Prove that the radius of the right circular cylinder of greatest curved surface area which can be inscribed in a given cone is half of that of the cone.

Explanation

`OE=x`
`OC=r`
`AO=h`
`/_ AOC cong /_ QEC`
`(AO)/(OC)=(QE)/(EC)`
`(QE)=(AO)/(OC)(EC)`
=`h/r(r-x)`
`deltax=2pix(h/r(r-x))`
=`(2pih)/r(rx-x^2)`
`delta'(x)=0`
`x=r/2`
derivating again
`delta''(r/2)=(2pih)/r(-2)<0`
`r/2` is at maxima
Hence, the radius of the right circular cylinder of the greatest curved surface area which can be inscribed in a given cone is half of that of the cone.