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.

  • $\begingroup$ $r$ of the cone and of the sphere are not necessarily equal. $\endgroup$
    – user376343
    Mar 8, 2019 at 12:03

1 Answer 1


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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

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


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


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

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.

  • $\begingroup$ thanks. Quite insightful $\endgroup$ Mar 8, 2019 at 12:27
  • $\begingroup$ Thank you, please accept the answer if you are happy with my argument :) $\endgroup$ Mar 8, 2019 at 12:30

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