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?