Cylinder inscribed in sphere

  • Thread starter Mindscrape
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  • #1
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The problem is to find the radius and height of the open right circular cylinder of largest surface area that can be inscribed in a sphere of radius a. What is the largest surface area?

The open cylinder's surface area will be
[tex] f(h,r) = 2 \pi r h [/tex]

I am not really sure about the sphere, because I'm not really sure about the constraints that would apply. It looks like it would just be
[tex] a^2 = r^2 + h^2 + r^2 = 2r^2 +h^2[/tex], but this would only be if the cylinder is centered about the origin. But I guess that since it is a sphere, and perfectly symmetrical, then trying to squeeze the cylinder in diagonally would be the same as along the origin. Right?
 

Answers and Replies

  • #2
radou
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Mindscrape said:
... But I guess that since it is a sphere, and perfectly symmetrical, then trying to squeeze the cylinder in diagonally would be the same as along the origin. Right?

Yes, it is the same. Combine the constraint into the function and try to differentiate.
 
  • #3
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Huh, I got something funny:

[tex] \nabla f = \lambda \nabla g [/tex]
[tex] \nabla f = (2 \pi r)i + (2 \pi h)j[/tex]
[tex] \nabla g = (4r)i + (2h)j [/tex]
so
[tex] 2 \pi r = \lambda 4r[/tex]
[tex] 2 \pi h = \lambda 2h[/tex]
which means
[tex]\lambda = \frac{ \pi}{2} = \pi[/tex]???
 
Last edited:
  • #4
HallsofIvy
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No, I'm pretty sure [itex]\pi\ne \frac{\pi}{2}[/itex]!
 
  • #5
radou
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I think the constraint you set is wrong. It should be [tex]a^2 = r^2 + \left(\frac{h}{2}\right)^2[/tex], unless I'm missing something.
 
  • #6
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Ah crap, you are right. I just realized that I built the constraint wrong. But this constraint doesn't work either. The only one that would work would be [tex] a^2 = r^2 + h^2[/tex].
 
Last edited:

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