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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 = πR^{2}x

= π(h - x)^{2} tan^{2}α. x

= π tan^{2}α (h - x)^{2}x

`dV/dx = π tan^{2}α (h2 - 4hx + 3x^{2})`

`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, V_{max} = π tan^{2}α (h - x)^{2}x

`= π tan^2α ( h - h/3 )^2 h/3`

= `(4πh)^3/27 tan^2 a`

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