Optimizations question with cylinder in sphere with a radius of 5

**s3a** · December 12th 2009, 07:01 AM

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

Answer: Volume ≈ 302

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!
Thanks in advance!

**shawsend** · December 12th 2009, 07:24 AM

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

**Danny** · December 12th 2009, 07:35 AM

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.

Answer: Volume ≈ 302

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!
Thanks in advance!

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

$<br /> V = 2\pi r^2 \sqrt{25-r^2}<br />$

**earboth** · December 12th 2009, 07:38 AM

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.

Answer: Volume ≈ 302

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!
Thanks in advance!

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 $V = \frac{500}9 \pi \sqrt{3}$

**Danny** · December 12th 2009, 08:53 AM

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.

**s3a** · December 12th 2009, 11:16 AM

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?

**earboth** · December 12th 2009, 10:10 PM

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?)

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