Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is Also show that the maximum volume of the cone is of the volume of the sphere.

Consider the sphere with centre O and having radius OC which is r

BC is the radius of the base of the right circular cone which is r_c and AB is the altitude or height of cone which is h

We need to maximize volume based on the height so. First, we will require some equation to look at how volume and height are related

The volume of cone let it be denoted by V_c is given by

Here we have got V_c in terms of h, but we have one more unknown parameter r_c which we should eliminate

We need r_c in terms of r or h

Consider ΔOCB

AB = h and OA is the radius of sphere = r

⇒ AO + OB = AB

⇒ r + OB = h

⇒ OB = h – r

As it is a right circular cone height is perpendicular to base hence ∠OBC = 90°

Using Pythagoras theorem

⇒ OC² = BC² + OB²

⇒ r² = r_c² + (h – r)²

⇒ r_c² = r² – (h – r)²

Using a² – b² = (a + b)(a – b)

⇒ r_c² = (r + h – r)(r – h + r)

⇒ r_c² = h(2r – h)

⇒ r_c² = 2rh – h² …(a)

Put this r_c² in V_c

Now consider this as where we have replaced V_c by y and h by x, we get maximum value of y when similarly here we will get the maximum value for volume V_c when

Hence differentiate with respect to h and equate to 0

⇒ 0 = 4rh – 3h²

⇒ h² = 4rh

Now, this h can be a point of minima or maxima. But if it would have been a point of minima the h should have been 0 because minimum volume can be 0. Hence h is the point of maxima

Hence for maximum volume of cone height should be

Now,

Now we will get maximum volume of the cone at because it is point of maxima

Substitute values of r_c² and h from (a) and (b)

And

Hence the maximum volume of the cone is of the volume of the sphere.