Show that the cone of the greatest volume which can be inscribed in a given sphere has an altitude equal to 2/3 of the diameter of the sphere.

Explanation

Let radius of sphere be `R`.

Let radius and height of cone be `r` and `h` respectively.

`implies h=R+sqrt{R^2-r^2}`

`implies r^2=2hR-h^2`

Volume,`V=1/3pir^2h`

`V=1/3pih(2hR-h^2}`

`{dV}/{dh}=pi/3(4hR-3h^2)`

Put `{dV}/{dr}=0`

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