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=\(2(\frac{50}{\pi})^{\frac{1}{3}} cm.\)