Show that the right circular cylinder of given surface and maximum volume
Question:

Show that the right circular cylinder of given surface and maximum volume is such that is heights is equal to the diameter of the base.

Solution:

Let r and h be the radius and height of the cylinder respectively.

Then, the surface area (S) of the cylinder is given by,

$\begin{aligned} S=& 2 \pi r^{2}+2 \pi r h \\ \Rightarrow h &=\frac{S-2 \pi r^{2}}{2 \pi r} \\ &=\frac{S}{2 \pi}\left(\frac{1}{r}\right)-r \end{aligned}$

Let V be the volume of the cylinder. Then,

$V=\pi r^{2} h=\pi r^{2}\left[\frac{S}{2 \pi}\left(\frac{1}{r}\right)-r\right]=\frac{S r}{2}-\pi r^{3}$

Then, $\frac{d V}{d r}=\frac{S}{2}-3 \pi r^{2}, \frac{d^{2} V}{d r^{2}}=-6 \pi r$

Now, $\frac{d V}{d r}=0 \Rightarrow \frac{S}{2}=3 \pi r^{2} \Rightarrow r^{2}=\frac{S}{6 \pi}$

When $r^{2}=\frac{S}{6 \pi}$, then $\frac{d^{2} V}{d r^{2}}=-6 \pi\left(\sqrt{\frac{S}{6 \pi}}\right)<0 .$

$\therefore$ By second derivative test, the volume is the maximum when $r^{2}=\frac{S}{6 \pi}$.

Now, when $r^{2}=\frac{S}{6 \pi}$, then $h=\frac{6 \pi r^{2}}{2 \pi}\left(\frac{1}{r}\right)-r=3 r-r=2 r$.

Hence, the volume is the maximum when the height is twice the radius i.e., when the height is equal to the diameter.

 

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