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manoranjan
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Why is it that circles have the largest enclosed area for a given perimeter?
 3 years ago
 3 years ago
manoranjan Group Title
Why is it that circles have the largest enclosed area for a given perimeter?
 3 years ago
 3 years ago

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someone1348 Group TitleBest ResponseYou've already chosen the best response.0
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.
 3 years ago

manoranjan Group TitleBest ResponseYou've already chosen the best response.0
very cute! i asked, WHY ?
 3 years ago

someone1348 Group TitleBest ResponseYou've already chosen the best response.0
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
 3 years ago

manoranjan Group TitleBest ResponseYou've already chosen the best response.0
it just happens to be ? just like that! nice! i dont have to do any proofs then!
 3 years ago

someone1348 Group TitleBest ResponseYou've already chosen the best response.0
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.
 3 years ago

abtrehearn Group TitleBest ResponseYou've already chosen the best response.2
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.
 3 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
http://en.wikipedia.org/wiki/Isoperimetric_inequality
 3 years ago
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