Question

Of all the closed cylindrical cans(right circular),of a given volume of 100cubic centimetres,find the dimensions of the can which has the minimum surface area?

Answer 1

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

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

$V=\pi r^{2}h=100(given)$

$∴h=\frac{100}{\pi r^{2}}$

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

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

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

$\frac{dS}{dr}=0⇒4\pi r=\frac{200}{r^{2}}$

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

$⇒r=(\frac{50}{\pi})^{\frac{1}{3}}$

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

By second derivative test,the surface area is the minimum when the radius of the

cylinder is$(\frac{50}{\pi })^{\frac{1}{3}}cm.$

When$r=(\frac{50}{\pi })^{\frac{1}{3}}$,$h=\frac{100}{\pi(\frac{50}{\pi})^{\frac{1}{3}}}$=\frac{2\times50}{(50\pi)\frac{2}{3}()1-\frac{2}{3}}$$=2(\frac{50}{\pi})^{\frac{1}{3}} cm.

Hence,the required dimensions of the can which has the minimum surface area is given

by radius=$(\frac{50}{\pi})^{\frac{1}{3}} cm.$ and height=$2(\frac{50}{\pi})^{\frac{1}{3}} cm.$