Of all the closed cylindrical cans (right circular),

Solution:

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

Volume $(V)$ of the cylinder $=\pi r^{2} h$

$\Rightarrow 100=\pi r^{2} h$

$\Rightarrow h=\frac{100}{\pi r^{2}}$

Surface area $(S)$ of the cylinder $=2 \pi r^{2}+2 \pi r h=2 \pi r^{2}+2 \pi r \times \frac{100}{\pi r^{2}}$

$\Rightarrow S=2 \pi r^{2}+\frac{200}{r}$

$\therefore \frac{d S}{d r}=4 \pi r-\frac{200}{r^{2}}$

For the maximum or minimum, we must have

$\frac{d S}{d r}=0$

$\Rightarrow 4 \pi r-\frac{200}{r^{2}}=0$

$\Rightarrow 4 \pi r^{3}=200$

$\Rightarrow r=\left(\frac{50}{\pi}\right)^{\frac{1}{3}}$

Now,

$\frac{\mathrm{d}^{2} \mathrm{~S}}{\mathrm{dr}^{2}}=4 \pi+\frac{400}{\mathrm{r}^{3}}$

$\Rightarrow \frac{d^{2} S}{d r^{2}}>0$ when $r=\left(\frac{50}{\pi}\right)^{\frac{1}{3}}$

Thus, the surface area is minimum when $r=\left(\frac{50}{\pi}\right)^{\frac{1}{3}}$.

At $r=\left(\frac{50}{\pi}\right)^{\frac{1}{3}}:$

$h=\frac{100}{\pi\left(\frac{50}{\pi}\right)^{\frac{2}{3}}}=2\left(\frac{50}{\pi}\right)^{\frac{1}{3}}$