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

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Discussion Overview

The discussion revolves around the isoperimetric problem in three dimensions, specifically focusing on the challenges of maximizing volume while maintaining a fixed surface area. Participants explore the application of mathematical techniques such as functional derivatives and Lagrange multipliers in this context.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant expresses difficulty in applying functional derivatives to the isoperimetric problem in R^3, noting that the resulting equations are complicated.
  • Another participant suggests that the original poster (OP) may be referring to Lagrange multipliers and asks for clarification on the specific issues encountered.
  • A different participant references a related problem in R^2 and implies that similar methods may not easily extend to R^3.
  • One participant questions whether the OP is asking how to solve the problem, suggesting that the difficulty may stem from the generalization of methods from lower dimensions.
  • Another participant mentions the Brunn–Minkowski theorem as a potential resource for proving the isoperimetric problem in arbitrary dimensions.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the applicability of methods from R^2 to R^3, and there is uncertainty regarding the effectiveness of Lagrange multipliers in this higher-dimensional context. The discussion remains unresolved regarding the best approach to the problem.

Contextual Notes

There are limitations in the discussion regarding the assumptions made about the generalization of techniques from lower dimensions and the specific mathematical steps involved in applying these methods in R^3.

graphking
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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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graphking said:
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

##\ ##
 
BvU said:
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
##
 
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?
 
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martinbn said:
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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