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I think the answer is a circle. If so, then what is the rigorous prove?

Self-Made Man
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4 Answers4

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user211299
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unclesteeg
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  • "Since $\epsilon$ can be anything, that implies ... that $r_{max}$ is constant". How so? This does not seem rigorous to me. – thedude Oct 10 '23 at 14:45
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You are correct. It follows from the Isoperimetric Inequality.

Michael Albanese
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Using polar coordinates and a Lagrange multiplier, you want to minimize the functional $$ \oint (\frac{r^2}{2}-\lambda r)d\theta.$$

The Euler-Lagrange equation is simply $$\frac{\partial}{\partial r}(\frac{r^2}{2}-\lambda r)=0$$ or $$ r=\lambda,$$ which is the equation of a circle.

thedude
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A circle gives the maximum area for a given perimeter.

So the triangle that gives the maximum area for a given perimeter is an equilateral triangle. The quadrilateral that gives the maximum area for a given perimeter is a square. The pentagon that gives the maximum area for a given perimeter is a regular pentagon. The n-gon that gives the maximum area for a given perimeter is a regular n-gon.

Additionally, as the number of sides, n, grows in a regular n-gon, the area will increase for a given perimeter. Thus of a square and an equilateral triangle with the same perimeter, the square will have more area. But of a square and a regular pentagon with the same perimeter, the pentagon will have the larger area.

So the most efficient shape for any perimeter is a circle. This is because you could imagine a circle as a regular n-gon with an infinite number of sides.

Plasman
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