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

In summary, the conversation discusses the difficulty of using functional derivative in R^3, specifically in solving the isoperimetric problem. The use of Lagrange multipliers is mentioned, but it is noted that it can only solve the problem in R^2. The conversation then shifts to finding a proof for the isoperimetric problem in R^3, with the realization that it is a difficult task. The suggestion of using the Brunn-Minkowski theorem is mentioned as a potential solution.
  • #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
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

##\ ##
 
  • #4
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
##
 
  • #5
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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  • #6
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
 

1. What is a 3-dimensional isoperimetric problem?

A 3-dimensional isoperimetric problem is a mathematical problem that involves finding the shape with the largest volume for a given surface area. In simpler terms, it is the problem of finding the most efficient way to enclose a certain amount of space.

2. What is the significance of a ball maximizing volume in a 3-dimensional isoperimetric problem?

The fact that a ball maximizes the volume in a 3-dimensional isoperimetric problem has important implications in various fields such as physics, engineering, and biology. It shows that nature tends to favor efficient and symmetrical shapes, and the sphere is the most efficient shape for enclosing a given volume.

3. How is the volume of a ball calculated in a 3-dimensional isoperimetric problem?

The volume of a ball can be calculated using the formula V = (4/3)πr^3, where r is the radius of the ball. This formula is derived from the fact that a ball is a three-dimensional shape with a constant radius, and the volume of a sphere is given by V = (4/3)πr^3.

4. Are there any real-life applications of the 3-dimensional isoperimetric problem?

Yes, the 3-dimensional isoperimetric problem has many real-life applications. For example, it can be used to determine the most efficient shape for a container, such as a water tank or a storage unit. It also has applications in biology, where it can be used to study the shape and structure of cells.

5. Is the ball the only shape that maximizes volume in a 3-dimensional isoperimetric problem?

No, the ball is not the only shape that maximizes volume in a 3-dimensional isoperimetric problem. Other shapes, such as the cube and the tetrahedron, also have the same volume-to-surface area ratio as a sphere. However, the sphere is the only shape that is perfectly symmetrical, making it the most efficient and preferred shape in many practical applications.

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