Precalculus (6th Edition) Blitzer

Let us assume $l$ be the length and $w$ be the width of the rectangle such that: \begin{align} & lw=108\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{I} \right) \\ & {{l}^{2}}+{{w}^{2}}={{15}^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{II} \right) \end{align} Solving equation (I): $l=\frac{108}{w}$ (III) Substitute the value of $l$ in equation (II). Then, \begin{align} & \frac{{{108}^{2}}}{{{w}^{2}}}+{{w}^{2}}=225 \\ & 11664+{{w}^{4}}-225{{w}^{2}}=0 \\ & \left( {{w}^{2}}-81 \right)\left( {{w}^{2}}-144 \right)=0 \end{align} Now, $\left( {{w}^{2}}-81 \right)=0\ \text{ or }\left( {{w}^{2}}-144 \right)=0$. All the possible values of $w$ are $w=\pm 9,\pm 12$. Negative value are not possible. Therefore, $w=9,12$ If $w=9$ then, \begin{align} & l=\frac{108}{9} \\ & =12 \end{align} If $w=12$ then, \begin{align} & l=\frac{108}{12} \\ & =9 \end{align} Thus, the dimensions are 12 feet by 9 feet.