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An old story I heard starts by two people that was arguing about how much land a man need. So they called to a young man, and said to him:

You are stating in that point, and start running all the day. All the area you cover before sunset will be yours. But you must return to the original point, otherwise you get nothing

So I considered what is the best route for him to get the maximal area.

More mathematical form:

Given a number d, what is the continues closed shape with the maximal area and perimeter d?

If we are limited to rectangles, the answer is "a square", what leads me to guess the optimal shape is a circle. Is it true and can it be proven?

What I can think about is that the shape must be convex, otherwise we can 'cut' it and save line. But I have no idea how to eliminate some strange elliptic shapes.

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marked as duplicate by hardmath, Lost1, TMM, Grigory M, Brian Rushton Jan 17 at 12:36

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
I think you mean "fixed perimeter," don't you? –  mjqxxxx Jan 16 at 19:06
    
@mjqxxxx Yes, thanks –  LeeNeverGup Jan 16 at 19:08
3  
It is the "circle", look at Isoperimetric inequality for more details. –  achille hui Jan 16 at 19:12
    
Surely this is a duplicate. (Posting from my phone.) –  hardmath Jan 17 at 1:46

2 Answers 2

up vote 2 down vote accepted

As @achille points out, the Isoperimetric inequality states that $$4\pi A\le L^2$$ Where $A$ denotes the area of a plane figure, and $L$ its perimeter. Equality holds only in the case of a circle.

Another consequence of this is that for shapes of a given area, a circle has the smallest perimeter. (the dual of your statement).

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It's not too hard to show the circle does better than the square for a given perimeter.

$ P_c = 2 \pi r $

$ P_{sq} = 4s $

The length of the run is equal in both cases

$2 \pi r = 4s $

$ r = \frac{2s}{\pi} $

$A_c = \pi r^2 = \pi (\frac{2s}{\pi})^2 = \frac{4s^2}{\pi} $

$ A_{sq} = s^2 $

Since $ \frac{4}{ \pi} > 1 $ we see the circle encloses a greater area for a given perimeter to the square.

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