Optimization, cylinder in sphere

  • Thread starter Tclack
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  • #1
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Homework Statement


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.



Homework 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])


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?
 

Answers and Replies

  • #2
Dick
Science Advisor
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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?
 
  • #3
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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:
  • #4
ideasrule
Homework Helper
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[tex]r^2(R^2-r^2)=R^4-4R^2r^2+4r^4[/tex]
[tex]5r^2R^2-3r^4=R^4[/tex]

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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