Question

A cylinderical barrel is to hold 210L of industrial chemicals. Determine the radius and height of the barrel with the minimum surface areato the nearest tenth of a centimeter. (Recall: 1L=100cm3)

Answer 1

Let r be the radius (in cm) of the barrel, and h be its height (in cm), then we can set the following equations:

$\begin{gathered} \pi r^2h=21000\text{.} \\ SA=2\pi rh+2\pi r^2\text{.} \end{gathered}$

Solving the first equation for h we get:

$h=\frac{21000}{\pi r^2}\text{.}$

Substituting the above equation in the second equation we get:

$\begin{gathered} SA=2\pi r\frac{21000}{\pi r^2}+2\pi r^2, \\ SA=\frac{42000}{r^{}}+2\pi r^2\text{.} \end{gathered}$

Now, we will use the first and second derivative criteria to obtain the minimum value for the above equation for the Surface Area.

The first and second derivatives of SA are:

$\begin{gathered} SA(r)^{\prime}=-\frac{42000}{r^2}+4\pi r, \\ SA(r)^{\doubleprime}=\frac{84000}{r^3}+4\pi\text{.} \end{gathered}$

Setting SA(r)'=0 and solving for r we get:

$\begin{gathered} 0=-\frac{42000}{r^2}+4\pi r, \\ \frac{42000}{r^2}=4\pi r, \\ 42000=4\pi r^3, \\ r^3=\frac{10500}{\pi}, \\ r\approx14.95. \end{gathered}$

Evaluating SA''(r) at r=14.95 we get:

$SA^{\doubleprime}(14.95)_{}=\frac{84000}{14.95^3}+4\pi>0.$

Therefore, SA(r) reaches a local minimum at r≈15.0.

Substituting r=15.0 in h(r) we get:

$\begin{gathered} h(15.0)=\frac{21000}{\pi\cdot15.0^2}, \\ SA(15.0)\approx29.7. \end{gathered}$

Answer: The radius is 15.0 cm and the height is 29.7 cm.