Question

Show that the right circular cylinder of given surface and maximum volume is such that is heights is equal to the diameter of the base.

Answer 1

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

Then,the surface area(S)of the cylinder is given by,

\(S=2\pi r^{2}+2\pi rh\)

\(⇒h=\frac{S-2\pi r^{2}}{2\pi r}\)

\(=\frac{S}{2\pi}(\frac{1}{r})-r\)

Let V be the volume of the cylinder.Then,

\(V=\pi r^{2}h\)=\(\pi r^{2}[\frac{S}{2\pi }(\frac{1}{r})-r]\)\(=\frac{Sr}{2}\)\(-\pi r^{3}\)

Then,\(\frac{dV}{dr}=\)\(\frac{S}{2}-3\pi r^{2},\frac{d^{2}V}{dr^{2}}=-6\pi r\)

Now,\(\frac{dV}{dr}=\)=0⇒\(\frac{S}{2}\)=\(3\pi r^{2}\)⇒\(r^{2}=\frac{S}{6\pi}\)

When \(r^{2}\)=\(\frac{S}{6\pi r}\),then \(\frac{d^{2}V}{dr^{2}}\)=-6π(√S/6πr)<0.

By second derivative test,the volume is the maximum when r2=S/6π.

Now,when \(r^{2}=\frac{S}{6\pi}\),then h=\(\frac{6\pi r^{2}}{2\pi }(\frac{1}{r})-r\)\(=3r-r=2r.\)

Hence,the volume is the maximum when the height is twice the radius i.e.,when the

height is equal to the diameter.