Question

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

Answer 1

Let r and h be the radius and height of the cylinder respectively.
Then, the surface area S o the cylinder is
$S=2\pi r^2+2\pi rh\left(given\right)\Rightarrow 2\pi rh=S-2\pi r^2\Rightarrow h=\frac{s-2\pi r^2}{2\pi r}\dots \left(i\right)$
Also $V=\pi r^{2}h\dots \left(ii\right)$
On putting the value of h from Eq. (i) in Eq (ii), we get
$V=\pi r^2\left(\frac{S-2\pi r^2}{2\pi r}\right)=\frac{Sr}{2}-\pi r^3\dots \left(iii\right)$
On differentiating Eq. (iii) w.r.t. r, we get

$\frac{dV}{dr}=\frac{S}{2}-3\pi r^2$
For maxima or minima put $\frac{dV}{dr}=0$
$\Rightarrow \frac{S}{2}-3\pi r^2=0\Rightarrow S=6\pi r^2\Rightarrow r^2=\frac{S}{6\pi }\dots \left(iv\right)$
Now, $\frac{d^{2}V}{d{r}^2}=-6\pi r$
At $r^2=\frac{S}{6\pi }{\left(\frac{d^{2}V}{d{r}^2}\right)}_{r=\sqrt{\frac{S}{6\pi }}}=-6\pi \left(\sqrt{\frac{S}{6\pi }}\right)<0$
By second derivative test, the volume is maximum when $r^2=\frac{S}{6\pi }$
When $r^2=\frac{S}{6\pi }orS=6\pi r^2$
Then, $h=\frac{6\pi r^2}{2\pi }\left(\frac{1}{r}\right)-r=3r-4=2r$
Hence, the volume is maximum when the height is twice the radius i.e., when the height is equal to the diameter.