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.