# Isoperimetric problem in the calculus of variations

I'm trying to solve the following isoperimetric problem:

A plane curve has length $l$ and 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 $y' = dy/ds$. 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 semicircle.

I have managed to show that the area is given by the above integral by writing $A = \int_0^a ydx$ and using the fact that $ds^2 = dx^2 + dy^2$. Using the Euler-Lagrange equation, I found that the function $y(s)$ is given by $$y(s) = c\mathrm{sin}\left(\frac{s}{c} + k\right),$$ where $c$ and $k$ are constants. Then applying the conditions $y(0) = 0$ and $y(l) = 0$, I get $k = 0$ and $c = \frac{l}{\pi}$, so that $$y(s) = \frac{l}{\pi}\mathrm{sin}\left(\frac{s\pi}{l}\right)$$

This looks ok so far, but now I'm not sure how to find "the value of $a$ which maximises $A$", and I also don't know how to show that the function $y(s)$ gives a semicircle in the $x -y$ plane. Substituting $y(s)$ into $A$ and evaluating, I get $A = \frac{l^2}{2\pi}$, which is what would be expected for a semi-circle, but I don't think that this helps. Am I supposed to use the fact that $l = \int_0^a \sqrt{1 + y'^2}dx$ or something? I think I must be missing something obvious here.

You know that $$\left(\frac{dx}{ds}\right)^2 + \left(\frac{dy}{ds}\right)^2 = 1$$ and you can easily calculate $y'(s)$.
Does that suggest a way of finding $x(s)$? It should be clear enough after that.