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manoranjan

  • 3 years ago

Why is it that circles have the largest enclosed area for a given perimeter?

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  1. someone1348
    • 3 years ago
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    You start by defining a two dimensional polygon of n sides. The minimum is n=3, which is for a triangle. You then define the series that defines the area in terms of n. Then you notice that when n tends to infinity the area tends to a given value that is the maximal possible area. And well, a polygon with infinite sides is a circle, if you can "see". The calculations a re a bit longer.

  2. manoranjan
    • 3 years ago
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    very cute! i asked, WHY ?

  3. someone1348
    • 3 years ago
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    Mathematically, because when n tends to infinity (when the polygon tends to a circle) the perimeter and the area tend to a value, and the value of the area happens to be maximal

  4. manoranjan
    • 3 years ago
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    it just happens to be ? just like that! nice! i dont have to do any proofs then!

  5. someone1348
    • 3 years ago
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    Hang on then.... Imagine a regular polygon of "radius" 1 and n sides, that is, the distance from the barycenter to each corner. The area of the polygon in terms of n is given by \[\frac{n}{2}\sin(\frac{2 \pi}{n}) \] Which tends to pi, the maximum value of the sequence, when n tends to infinity, that is when the polygon tends to a circle. Happy now ? As for getting to that formula, you can do it yourself, this site is for help, not for people doing all of your schoolwork.

  6. abtrehearn
    • 3 years ago
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    Polygons are not the only thing to consider. We must look at all simple closed curves, also known as Jordan curves. These curves can be smooth throughout, like an ellipse. The proof you seek that the circle fits the bill is quite subtle. Look up the isoperimetric problem for more details. One place to start is mathworld.wolfram.com.

  7. estudier
    • 3 years ago
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    http://en.wikipedia.org/wiki/Isoperimetric_inequality

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