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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 the cylinder is `(4)/(27) pi"h"^3 tan^2 α`.
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Solution
Let CD = R, AD = x
⇒ OD = h - x
∵ ODC ∼ ΔOAB
⇒ `(h - x)/h = R/"AB" ⇒ (h - x)/h = R/(h tan α)`
⇒ R = (h - x) tan α
V = πR2x
= π(h - x)2 tan2α. x
= π tan2α (h - x)2x
`dV/dx = π tan2α (h2 - 4hx + 3x2)`
`dV/dx = 0 = h^2 - 4hx + 3x^2 = 0`
⇒ (h - x)(h - 3x) = 0
⇒ x = h (not possible ) or x =` h/3`
`(d^2V)/(dx^2) = π tan^2α (-4h + 6x)`
`((d^2V)/(dx^2))_(x = h/3) = π tan^2α (-2h) < 0`
⇒ V is maximum for x = `h/3`.
So, Vmax = π tan2α (h - x)2x
`= π tan^2α ( h - h/3 )^2 h/3`
= `(4πh)^3/27 tan^2 a`
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