The Dido problem with an arclength constraint

Question

It is well known that the solution to the classical Dido problem is a semicircle, and that the solution to the classical isoperimetric problem is a circle. It's also reasonably obvious that the solution to the following variant is a circular arc:

Let $A$ and $B$ be fixed points on a plane, and let $l$ be a length greater than $\overline{AB}$. Which (smooth) curve through $A$ and $B$, of length $l$, maximises the area between itself and the line $AB$?

It's a straightforward exercise to verify extremality using the calculus of variations, but are there alternative proofs that do not invoke e.g. the Euler-Lagrange equations? This was originally a homework problem with the isoperimetric inequality given as a hint, and I'm just wondering what the intended solution was...

Answer 1

This diagram might help:

For any length $l > \overline{AB}$, we can find a circle passing through $AB$ such that the length of a circular arc between $A$ and $B$ is equal to $l$. In the diagram above, suppose that the red, dotted arc $ADB$ and the upper part of the circle both have length $l$. Note that the region bounded by $ADBC$ (i.e.: the pink area) has the same perimeter length as the circle.

Answer 2

While I'm not sure that this will perfectly solve the matter for arbitrary $l$, it's easy to get this in special cases, in a way that may point the way to a general solution (in particular, by continuity) : build an isosceles triangle on $AB$ with the other two sides both equal to the radius of the appropriate circle, and consider the 'ice-cream-cone' with your curve and that the two radii, and piece together however many of these it takes to form a circle (note that this only works for arcs that are integral divisions of the appropriate circle!). The isoperimetric inequality then says that the total area-maximizing curve is a circle, which implies that the section of curve between $AB$ that maximizes the wedge area is a circular arc, and since the area of the triangle is constant you can just subtract it out...

Answer 3

A proof that doesn't necessarily invoke the Euler-Lagrange equations was given by E. Schmidt. He uses two algebraic inequalities, the Green's theorem and property of parameterised curves as the only pre-requisites for the proof.

Another simple proof was given was Peter D. Lax who again doesn't use Euler Lagrange equations to prove the Isoperimetric Inequality.

A paper which discusses the Isoperimetric Inequality extensively including the history of a lot of proofs was presented by Robert Osserman.

One of the first proofs was presented by Jakob Steiner, but his proof was deemed to be incomplete by Weierstrass, who by a new formalized mathematical system showed the proof was not rigorous, in the sense that Steiner assumed the existence of a curve of the maximum area without actually proving it. Weierstrass himself then provided the first rigorous proof as a corollary to his Theory of Calculus of Several Variables in 1870.

References:

[1] Hehl, A., The Isoperimetric Inequality, presented at Eberhard-Karls Universitaet Tue- Bingen, 9th February 2013.

[2] Lax, P. D., A Short Path to the Shortest Path, The American Mathematical Monthly, 102:2(1995), 158{159.

[3] Siegel, A., A Historical Review of the Isoperimetric Theorem in 2-D, and its place in Elementary Plane Geometry, Retrieved April 1, 2012, from http://www.cs.nyu.edu/faculty/siegel/SCIAM.pd