Answer to Question #174209 in Calculus for Holden Giles Cabrito

Question

A steel storage tank for propane gas is to be constructed in the shape 

of a right circular cylinder with a hemisphere at both ends. If the 

desired capacity is 100 cu. ft., what dimensions will require the least 

amount of steel ?

Answer 1

Consider the right circular cylindrical storage tank with a hemisphere at both ends.

Let $h$ be the height of the cylindrical section and $r$ be the radius of cylinder.

Obviously, the radius of hemisphere will also be $r$

The total volume of the storage tank is $100 ft^3$

So,

$\pi r^2 h+2(\frac{2}{3} \pi r^3)=100$

$\pi r^2 h+\frac{4}{3} \pi r^3=100$

$3\pi r^2 h+4 \pi r^3=300$

$3\pi r^2 h=300-4 \pi r^3$

$h=\frac{300-4 \pi r^3}{3\pi r^2}$

Now, the total surface area of the tank is,

$S=2 \pi r h+4 \pi r^2$

Substitute $\frac{300-4 \pi r^3}{3\pi r^2}$ for $h$ into $S=2 \pi r h+4 \pi r^2$ to obtain the total surface area in a single variable $r$ as,

$S(r)=2 \pi r (\frac{300-4 \pi r^3}{3\pi r^2})+4 \pi r^2$

$S(r)=\frac{200}{r}-\frac{8 \pi}{3}r^2+4 \pi r^2$

$S(r)=\frac{200}{r}+\frac{4 \pi}{3}r^2$

Differentiate $S$ with respect to $r$ and set it equal to zero as,

$S'(r)=0$

$-\frac{200}{r^2}+\frac{8 \pi}{3}r=0$

$\frac{8 \pi}{3}r=\frac{200}{r^2}$

$r^3=\frac{75}{\pi}$

$r=\sqrt[3]{\frac{75}{\pi}}$

Here, "S"(r)=\\frac{400}{r^3}+\\frac{8 \\pi}{3}>0," therefore, $S$ is minimum.

Plug $r=\sqrt[3]{\frac{75}{\pi}}$ into $h=\frac{300-4 \pi r^3}{3\pi r^2}$ and simplify for $h$ as,

$h=\frac{300-4 \pi (\frac{75}{\pi})}{3\pi (\sqrt[3]{\frac{75}{\pi}})^2}=0$

Therefore, the dimensions are: $r=\sqrt[3]{\frac{75}{\pi}}$ and $h=0$ .