# Simple application of differential calculus

A right circular cone is inscribed in a sphere. Prove that the volume of the cone cannot exceed ⁸/₂₇ of the volume of the sphere.

One would have asked what I have done on this question but no matter what I did I wasn't making headways I was only moving back and forth.

$$V = 4/3πr^3$$ , $$(πr^2*h )/3$$ For sphere and cone.. I have differentiated trying to make some substitutions. ..tried considering those points where the cone touches the sphere. Any clue to this will be appreciated.

• $r$ of the cone and of the sphere are not necessarily equal. – user376343 Mar 8 '19 at 12:03

In my diagram the sphere has centre $$C$$ and radius $$R$$. The cone has base radius $$r$$ and height $$h$$.

The volume of a right circular cone is $$V = \frac{π r^2 h}{3}$$. To apply the calculus you know you need to express this volume as a function of one variable. The right triangle ABC gives the information you need.

By Pythagoras theorem

$$|CA|^2 + |AB|^2 = |BC2|^2$$

thus

$$(h - R)^2 + r^2 = R^2$$

or

$$r^2 = R^2 - (h - R)^2.$$

Substitute $$r^2$$ into the expression for the volume of the cone and you have then a function of one variable $$h$$. Use your calculus to find the value of $$h$$ that maximized the volume.

It is then trivial to prove.

• thanks. Quite insightful – Orestes Dante Mar 8 '19 at 12:27
• Thank you, please accept the answer if you are happy with my argument :) – rami_salazar Mar 8 '19 at 12:30