In a 3-dimension isoperimetric problem, a ball maximizes the volume

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  • #1
graphking
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TL;DR Summary
isoperimetric problem: in R^n, fix the surface area, when the volume can be max? the answer is ball, like B(0,1). in R^2 you can find many ways, such as using the variation of fixed end curve, functional derivative. But in R^3 I found it hard to use functional derivative (the equation get from derivative=0 is complicated, I can't further get to B(0,1))
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  • #2
BvU
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But in R^3 I found it hard to use functional derivative (the equation get from derivative=0 is complicated, I can't further get to B(0,1)
I suppose you are referring to Lagrange multipliers ?
If so, what's the problem ?

If not, please post your work

##\ ##
 
  • #4
graphking
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I suppose you are referring to Lagrange multipliers ?
If so, what's the problem ?

If not, please post your work

##\ ##
Using the lagrange multiplier is a way can only solve the isoperimetric problem in R^2, I show you the result you get in R^3:
assuming the surface is z(x,y), with fixed boundary in XOY plane, then
##(z_x/(1+z^2_x+z^2_y)^{1/2})_x+(z_y/(1+z^2_x+z^2_y)^{1/2})_y \equiv -1/\lambda
##
 
  • #5
martinbn
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Is there a question here? If i understand you correctly, you tried to use the ideas of one of the solutions to the problem in dimension two to solved it in higher dimensions, but you couldn't. So? The problem is not easy. May be this approach doesn't generalize or may be it does and you couldn't do it. Are you asking how it is done?
 
  • #6
graphking
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Is there a question here? If i understand you correctly, you tried to use the ideas of one of the solutions to the problem in dimension two to solved it in higher dimensions, but you couldn't. So? The problem is not easy. May be this approach doesn't generalize or may be it does and you couldn't do it. Are you asking how it is done?
please teach me a good way to proof the isoperimetric problem in R^3
 

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