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Homework Help: Optimization, cylinder in sphere

  1. Oct 13, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the dimensions(r and h) of the right circular cylinder of greatest Surface Area that can be inscribed in a sphere of radius R.

    2. Relevant equations
    [tex]SA=2\pi r^2+2\pi rh[/tex]
    [tex]r^2 + (\frac{h}{2})^2 = R^2[/tex] (from imagining it, I could also relate radius and height with [tex]r^2 = h^2 +2R^2[/tex])

    3. The attempt at a solution
    [tex]SA=2\pi r^2+2\pi rh[/tex]

    [tex]r^2 + (\frac{h}{2})^2 = R^2[/tex]


    [tex]SA=2\pi r^2+4\pi r\sqrt{R^2-r^2}[/tex]

    [tex]\frac{dSA}{dr}=4\pi r+4\pi (\sqrt{R^2-r^2}+\frac{-2r^2}{2\sqrt{R^2-r^2}})[/tex]

    I tried setting that equal to zero, but I wasn't coming up with the right answer

    The answer in the book(not mine): [tex]r=\sqrt{\frac{5+\sqrt{5}}{10}}R[/tex]

    Can anyone see my error, or did I make one?
  2. jcsd
  3. Oct 13, 2009 #2


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    Homework Helper

    You haven't made any errors yet. I guess the error is in the part you didn't show us. How did you get a different answer from the book?
  4. Oct 14, 2009 #3
    Well, from
    [tex]\frac{dSA}{dr}=4\pi r+4\pi (\sqrt{R^2-r^2}+\frac{-2r^2}{2\sqrt{R^2-r^2}})=0[/tex]
    [tex]4\pi r+4\pi (\sqrt{R^2-r^2}-\frac{r^2}{\sqrt{R^2-r^2}})=0[/tex]
    [tex]4\pi r+4\pi (\frac{R^2-r^2-r^2}{\sqrt{R^2-r^2}})=0[/tex]
    [tex]-4\pir=4\pi (\frac{R^2-2r^2}{\sqrt{R^2-r^2}})[/tex]

    I seem to be going nowhere, I could square both sides

    But it doesn't clarify anything. Plus even, If I plug in [tex]\sqrt{\frac{5+\sqrt{5}}{10}}R[/tex] for r, I'm getting nothing.
    Last edited: Oct 14, 2009
  5. Oct 14, 2009 #4


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    Homework Helper

    You didn't simplify correctly. Aside from that, all you need to do now is use the quadratic equation to calculate r^2 (since r^2 = r^2 squared).
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