && \text{model}\\ \\ & \text{or}\\ \\ x &= \sqrt[3]{\frac{V}{2}} \end{aligned} \end{equation}$