From: Fred W. Helenius <fredh@ix.netcom.com>
Subject: Re: Maximal area for a given perimeter
Date: Thu, 30 Nov 2000 23:55:47 -0500
Newsgroups: sci.math
Summary: [missing]

dw22@netvis.com (Dan Wolf) wrote:

>How can one prove that of all sets in the plane with a given
>perimeter the circle has the maximal area ?

This theorem was first established by Jakob Steiner, who proved it
in several different ways.  I'll describe one of his proofs.  It
would probably be easier to follow with some diagrams; try to draw
them yourself.

What properties must the largest region enclosed by a given
perimeter have?  First, it must be convex.  If it were not convex,
there would be a line segment between two points on the perimeter
lying outside the enclosed region; replacing the cut-off section
of the perimeter with its reflection in the segment would increase
the area enclosed without changing the perimeter.

Choose any point A on the perimeter, and draw the line between it
and the corresponding point B halfway around the perimeter.  Line
AB divides the region into two subregions, each bounded by a line
segment and a curve whose length is half the given perimeter.
Each subregion must have the maximum possible area for a region
bounded by a line segment and this half-perimeter; if not, we
could construct a larger region with the original perimeter by
pasting two maximal subregions together.

Considering one of the subregions, choose a point C on the curved
part of its boundary.  The region can now be divided into three
smaller pieces:  the triangle ABC, the region R bounded by part of
the curve and segment AC, and the region S bounded by the rest of
the curve and segment BC.  A family of regions can be formed using
the same regions R and S, but with different angles at C; each is
bounded by a line segment and a half-perimeter curve, but may have
a different area.  Since the region we have is supposed to be
maximal, angle C must take on the value that maximizes the area of
this family of regions.  Regions R and S do not change, so only the
area of triangle ABC needs to be maximized.  Its area is given by
1/2 a b sin C, where a = BC and b = AC.  As a and b are constant,
the area is maximized when C is a right angle.

Thus, for any point C on the curve between A and B, angle ACB is a
right angle; therefore the curve is a semicircle.  Finally, the
boundary of the original region consists of two semicircles joined
along their diameters; in other words, a circle.

-- 
Fred W. Helenius	<fredh@ix.netcom.com>