Of all the closed cylindrical cans (right circular),

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

Then, volume (V) of the cylinder is given by,

$V=\pi r^{2} h=100$  (given)

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

Surface area (S) of the cylinder is given by,

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

$\therefore \frac{d S}{d r}=4 \pi r-\frac{200}{r^{2}}, \frac{d^{2} S}{d r^{2}}=4 \pi+\frac{400}{r^{3}}$

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

$\Rightarrow r^{3}=\frac{200}{4 \pi}=\frac{50}{\pi}$

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

Now, it is observed that when $r=\left(\frac{50}{\pi}\right)^{\frac{1}{3}}, \frac{d^{2} \mathrm{~S}}{d r^{2}}>0$.

$\therefore$ By second derivative test, the surface area is the minimum when the radius of the cylinder is $\left(\frac{50}{\pi}\right)^{\frac{1}{3}} \mathrm{~cm}$.

When $r=\left(\frac{50}{\pi}\right)^{\frac{1}{3}}, h=\frac{100}{\pi\left(\frac{50}{\pi}\right)^{\frac{2}{3}}}=\frac{2 \times 50}{(50)^{\frac{2}{3}}(\pi)^{1-\frac{2}{3}}}=2\left(\frac{50}{\pi}\right)^{\frac{1}{3}} \mathrm{~cm}$

Hence, the required dimensions of the can which has the minimum surface area is given by radius $=\left(\frac{50}{\pi}\right)^{\frac{1}{3}} \mathrm{~cm}$ and height $=2\left(\frac{50}{\pi}\right)^{\frac{1}{3}} \mathrm{~cm}$.