1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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]h=2\sqrt{R^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]
    [tex]h=2\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

    Dick

    User Avatar
    Science Advisor
    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]
    [tex]-r{\sqrt{R^2-r^2}={R^2-2r^2}[/tex]

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

    [tex]r^2(R^2-r^2)=R^4-4R^2r^2+4r^4[/tex]
    [tex]5r^2R^2-3r^4=R^4[/tex]
    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

    ideasrule

    User Avatar
    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).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Optimization, cylinder in sphere
  1. Cylinder in a sphere (Replies: 4)

Loading...