Question

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

Answer 1

Let $S$ and $V$ dente the surface and volume of the right circular cylinder of height $h$ and base radius $r$. Then

$S=2\pi rh+2\pi r^2⟶\left(1\right)$

$V={\pi r}^{2}h$

$={\pi r}^2\left(\frac{S-2\pi r^2}{2\pi r}\right)$

$=\frac{r}{2}(S-2\pi r^2)=\frac{Sr}{2}-\pi r^3$

$\frac{dV}{dr}=\frac{S}{2}-3\pi r^2=0$ for max or min $S=6\pi r^2$

Substituting $S=6\pi r^2$ in equation $\left(1\right)$

${6\pi r}^2=2\pi r^2+2\pi rh$

$\Rightarrow 2r=h$

Also $\frac{d^{2}V}{{dr}^2}=-6\pi r$

$\frac{d^{2}V}{{dr}^2}$ at $r=h/2=-3\pi h<0$

Hence volume is maximum when height is equal to diameter of the base.