## Precalculus (6th Edition) Blitzer

Let us assume $l$ to be the length and $w$ to be the width of the rectangle such that: \begin{align} & 2l+2w=36\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{I} \right) \\ & lw=77\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{II} \right) \end{align} Divide equation (I) by $2$ and the value of $l$ is, \begin{align} & l+w=18 \\ & l=18-w\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{III} \right) \\ \end{align} Substitute the value of (I) in equation (II). Then, \begin{align} & \left( 18-w \right)w=77 \\ & 18w-{{w}^{2}}=77 \\ & {{w}^{2}}-18w+77=0 \end{align} Now, solve for $w$. Therefore, \begin{align} & \left( {{w}^{2}}-11 \right)w-7w+77=0 \\ & \left( w-11 \right)w-7\left( w-11 \right)=0 \\ & w=11,7 \end{align} Substitute the value of $w=11,7$ into equation (III): If $w=11$ then, \begin{align} & l=18-11 \\ & =7 \end{align} If $w=7$ then, \begin{align} & l=18-7 \\ & =11 \end{align} Thus, the dimensions of the rectangle are 11 feet by 7 feet.