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 AD =x be the height of the cone ABC inscribed in a sphere of radius a Therefore
OD =x-a

Then radius of its base (r ) =`CD=sqrt((OC^(2)-OD^(2))`
`=sqrt[(a^(2)-(x-a)^(2)])`
Thus volume V of the cone is given by
`V=1/3 pir^(2)x=1/3pi(2ax-x^(2))x=1/3pi(2ax^(2)-x^(3))`
`therefore (dV)/(dx)=1/3(4ax-3x^(2))and (d^(2)V)/(dx)=1/3pi(4a-6x)`
for max or min of V dV/dx=0 or `x =4a//3`
For this value of V, `(d^(2)V)/(dx^(2))=-(4pia//3)=(-ve)`
Thus V is maximum (i.e greatesty) when `x =4a//3 = (2//3)(2a)`
i.e when the height of cone is `(2//3)rd` of the diameter of sphere