MHB Calculus of Variation: Maximizing Volume & Min Area

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The discussion centers on determining the geometric surface that maximizes volume while minimizing surface area, specifically under the constraint of a fixed surface area. It is suggested to rephrase the problem to focus on maximizing volume with a given surface area, emphasizing that a sphere achieves this optimal configuration. The conversation highlights that the relationship between volume and surface area is complex and involves advanced mathematical concepts, including non-linear partial differential equations. Additionally, it is noted that soap bubbles naturally form spherical shapes, illustrating the principle that the smallest surface enclosing a given volume is indeed a sphere. The problem remains challenging and poorly phrased, indicating a need for clarity in mathematical formulation.
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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.
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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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