Sphere in a cone?

Geometry Level 3

What is the minimum volume of a cone needed to fully contain a sphere? Report your answer as a ratio of volume of cone to that of sphere.

V c o n e / V s p h e r e {V_{cone}} / {V_{sphere}} = ?


The answer is 2.

Krishnaraj Sambath by Krishnaraj Sambath , 35, USA

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2 solutions

4 Helpful 1 Interesting 6 Brilliant 0 Confused
David Vreken
Feb 25, 2019

Let R R be the radius of the sphere tangent to the cone, r r be the radius of the cone, and h h be the height of the cone.

Then by similar triangles, R h − R = r r 2 + h 2 \frac{R}{h - R} = \frac{r}{\sqrt{r^2 + h^2}} , which solves to r 2 = R 2 h 2 h 2 − 2 R h r^2 = \frac{R^2h^2}{h^2 - 2Rh} , which means the volume of the cone is V c o n e = 1 3 π R 2 h 3 h 2 − 2 R h V_{cone} = \frac{1}{3}\pi \frac{R^2h^3}{h^2 - 2Rh} .

For the minimum V c o n e V_{cone} , the derivative V c o n e ′ = 0 V_{cone}' = 0 , which solves to h = 4 R h = 4R .

Therefore, r 2 = 2 R 2 r^2 = 2R^2 , and V c o n e = 8 3 π R 3 = 2 V s p h e r e V_{cone} = \frac{8}{3}\pi R^3 = 2V_{sphere} , so V c o n e V s p h e r e = 2 \frac{V_{cone}}{V_{sphere}} = \boxed{2} .

1 Helpful 0 Interesting 5 Brilliant 0 Confused

This volume function was much easier to minimize than mine!

On the left hand side of your first equation, you want a big R R in the denominator, not a small r r . R h − R = r r 2 + h 2 \frac{R}{h-R} = \frac {r}{\sqrt {r^2+h^2}}

Matthew Feig - 2 years, 3 months ago

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Thanks! I edited it.

David Vreken - 2 years, 3 months ago

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