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` .

Explanation

Let a cone of base radius r and height h is inscribed in a sphere of centre O and radius R. The centre O of the sphere will be on the height AC of the cone.
In ` Delta OAB`,
`OB^(2)=OA^(2)+AB^(2)`
` R^(2) = (h-R)^(2)+r^(2)`
` rArr 0 = h^(2)-2hR+r^(2)` ...(1)
Now volume of cone
`V = 1/3 pi r^(2)h`
` = 1/3 pi h(2hR-h^(2))` From eqn. (1)
` = 1/3 pi (2h^(2)R-h^(3))`
` (dV)/(dh) = 1/3 pi (4hR-3h^(2))`
and `(d^(2)V)/(dh^(2))= 1/3 pi (4R - 6h)`
For maxima/minima
` (dV)/(dh) = 0`
` rArr hR-3h^(2)= 0`
` rArr 4R = 3h`
` rArr h/R = 4/3 = 4:3`
at ` h=(4R)/3`
`(d^(2)V)/(dh^(2))=1/3 pi (4R - 8R) =- 1/3 pi (4R) lt 0`
` rArr` Volume is maximum.
Therefore, for maximum volume, the ratio of the height of cone and radius of sphere is `4 : 3`.