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

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SUMMARY

The discussion centers on the challenges of solving the isoperimetric problem in three dimensions (R^3) using Lagrange multipliers. Participants highlight the complexity of deriving the functional derivative and express the difficulty in extending methods from R^2 to R^3. A reference to the Brunn–Minkowski theorem is provided as a potential resource for proofs applicable to arbitrary dimensions. The conversation emphasizes the need for a deeper understanding of the mathematical principles involved in higher-dimensional geometry.

PREREQUISITES
  • Understanding of Lagrange multipliers in optimization
  • Familiarity with functional derivatives
  • Knowledge of the isoperimetric problem in R^2
  • Basic concepts of differential geometry
NEXT STEPS
  • Study the application of Lagrange multipliers in higher dimensions
  • Research the Brunn–Minkowski theorem and its implications
  • Explore advanced techniques in functional analysis
  • Investigate existing proofs of the isoperimetric problem in R^3
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Mathematicians, researchers in geometry, and students studying optimization techniques in higher dimensions will benefit from this discussion.

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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