सिद्ध कीजिए कि एक शंकु के अंतर्गत महत्तम वर्कपृष्ठ वाले लंब वृत्तीय बेलन की त्रिज्या शंकु की त्रिज्या की आधी होती है

Explanation

Let radius of cone OC=R
Height of cone OA=H
Let a cylinder of radius OE=r
is constructed inside the cone. The height of cylinder will be given by using similarly of triangle in `DeltaAOC and DeltaQCE` .
`(QE)/(OA)=(CE)/(OC)`
`(h)/(H)=(R-r)/(R)`
`h=((R-r)/(R))H`
Curved surface area of cylinder
`S(r)=2pir ((R-r)/(R))H`
`=(2pi)/(R)(rR-r^(2))H`
`implies S'(r)=(2pi)/(R)(R-2r)H`
`implies S''(r)=-(4piH)/(R)`
For point of maxima and minima
S'(r)=0 ltbgt R=2r
Since ,`S''(r) lt 0`
So, R=2r
`implies r=(R)/(2)` is point of maxima of S(r)
Hence , the radius of cylinder of greatest curved surface area which cane be inscribed in a given cone is half of the cone .