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Cylinder inscribed in sphere

  1. Oct 16, 2006 #1
    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?
     
  2. jcsd
  3. Oct 16, 2006 #2

    radou

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    Yes, it is the same. Combine the constraint into the function and try to differentiate.
     
  4. Oct 16, 2006 #3
    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: Oct 16, 2006
  5. Oct 16, 2006 #4

    HallsofIvy

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    No, I'm pretty sure [itex]\pi\ne \frac{\pi}{2}[/itex]!
     
  6. Oct 16, 2006 #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.
     
  7. Oct 16, 2006 #6
    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: Oct 16, 2006
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