Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 7 - Section 7.4 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 852: 62

Answer

The length is 18 inches and the width is 12 inches.

Work Step by Step

We start with: $\begin{align} & lw=216\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{I} \right) \\ & 2\left( l-4 \right)\left( w-4 \right)=224\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{II} \right) \end{align}$ Solving equation (I) Then, $ l=\frac{216}{w}$ (III) Substitute the value of l in equation (II): $\begin{align} & \frac{216}{w}w-4\frac{216}{w}-4w-96=0 \\ & 216+\frac{864}{w}-4w-96=0 \\ & 4{{w}^{2}}+864-120w=0 \\ & {{w}^{2}}-30w+216=0 \end{align}$ Now, solve: $\begin{align} & \left( w-12 \right)\left( w-18 \right)=0 \\ & w=12,\ 18 \end{align}$ Substitute the value of $ w=12,18$ into equation (III). Then, $\begin{align} & l=\frac{216}{12} \\ & =18 \end{align}$ And $\begin{align} & l=\frac{216}{18} \\ & =12 \end{align}$ Hence, the length is 18 inches and the width is 12 inches.
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