# Thread: Optimizations question with cylinder in sphere with a radius of 5

1. ## Optimizations question with cylinder in sphere with a radius of 5

Question: A right cylinder is inscribed in a sphere with a radius of 5. Determine the maximum volume of the cylinder.

I'm having problems visualizing and even drawing this thing in the first place. The diameter of the sphere is not equal to the height cylinder, right? Apparently, I need to construct a right angle triangle in there but what do I do then?

Any help would be GREATLY appreciated!

2. This one is kinda' related. Read through it and see if you can construct a route to your problem:

http://www.mathhelpforum.com/math-he...al-volume.html

3. Originally Posted by s3a
Question: A right cylinder is inscribed in a sphere with a radius of 5. Determine the maximum volume of the cylinder.

I'm having problems visualizing and even drawing this thing in the first place. The diameter of the sphere is not equal to the height cylinder, right? Apparently, I need to construct a right angle triangle in there but what do I do then?

Any help would be GREATLY appreciated!
If you use symmetry then you can visualize it as a rectangle inscribed in a circle, the revolve the about the $z$-axis.

For a certain radius of cylinder r, the height will touch the sphere $r^2 + z^2 = 25 \; \text{or}\; z = \sqrt{25-r^2}$ (X2 for the full height)

Thus, what you wish to maximize is

$
V = 2\pi r^2 \sqrt{25-r^2}
$

4. Originally Posted by s3a
Question: A right cylinder is inscribed in a sphere with a radius of 5. Determine the maximum volume of the cylinder.

I'm having problems visualizing and even drawing this thing in the first place. The diameter of the sphere is not equal to the height cylinder, right? Apparently, I need to construct a right angle triangle in there but what do I do then?

Any help would be GREATLY appreciated!
1. See attachment.

2. With your question R = 5.

3. The volume of a cylinder is $V = \pi r^2 \cdot h$

$r^2 + \frac14 h^2 = R^2~\implies~r^2 = R^2 - \frac14 h^2$

4. You'll get the volume of the cylinder as a function wrt h:

$V(h) = \pi R^2 \cdot h - \frac14 \pi h^3$

5. Differentiate V(h) wrt h and solve the equation V'(h) = 0 for h. I've got $h = \frac23 \sqrt{3} \cdot R$ (Of course there is a negative solution too, but a negative radius of a sphere isn't very plausible)

6. Plug in this value into the equation of the function:

$V\left(\frac23 \sqrt{3} \cdot R\right) = \pi R^2 \cdot \frac23 \sqrt{3} \cdot R - \frac14 \left( \frac23 \sqrt{3} \cdot R \right)^3$

7. The exact value is $V = \frac{500}9 \pi \sqrt{3}$

5. Originally Posted by earboth
1. See attachment.

2. With your question R = 5.

3. The volume of a cylinder is $V = \pi r^2 \cdot h$

$r^2 + \frac14 h^2 = R^2~\implies~r^2 = R^2 - \frac14 h^2$

4. You'll get the volume of the cylinder as a function wrt h:

$V(h) = \pi R^2 \cdot h - \frac14 \pi h^3$

5. Differentiate V(h) wrt h and solve the equation V'(h) = 0 for h. I've got $h = \frac23 \sqrt{3} \cdot R$ (Of course there is a negative solution too, but a negative radius of a sphere isn't very plausible)

6. Plug in this value into the equation of the function:

$V\left(\frac23 \sqrt{3} \cdot R\right) = \pi R^2 \cdot \frac23 \sqrt{3} \cdot R - \frac14 \left( \frac23 \sqrt{3} \cdot R \right)^3$

7. The exact value is $V = \frac{500}9 \pi \sqrt{3}$
This is much easier than mine same answer though.

6. Now, I get how to do it and I get the final answer! But...what is the restriction on r (cylinder's radius) to prove that it is indeed the maximum. Would the 1st and second derivative tests work as well? Or would those just tell me that it is a local maximum or minimum and not if it is the absolute maximum or minimum?

7. Originally Posted by s3a
Now, I get how to do it and I get the final answer! But...what is the restriction on r (cylinder's radius) to prove that it is indeed the maximum. Would the 1st and second derivative tests work as well? Or would those just tell me that it is a local maximum or minimum and not if it is the absolute maximum or minimum?
1. The 2nd derivative is negative for all positive values of h. Thus you have found a maximum.

2. The radius of the cylinder varies between 0 and R: If r = 0 then the volume of the cylinder is 0, which is quite a minimal value for a volume. If r = R then the volume of the cylinder is 0 too, which is another minimal value for a volume. Since the V-function is continuous and differentiable you know that between 2 minimums (is that the correct expression? It sounds like a very small mother) is a maximum (big mother?)