College Algebra, Chapter 1, 1.5, Section 1.5, Problem 82

Suppose that a large plywood box has a volume of $180 ft^3$. Its length is $9 ft$ greater than its height, and its width is $4ft$ less than its height. Find the dimensions of the box.

Recall that the volume of the box is


$\begin{equation} \begin{aligned} V =& L x W x H, && \\ \\ \text{ where } L =& 9 + W && \\ \\ W =& H - 4 && \\ \\ V =& (9 + W)(H - 4)(H) && \text{Substitute } W = H - 4 \\ \\ V =& (9 + (H - 4))(H - 4)(H) && \\ \\ V =& (H + 5)(H - 4)(H) && \text{Expand} \\ \\ V =& (H^2 + 5H - 20)H && \text{Substitute the given} \\ \\ 180 =& H^3 + 5H^2 - 20H && \text{Subtract } 180 \\ \\ 0 =& H^3 + 5H^2 - 20H - 180 && \text{Factor} \\ \\ 0 =& (H + 6)(H - 6)(H + 5) && \text{Zero Product Property} \\ \\ H + 6 =& 0, H - 6 = 0 \text{ and } H + 5 = 0 && \text{Solve for } H \\ \\ H =& -6, H = 6 \text{ and } H = -5 && \text{Choose } H > 0 \\ \\ \text{So if } H =& 6 ft, \text{ then} \\ \\ W =& H - 4 = 6- 4 = 2 ft \\ \\ L =& 9 + W = 9 + 2 = 11 ft \end{aligned} \end{equation}$