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

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graphking
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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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 ?

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

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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
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