anonymous
  • anonymous
Why is it that circles have the largest enclosed area for a given perimeter?
Mathematics
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
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.
anonymous
  • anonymous
very cute! i asked, WHY ?
anonymous
  • anonymous
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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
it just happens to be ? just like that! nice! i dont have to do any proofs then!
anonymous
  • anonymous
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.
anonymous
  • anonymous
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.
anonymous
  • anonymous
http://en.wikipedia.org/wiki/Isoperimetric_inequality

Looking for something else?

Not the answer you are looking for? Search for more explanations.