Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.7 - Optimization Problems - 4.7 Exercises - Page 338: 31

Answer

The greatest possible volume of the cylinder must be $\begin{equation} \frac{4\pi r^{3}}{3\sqrt{3}} \end{equation}$, or $\begin{equation}\frac{1}{\sqrt{3}}\end{equation}$ times the volume of the sphere.

Work Step by Step

Let the letter R represent the radius of the cylinder, and let the letter h represent its height. $\begin{equation} V_{cylinder}=\pi R^{2}h\\ 0\geq R=h\geq r \end{equation}$ The radius of the sphere and the radius of the cylinder and half the height of the cylinder form a right triangle, with the radius of the sphere being the hypotenuse. So $\begin{equation} R^{2}+\left(\frac{h}{2}\right)^{2}=r^{2}\\ R^{2}=r^{2}-\frac{h^{2}}{4}. \end{equation}$ Therefore the volume of the cylinder is, by substitution, $\begin{equation} V_{cylinder}=\pi\left(r^{2}-\frac{h^{2}}{4}\right)h=\pi r^{2}h-\frac{\pi h^{3}}{4}=\frac{4\pi r^{2}h-\pi h^{3}}{4} \end{equation}$ Since r is a given in the problem, the volume is now expressed as a function of h. We now take the derivative to find critical values. $\begin{equation} \frac{dV}{dh}=\frac{4\pi r^{2}-3\pi h^{2}}{4} \end{equation}$ Set to 0 to find critical values. $\begin{equation} 0=\frac{4\pi r^{2}-3\pi h^{2}}{4}\\ 0=4\pi r^{2}-3\pi h^{2}\\ 3\pi h^{2}=4\pi r^{2}\\ h^{2}=\frac{4r^{2}}{3}\\ h=\sqrt{\frac{4r^{2}}{3}}=\frac{2r}{\sqrt{3}} \end{equation}$ Since both of the end point values lead to a total volume of 0, 2r/√3 must be the value of h when the volume is greatest. $\begin{equation} V_{cylinder}=\frac{4\pi r^{2}\left(\frac{2r}{\sqrt{3}}\right)-\pi \left(\frac{2r}{\sqrt{3}}\right)^{3}}{4}=\frac{\frac{8\pi r^{3}}{\sqrt{3}}-\frac{8\pi r^{3}}{3\sqrt{3}}}{4}=\frac{24\pi r^{3}-8\pi r^{3}}{12\sqrt{3}}=\frac{16\pi r^{3}}{12\sqrt{3}}=\frac{4\pi r^{3}}{3\sqrt{3}} \end{equation}$
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