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Question
Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height h and semi vertical angle α is one-third that of the cone and the greatest volume of cylinder is `4/27 pih^3` tan2α.
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Solution
Let VAB be a cone.
Height of the cone = h
Semivertex angle = α
Cylinder A'B'DC is inscribed in a cone whose radius = x.
OO' = Height of cylinder = VO - VO'
= h - x cot α
V, volume of the cylinder = πx2 (h - x cot α)
On differentiating with respect to x,
`(dV)/dx = 2 pi xh - 3 pix^2 cot alpha`
For maximum and minimum, `(dV)/dx = 0`
⇒ 2πxh - 3πx2 cot α = 0
⇒ πx (2h - 3x cot α) = 0
⇒ 2h - 3 x cot α = 0
⇒ 3x cot α = 2h
`therefore x = (2h)/3 tan alpha ...[x ne 0]`
Now, `(d^2 v)/(dx^2) = 2 pih - 6 pih cot alpha`
When x = `(2h)/3 tan alpha`
`therefore (d^2V)/dx^2 = 2pih - 6pi (2h)/3 tan alpha cot alpha`
`= 2pih - 4pih`
`= pi(2h - 4h)`
`= - 2 pih < 0`
⇒ V is maximum when x = `(2h)/3 tan alpha`
Height of cylinder, OO' = VO - VO' = h - x cot α
`= h - ((2h)/3 tan alpha) cot alpha ...[because x = (2h)/3 tan alpha]`
`= h - (2h)/3 = h/3`
`= 1/3` height of the cone
Maximum volume of cylinder = πx2 (h - x cot α)
`= pi - ((2h)/3 tan alpha)^2 * (h - (2h)/3)`
`= pi((2h)/3 tan alpha)^2 xx h/3`
`= 4/27 pih^3 tan^2 alpha`
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