# Maximum Volume

• Nov 12th 2009, 10:21 AM
iheartthemusic29
Maximum Volume
A hemisphere of radius 7 sits on a horizontal plane. A cylinder stands with its axis vertical, the center of its base at the center of the sphere, and its top circular rim touching the hemisphere. Find the radius and height of the cylinder of maximum volume.

I have no clue how to go about solving this. Help, please?!?
• Nov 12th 2009, 10:36 AM
galactus
Draw a right triangle inside the sphere, then we can see that

$r^{2}+h^{2}=7^{2}$, where R is the radius of the sphere

and r and h are the radius and height of the cylinder, respectively.

$r^{2}=49-h^{2}$

The volume of the cylinder is $V={\pi}(49-h^{2})h$

$\frac{dV}{dh}={\pi}(49-3h^{2})$

Set to 0 and solve for h. The rest will follow.
• Nov 12th 2009, 10:48 AM
iheartthemusic29
Quote:

Originally Posted by galactus
Draw a right triangle inside the sphere, then we can see that

$r^{2}+h^{2}=7^{2}$, where R is the radius of the sphere

and r and h are the radius and height of the cylinder, respectively.

$r^{2}=49-h^{2}$

The volume of the cylinder is $V={\pi}(49-h^{2})h$

$\frac{dV}{dh}={\pi}(49-3h^{2})$

Set to 0 and solve for h. The rest will follow.

I tried what you said, and got 7(3^(1/2))/3 for h, which is incorrect, thereby making my value for r incorrect.
What did I do incorrectly?
• Nov 13th 2009, 09:30 AM
galactus
Well, solving dV/dh=0, we get $h=\frac{7}{\sqrt{3}}$

This gives $r=\frac{7\sqrt{6}}{3}$

That appears correct to me.

You have $h=\frac{7\sqrt{3}}{3}$

You do realize this is the same thing as $\frac{7}{\sqrt{3}}$.

The denominator has been rationalized in your version, but they are equivalent.