Chapter 5, Problem 8TP

Solution

To find: The approximate radius of the cylinder.

Option A is correct.

Given:

The cylinder has the same volume as the rectangular prism below and has the least surface area.

  

Calculation:

The volume of the given prism is,

  $18.5\times 3.2\times 5=296{\text{cm}}^{\text{3}}$

Since the cylinder has the same volume as the prism, therefore

  $\begin{array}{c}V=296\\ \pi r^{2}h=296\\ h=\frac{296}{\pi r^2}\end{array}$

Now consider the surface area of the cylinder,

  $\begin{array}{c}S=2\pi r\left(h+r\right)\\ S=2\pi r\left(\frac{296}{\pi r^2}+r\right)\\ S=2\pi r\left(\frac{296+\pi r^3}{\pi r^2}\right)\\ S=\frac{592+2\pi r^3}{r}\\ S=\frac{592}{r}+2\pi r^2\end{array}$

Now by differentiating with respect to $r$ , it follows that

  $\begin{array}{c}\frac{dS}{dr}=-\frac{592}{r^2}+2\pi \left(2r\right)\\ =-\frac{592}{r^2}+4\pi r\end{array}$

Now to find the critical values, solve the first derivative by equating it to zero. That is,

  $\begin{array}{c}\frac{dS}{dr}=0\\ -\frac{592}{r^2}+4\pi r=0\\ \frac{592}{r^2}=4\pi r\\ r^3=\frac{592}{4\pi }\\ r\approx 3.61\text{cm}\end{array}$

Therefore, the approximate radius of the cylinder is 3.61 cm.

Hence option A is correct.