A right circular cylinder is inscribed in a cone. Show that the curved surface area of the cylinder is maximum when the diameter of the cylinder is equal to the radius of the base of the cone.

Explanation

Let `r_(1)` be the radius of the cone and `h_(1)` be its height and let r be the radius and h the height of the inscribed cylinder.
Clearly, `traingleOAD and triangle OBC` are similar.
`:. (r)/(r_(1)) = (h_(1) - h)/(h_(1)) rArr r = (r_(1))/(h_(1)) (h_(1) - h)`...(i)
Let S be the curved surface area of the cylinder
Then, `S = 2 pih rArr S = (2pi r_(1))/(h_(1)) (h_(1) - h)h`
`rArr (dS)/(dh) = (2pi r_(1))/(h_(1)) (h_(1) - 2h) and (d^(2)S)/(dh^(2)) = (-4 pi r_(1))/(h_(1)) lt 0`
Now, `(dS)/(dh) = 0 hArr h_(1) = 2h`. Also, `(d^(2)S)/(dh^(2)) lt 0`
So, S is maximum when `h_(1) = 2h`
`:. (r)/(r_(1)) = (2h - h)/(2h) = (1)/(2) hArr r_(1) = 2r` = diameter of the cylinder