Of all the closed right circular cylindrical cans of volume 128 cm^3,…

The correct answer is Option (1) → $r=4\,cm,h=8\,cm$

Let r cm be the radius of the circular base and h cm be the height of the cylindrical can, then

V (volume) = $\pi г^2h = 128π$ (given)

$⇒h =\frac{128}{r^2}$   ...(i)

Let S be the (total) surface area of the cylinder,

then $S = 2πrh + 2πг^2$

$= 2πг. \frac{128}{r^2}+ 2πг^2$  (using (i))

$= 2π\left(\frac{128}{r}+r^2\right)$

Differentiating it w.r.t. r, we get

$\frac{dS}{dr}=2π\left(-\frac{128}{r}+2r\right)$ and $\frac{d^2S}{dr^2}=2π\left(\frac{128×2}{r^3}+2\right)$.

Now $\frac{dS}{dr}=0⇒2π\left(\frac{256}{64}+2\right)=2π(4+2)=12π>0$

⇒ S is minimum when $r=4$.

From (i), $h=\frac{128}{4^2}=\frac{128}{16}=8$

Hence, the surface area of the cylindrical can is minimum when radius = 4 cm and height = 8 cm.