Question

. Find the dimensions of the right circular cylinder of greatest volume which can be

inscribed in a right circular cone with radius of 5 in and height of 12 in.

Answer 1

Using similar trianglr

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

Volume of cylinder =$\pi r^2h$

${ 12 \over 5}={h \over 5-r}$

$h={60-12r \over 5}$

Put the value of the height into the formula for the volume of cylinder.

$V=({60-12r \over 5})\pi r^2$

$V=12\pi r^2-{{12\pi r^3} \over 5}$

differentiating V w.r.t r

$V'=24\pi r - {36\pi r^2 \over 5}$

Put V'=0

Then,

$24\pi r - {36\pi r^2 \over 5}=0$

$120 - 36r=0$

$r={120 \over 36}={10 \over 3}$ in

Since we have gotten r, then

From $h={60-12r \over 5}=12-{120 \over 15}=12-8=4 in$

Greatest volume=$\pi r^2h={400 \over 9}$ in³