# Volume

• April 10th 2008, 09:58 PM
SengNee
Volume
A right circular cone is inscribed in a sphere of radius $a$ with the vertex and the circumference of the base of the cone touching the surface of the sphere.
Show that the volume of the cone is given by $I=\frac{1}{3}\pi (2ah^2-h^3)$ where $h$ is the height of the cone.
Hence, find the height of the cone when its volume is maximum.
• April 10th 2008, 10:27 PM
earboth
Quote:

Originally Posted by SengNee
A right circular cone is inscribed in a sphere of radius $a$ with the vertex and the circumference of the base of the cone touching the surface of the sphere.
Show that the volume of the cone is given by $I=\frac{1}{3}\pi (2ah^2-h^3)$ where $h$ is the height of the cone.
Hence, find the height of the cone when its volume is maximum.

In a right triangle the product of the two hypotenuse segments equals the square of the height. Euclid's theorem (That's the name how this theorem is called in Germany - maybe you know this theorem under a different name)

In comparison with the two sketches you see:

q = h ..... of the cone
p = 2a - h ..... of the cone

The volume of a cone is:

$V=\frac13 \pi r^2 \cdot h$

Therefore you get:

$V=\frac13 \pi (h(2a-h)) \cdot h~\implies~V=\frac13 \pi\left(2ah^2 - h^3\right)$

To find the maximum volume calculate

$V'(h) = \frac13 \pi \left(2ah - 3h^2\right)$

$V'(h) = 0~\implies~h = 0~\vee~\boxed{h = \frac23 a}$