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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.
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Solution
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.
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