Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is `8/(27)` of the volume of the sphere.

Explanation

Let the center of the sphere be `O` and radius be `R`. let the height and radius of the variable cone inside the sphere be `h` and `r` respectively
So, in the diagram, `OA=OB=R, AD=h, BD=r`
`OD=AD-OA=h-R`
Using pythagoras theorem in `triangle OBD`
`=OB^2=OD^2+BD^2`
`=>R^2=(h-R)^2+r^2`
`=>R^2=h^2+R^2-2hR+r^2`
`=>r^2=2hR-h^2`
Volume of the cone `V=1/3(pir^2h)`
`=1/3(pi)(2hR-h^2)h`
`=(2pih^2R)/3-(pih^3)/3`
For maximum volume `(dV)/(dh)=0`
`=>0=(4pihR)/3-pih^2`
`=>h=(4R)/3`
`:. V=(2piR)/3(16R^2)/9-(64piR^3)/81`
`=((96-64)piR^3)/81`
`=(32piR^3)/81`
`=8/(27)xx(4/3)piR^3`
We know that the volume of the sphere is `V_s=4/3(piR^3)`
Therefore `V=8/27(V_s)`