Precalculus (6th Edition) Blitzer

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

Chapter 1 - Section 1.10 - Modeling with Functions - Exercise Set - Page 295: 47


If a cylindrical soft drink has a volume of $12$ fluid ounces, approximately $22\text{ cubic}\text{ }\text{inches}$, then the surface area of the can, $A$ in square inches as a function of its radius $r$ is given as: $A\left( r \right)=2\pi {{r}^{2}}+\frac{44}{r}$.

Work Step by Step

We have a cylindrical can. The volume of the can is: $\text{V=}\pi {{r}^{2}}h$ And the volume is $22\text{~cubic}\text{ }\text{inches}$ Therefore, $\begin{align} & \text{22=}\pi {{r}^{2}}h \\ & h=\frac{22}{\pi {{r}^{2}}} \end{align}$ The area of the can: $A=2\pi {{r}^{2}}+2\pi rh$ Now put the value of $h$ in the above equation, $\begin{align} & A=2\pi {{r}^{2}}+2\pi rh \\ & =2\pi {{r}^{2}}+2\pi r\left( \frac{22}{\pi {{r}^{2}}} \right) \\ & =2\pi {{r}^{2}}+\frac{44}{r} \end{align}$ Here, $A$ is the function of $r$ (can radius), Hence $A$ is expressed as, $A\left( r \right)=2\pi {{r}^{2}}+\frac{44}{r}$
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