Precalculus (6th Edition) Blitzer

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

Chapter 1 - Section 1.9 - Distance and Midpoint Formulas; Circles - Exercise Set - Page 282: 103

Answer

The required value is $h=\frac{22}{\pi {{r}^{2}}}$ and the expression in terms of $r$ is $2\pi {{r}^{2}}+\frac{44}{r}$.

Work Step by Step

Consider the equation $\pi {{r}^{2}}h=22$. Solve for $h$. Divide both sides $\pi {{r}^{2}}$, $\begin{align} & \pi {{r}^{2}}h=22 \\ & \frac{\pi {{r}^{2}}h}{\pi {{r}^{2}}}=\frac{22}{\pi {{r}^{2}}} \\ & h=\frac{22}{\pi {{r}^{2}}} \end{align}$ Therefore, the required value $h=\frac{22}{\pi {{r}^{2}}}$. Now, re-write the expression $2\pi {{r}^{2}}+2\pi rh$ in terms of $r$. Substitute $h=\frac{22}{\pi {{r}^{2}}}$, $\begin{align} & 2\pi {{r}^{2}}+2\pi rh=2\pi {{r}^{2}}+2\pi r\left( \frac{22}{\pi {{r}^{2}}} \right) \\ & =2\pi {{r}^{2}}+2\left( \frac{22}{r} \right) \\ & =2\pi {{r}^{2}}+\frac{44}{r} \end{align}$ Therefore, the expression in terms of $r$ is $2\pi {{r}^{2}}+\frac{44}{r}$.
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