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Question
Solve the following : Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is `(4r)/(3)`.
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Solution
Let x be the radius of base and h be the height of the cone which is inscribed in a sphere of radius r.
In the figure, AD = h and CD = x = BD
Since, ΔABD and ΔBDE are similar,
`"AD"/"BD" = "BD"/"DE"`
∴ BD2 = AD.DE = AD.(AE – AD)
∴ x2 = h(2r – h) ...(1)
Let V be the volume of the cone.
Then V = `(1)/(3)pix^2h`
= `pi/(3)h(2r - h)h` ...[By (1)]
∴ V = `pi/(3)(2rh^2 - h^3)`
∴ `"dV"/"dh" = pi/(3) d/"dh"(2rh^2 - h^3)`
= `pi/(3)(2r xx 2h - 3h^2)`
= `pi/(3)(4rh - 3h^2)`
and
`(d^2V)/(dh^2) = pi/(3).d/"dh"(4rh - 3h^2)`
= `pi/(3)(4r xx 1 - 3 xx 2h)`
= `pi/(3)(4r - 6h)`
For maximm volume, `"dV"/"dh"` = 0
∴ `pi/(3)(4rh - 3h^2)` = 0
∴ `4rh = 3h^2`
∴ h = `(4r)/(3)` ...[∵ h ≠ 0]
and
`((d^2V)/"dh"^2)_("at" h = (4r)/(3)`
= `pi/(3)(4r - 6 xx (4r)/3)`
= `pi/(3)(4r - 8r)`
= `-(4pir)/(3) < 0`
∴ V is maximum when h = `(4r)/(3)`
Hence, the attitude (i.e. height) of the right circular cone of maximum volume = `(4r)/(3)`.
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