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

Explanation

Let ` OC=x, CQ = r`
Now OA=R (given)
height of cone = h = x + R …(1)
`:." Volume of " e=V= 1/3 pi r^(2)h`
Now in right ` Delta OCQ,`
`OC^(2)+CQ^(2)=CQ^(2)`
`rArr x^(2)+r^(2)=R^(2)`
` rArr r^(2)=R^(2)-x^(2)` ....(2)
put the values of h and r from equations (1) and (2) in V,
` V= 1/3 pi [(R^(2)-x^(2))-2x(x+R)]`
Differentiate equation (3) w.r.t. x,
` (dV)/(dx) = 1/3 pi [R^(2)-x^(2))-2x(x+R)]`
` rArr (dV)/(dx) = pi/3 [R^(2)-x^(2)-2x^(2)-2xR]`
` rArr (dV)/(dx) = pi/3 (R^(2)-2xR-3x^(2))`
` rArr (dV)/(dx) = pi/3 (R-3x)(R+x)` ....(4)
For maximum value ` (dV)/(dx) = 0`
` rArr pi/3 (R-3x)(R+x)=0`
` rArr x = R/3 pr x =- R rArr x = R/3 `
`(.:' ` x cannot be negative)
Differentiate equation (4) w.r.t. x,
` (d^(2)V)/(dx^(2)) = pi/3 [(-3)(R+x)+(R-3x)]`
` pi/3 (-2R-6x) =- pi/3 (2R+6x)`
at ` x = R/3 , (d^(2)V)/(dr^(2)) = (-pi)/3 (2R+ (6R)/3) =- (4pi)/3 R lt 0 `
` :." local maximum value of V is at " x = R/3`.
Now, put the value of x in equation (3),
` V= pi/3 (R^(2)-R^(2)/9)(R+R/3)`
` pi/3 * (8R^(2))/9 * (4R)/3 = 8/27 (4/3 pi R^(3))`
` rArr V= 8/27 xx ` Volume of sphare .