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)`.

Answer 1

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)`.