(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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

2. Relevant equations

The Euler-Lagrange equation

3. The attempt at a solution

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

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

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# Homework Help: Isoperimetric problem

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