Chapter 5.4, Problem 22E

To determine

Find the maximum volume enclosed by the given curves.

Answer to Problem 22E

The maximum surface area, the dimensions are radius is $8.16$ ,the height is $11.55$ cm and volume is $2418.40c{m}^3$

Explanation of Solution

Given information: Given Label the sketch as follows where $2h$ is the height of the cylinder and $h$ is the height of the triangle.

Calculation:

Using the right triangle ,

  $\begin{array}{l}{r}^2=R^2-h^2\\ =100-h^2\end{array}$

  

The volume of the right circular cylinder,

  $\begin{array}{l}V=\pi r^2\left(2h\right)\\ =2\pi \left(100-h^2\right)h\\ =-2\pi h^{{}^3}+200\pi h\\ \text{the derivative of }V\\ \frac{dV}{dh}=-6\pi h^{{}^2}+200\pi \\ \frac{dV}{dh}=0\text{ when}\\ -6\pi h^{{}^2}+200\pi =0\\ h=\sqrt{\frac{100}{3}}\\ =\frac{10}{\sqrt{3}}\\ \frac{d^{2}V}{d{h}^2}=-12\pi h\\ \text{Note that }\frac{d^{2}V}{d{h}^2}<0\text{ for }h=\frac{10}{\sqrt{3}}.\end{array}$

Height of cylinder:

  $\begin{array}{l}2h=\frac{20}{\sqrt{3}}\\ r^2=100-h^2\\ =100-{\left(\frac{10}{\sqrt{3}}\right)}^2\\ =100-\frac{100}{3}\\ =\frac{200}{3}\end{array}$

Radius of the cylinder:

  $\begin{array}{l}r=\sqrt{\frac{200}{3}}\\ =\frac{10\sqrt{2}}{\sqrt{3}}\\ V=\pi r^2\left(2h\right)\\ \pi \cdot \frac{200}{3}\cdot \frac{20}{\sqrt{3}}\\ =\frac{4000\pi }{3\sqrt{3}}\\ =2418.40\end{array}$

Thus, the maximum surface area, the dimensions are radius is $8.16$ ,the height is $11.55$ cm and volume is $2418.40c{m}^3$ .