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

we have to point out 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.
Suppose the radius and height of cone be r and h respectively.
Radius of the sphere `= R `
`R^2=r^2+(h-R)^2`
`R^2=r^2+h^2+R^2-2hR`
`r^2=2hR-h^2 ......(1)`
Assume the volume of cone be `V`
Volume of the cone
`V=1/3 pi(2hR-h^2)h`
from equation `(1)`
`V=1/3 pi(2h^2R-h^3)`
Condition for the maxima and minima is
`(dV)/(dh)=0`
`=>1/3 pi(4hR-3h^2)=0`
`4hR-3h^2=0`
`h=(4R)/3`
for `h=(4R)/3`,`(d^2V)/(dh^2)<0`
`=>V `will be maximum for` h=(4R)/3`
`h=2/3(2R)`
Hence proved