Question

Determine the largest capacity of a cylindrical tank if its surface area (without lid) should be equal to S

Answer 1

The surface area of the cylinder without the lid is $S=2\pi rh+\pi r^2$ and the maximum volume will be $V=\pi r^2 h$. From the first expression, we find $h=\cfrac{S}{2\pi r}-\cfrac{r}{2}$ and we substitute this on the formula for S to find:

$V=\pi r^2 (\frac{S}{2\pi r}-\frac{r}{2} )=\cfrac{S}{2}\cdot r-\cfrac{\pi}{2}\cdot r^3$

We proceed to find the maximum for V in terms of r when V'(r) = 0:

$V'_{(r)}=\cfrac{d}{dr} \Bigg( \cfrac{S}{2}\cdot r-\cfrac{\pi}{2}\cdot r^3 \Bigg) \\ V'_{(r)}= \cfrac{S}{2}-\cfrac{3\pi}{2}\cdot r^2=0 \\ \therefore V''_{(r)}= -3\pi\cdot r \\ V'_{(r)}= \cfrac{S}{2}-\cfrac{3\pi}{2}\cdot r^2=0 \\ \cfrac{S}{2}=\cfrac{3\pi}{2}\cdot r^2 \implies r=\sqrt{\cfrac{S}{3\pi}}$

Once we find out the value for r, we proceed to substitute and find

$h= \cfrac{S}{2\pi r}-\cfrac{r}{2} = \cfrac{S}{2\pi \sqrt{\cfrac{S}{3\pi}}}-\cfrac{\sqrt{\cfrac{S}{3\pi}}}{2} \\ \therefore h=\cfrac{1}{2} \Bigg( \sqrt{\cfrac{3S}{\pi}}-\sqrt{\cfrac{S}{3\pi}} \, \Bigg)=\sqrt{\cfrac{S}{3\pi}} \Bigg( \cfrac{3-\sqrt{3}}{2} \, \Bigg)$

The fist result is valid because V''(r) is negative and this occurs when V(r) is at a maximum. In conclusion, the dimensions for the cylinder to have the maximum volume (with a surface area S) are $h=\sqrt{\cfrac{S}{3\pi}} \Bigg( \cfrac{3-\sqrt{3}}{2} \, \Bigg)$ and $r=\sqrt{\cfrac{S}{3\pi}}$.

Reference:

• Thomas, G. B., & Finney, R. L. (1961). Calculus. Addison-Wesley Publishing Company.