Calculus of Variation: Maximizing Volume & Min Area

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SUMMARY

The discussion centers on the calculus of variation, specifically addressing the problem of determining the geometric surface that maximizes volume while minimizing surface area. The key formulation involves maximizing the integral of a bounded region $\Omega$ in $\mathbb{R}^3$, subject to the constraint that the surface area of the boundary $\partial \Omega$ equals one unit. The conclusion drawn is that the optimal shape is a sphere, as it minimizes surface area for a given volume, a fact supported by the behavior of soap bubbles forming spherical shapes under similar constraints.

PREREQUISITES
  • Understanding of calculus of variations
  • Familiarity with geometric properties of spheres
  • Knowledge of non-linear partial differential equations
  • Basic concepts of measurable sets in $\mathbb{R}^3$
NEXT STEPS
  • Study the principles of calculus of variations in depth
  • Explore the derivation of the isoperimetric inequality
  • Learn about non-linear partial differential equations and their applications
  • Investigate the physical principles behind soap bubble formation
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Mathematicians, physicists, and engineers interested in optimization problems, particularly those involving geometric shapes and surface area considerations.

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How do I set up the following problem?

What geometric surface encloses the maximum volume with the minimum surface area?
 
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dwsmith said:
How do I set up the following problem?

What geometric surface encloses the maximum volume with the minimum surface area?

I think you want to rephrase it as, "What geometric surface encloses the maximum volume with a given surface area". In other words, you have 1 unit-squared of material and you want to enclose a surface to yields maximum volume.

Let $\Omega$ be bounded region in $\mathbb{R}^3$. You want to maximize,
$$ \iiint_\Omega 1 $$
Given the condition that,
$$ \iint_{\partial \Omega} 1 ~ ds = 1 $$

If we care about small details this question is a lot more complicated to phrase. Then we need to restrict ourselves to measurable sets such that .. blah blah blah.
------------

Here is another way to phrase it: Suppose $\Omega$ is a bounded (measurable) region in space with $\partial \Omega$ (boundary of $\Omega$) being a surface. Then,
$$ \text{Volume}(\Omega) \leq \frac{1}{6\sqrt{\pi}} \text{Area}(\partial \Omega)^{3/2} $$
 
Last edited:
ThePerfectHacker said:
I think you want to rephrase it as, "What geometric surface encloses the maximum volume with a given surface area". In other words, you have 1 unit-squared of material and you want to enclose a surface to yields maximum volume.

Let $\Omega$ be bounded region in $\mathbb{R}^3$. You want to maximize,
$$ \iiint_\Omega 1 $$
Given the condition that,
$$ \iint_{\partial \Omega} 1 ~ ds = 1 $$

If we care about small details this question is a lot more complicated to phrase. Then we need to restrict ourselves to measurable sets such that .. blah blah blah.

I don't think I want to rephrase it. The question is number 13 here
 
That's still a very badly phrased question. I think they expect you to combine the facts that the smallest surface containing a given volume is a sphere and that the largest area for a given size surface is a sphere.
 
HallsofIvy said:
That's still a very badly phrased question. I think they expect you to combine the facts that the smallest surface containing a given volume is a sphere and that the largest area for a given size surface is a sphere.

Then how would I show a sphere is the smallest surface containing a given volume?
 
This is a very complicated question without any "simple" answer, as the solution involves solving a non-linear partial differential equation of order 2.

The "best" answer that isn't too complicated I can think of is: soap bubbles are smarter than us, and they form spherical shells under such constraints.
 

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