From: Jamie Radcliffe <jradclif@math.unl.edu>
Subject: Re: Volume and surface
Date: Fri, 20 Aug 1999 10:31:20 -0500
Newsgroups: sci.math
Keywords: Isoperimetric inequality: definitions, pointers

The question of which convex body has the largest surface area for 
its volume can be given at least one sensible meaning, for which
the answer is indeed the tetrahedron [and more generally a simplex
in n dimensions].

To eliminate the "flat" examples we define the >shape< of a convex body
K to be the collection of all images of K under linear maps of 
determinant +1. These clearly all have the same volume as K. Now 
define 
	s(K) = min{ surface-area(H) : H in shape(K) }

This quantity is still measuring something about the surface
area of K, but without allowing examples which are merely squashed --
the definition unsquashes them.

Keith Ball, in Volume ratios and a reverse isoperimetric inequality. J. London
Math. Soc. (2) 44 (1991), no. 2, 351--359, proved that 
among all convex bodies of volume 1 the quantity s(K) is maximized 
by (any) simplex.

Cheers,
	Jamie Radcliffe
==============================================================================

From: cet1@cus.cam.ac.uk (Chris Thompson)
Subject: Re: Circle is maximum area for fixed perimeter?
Date: 9 Oct 1999 17:45:48 GMT
Newsgroups: sci.math

In article <7tnrks$fpt$1@pegasus.csx.cam.ac.uk>,
Dan Goodman <dog@fcbobDOTdemon.co.uk> wrote:

>I think that a circle is the shape which has maximum area for a fixed
>perimeter, but I have no idea how one might go about proving this, is there
>a simple proof, or does it require a great load of theory? If there is a
>simple proof, please could you post it. Thankyou very much in advance,

For an elementary proof that if there is a maximum, it must be the circle,
see Rademacher & Toeplitz "The Enjoyment of Mathematics" (Princeton 1957, 1966,
...; Dover 1990) [originally "Von Zahlen und Figuren"].

For a watertight proof that a maximum exists, you'll need a bit more 
background, but it's not all that deep.

Chris Thompson
Email: cet1@cam.ac.uk
==============================================================================

From: Fred Galvin <galvin@math.ukans.edu>
Subject: Re: Circle is maximum area for fixed perimeter?
Date: Tue, 12 Oct 1999 00:04:47 -0500
Newsgroups: sci.math
Keywords: proofs of isoperimetric inequality

On 9 Oct 1999, Chris Thompson wrote:

> In article <7tnrks$fpt$1@pegasus.csx.cam.ac.uk>,
> Dan Goodman <dog@fcbobDOTdemon.co.uk> wrote:
> 
> >I think that a circle is the shape which has maximum area for a fixed
> >perimeter, but I have no idea how one might go about proving this, is there
> >a simple proof, or does it require a great load of theory? If there is a
> >simple proof, please could you post it. Thankyou very much in advance,
> 
> For an elementary proof that if there is a maximum, it must be the
> circle, see Rademacher & Toeplitz "The Enjoyment of Mathematics"
> (Princeton 1957, 1966, ...; Dover 1990) [originally "Von Zahlen und
> Figuren"].
> 
> For a watertight proof that a maximum exists, you'll need a bit more 
> background, but it's not all that deep.

Gary Lawlor, "A new area-maximization proof for the circle", The
Mathematical Intelligencer vol. 20 no. 1 (Winter 1998), 29-31.
==============================================================================

From: Allan Adler <ara@zohar.ai.mit.edu>
Subject: Re: Circle is maximum area for fixed perimeter?
Date: 12 Oct 1999 14:51:31 -0400
Newsgroups: sci.math

There is a very nice proof in Hurwitz' collected works. He notes that
the closed curve will be the image of the circle and can be viewed as
a periodic function. So he starts with the fourier expansion of the
periodic function and expresses the perimeter and area in terms of
the Fourier coefficients. It then becomes obvious from these expressions
that the circle is the extremum.

I never tried to imitate this using spherical harmonics in higher
dimensions.

Allan Adler
ara@altdorf.ai.mit.edu

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