Maximizing Area: Solving Isoperimetric Problem for Plane Curves

[Adorno]

Homework Statement

A plane curve with length l has its end points at (0, 0) and (a, 0) on the positive x-axis. Show that the area A under this curve is given by A = \int_0^l y \sqrt{1 - y'^2}ds, where y' = dy/dx, Find the function y(s) and the value of a which maximises A and in turn determine that the curve in the x - y plane is a semi-circle.

Homework Equations

The Euler-Lagrange equation

The Attempt at a Solution

I have managed to show that the area is given by that integral (by making the substitution ds^2 = dx^2 + dy^2). I applied the Euler-Lagrange equation to the integrand and I got the solution y(s) = c\mathrm{sin}(\frac{s}{c} + k), where c and k are arbitrary constants.

I think this is correct so far, but I'm unsure how to find the "value of a which maximises A", and I don't know how to express the curve in terms of x. Do we need to use some relationship between s and x?

 

[Kreizhn]

Well, you know that this curves goes between (0,0) and (a,0) right? So in particular, y(0)=0 and y(l) = 0. Those should allow you to get rid of some constants.

 

[Adorno]

Ok, so y(0) = 0 implies k = 0 and y(l) = 0 implies l/c = π, so c = l/π and y(s) = (l/π)sin(πs/l).

I'm still not entirely sure how I'm supposed to find the value of a that maximises the area or the equation in terms of x and y.

 

[Kreizhn]

It seems to me that you should be able to calculate l using a.

 

[Adorno]

Well, I know that l = \int_0^a \sqrt{1 + y'^2}dx but in order to evaluate this integral I would need to know y in terms of x, which is what I'm trying to find.

 

[Adorno]

Does anyone else have any ideas?