Question

Solve the following : Show that the height of a right circular cylinder of greatest volume that can be inscribed in a right circular cone is one-third of that of the cone.

Answer 1

Given the right circular cone of fixed height h and semi-vertical angle `oo`.
Let R be the radius of the base and H be the height of right circular cylinder that can be inscribed in the right circular cone.
In the figure, ∠GAO = ∞, OG = r, OA = h, OE = R, CE = H.

We have `r/h = tanx`

∴ r = h tan x                             ...(1)

Since ΔAOG anf ΔCEG are similar.

∴ `"AO"/"OG" = "CE"/"EG" = "CE"/"OG - OE"`

∴ `h/r = "H"/"r - R"`

∴ H = `h/r(r - R)`

= `"h"/"h tan oo"(h tan oo- R)`           ...[By (1)]

∴ H = `(1)/tanoo(h tan oo - R)`      ...(2)

Let V be the volume of the cylinder

Then V = `piR^2H = (piR^2)/tanoo(h tan oo - R)`

∴ V = `piR2h - (piR^3)/tan oo`

∴ `"dV"/"dR" = d/"dR"(piR^2h - (piR^3)/tan oo)`

= `piR xx 2R - pi/tan oo xx 3R^2`

= `2piRh - (3piR^2)/tan oo`
and
`(d^2V)/(dR^2) = d/"dR"(2piRh - (3pir^2)/tan oo)`

= `2pih xx 1 - (3pi)/tanoo xx 2R`

= `2pih - (6piR)/tanoo`

For maximum volume, `"dV"/"dR"` = 0

∴ `2piRh - (3piR^2)/tanoo` = 0

∴ `(3piR2)/tanoo = 2piRh`

∴ R = `(2h)/(3) tanoo`                ...[∵ R ≠ 0]
and
`((d^2V)/(dR^2))_("at"  R = (2h)/(3)tanx`

= `2pih - (6pi)/tanoo xx (2h)/(3) tan oo`

= 2πh – 4πh = – 2πh < 0

∴ V is maximum when R = `(2h)/(3)tanoo`

When R = `(2h)/(3)tanoo`, then from (2), we get

H = `(1)/tanoo(h tan oo - (2h)/3 tann oo) = h/(3)`

Hence, the height of the right circular cylinder is one- third of that of the cone.