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The problem reads: " You are given a string of fixed length $\ell$ with one end fastened at the origin $O$, and you are to place the string in the $\left(x, y\right)$ plane with its other end on the x axis in such a way as to maximise the area between the string and the x axis. Show that the required shape is a semicircle ".

Hint: Show that the area can be written as $$J(y) = \frac{1}{c}\int_{0}^{\ell} y(s)\sqrt{1 + (y'(s))^2}ds$$ where $s$ is the arc-length along the string, and then use the following first integral of the $\mbox{Euler-Lagrange}$ equation $$ F - y'\,\frac{\partial F}{\partial y'} = C\quad\mbox{where}\quad y' = dy/ds\quad\mbox{and}\quad F = y\sqrt{1 + (y')^2} $$

My attempt at a solution:

I know that the rules of this site state that saying I really don't have a clue what to do is not allowed but I genuinely have been stuck on this for all of today and for hours last night as well. I know that if y were a function of x, then the following would work $$\frac{\triangle{s^2}}{\triangle{x^2}} = 1 + \frac{\triangle{y^2}}{\triangle{x^2}}$$ and then taking the limit as x tends to 0 would give $$ds = \sqrt{1 + (y')^2}dx$$

But due to the question stating that $y$ is a function of $s$, the arc-length. I have absolutely no clue where to go, I simply can not see how it is shown. Any help would be much appreciated, thank you !.

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  • $\begingroup$ $\mathrm{J}\left[y\right]$ doesn't enforce the restriction ( the string length $\ell$ is a constant ). $\endgroup$ – Felix Marin Oct 27 '16 at 1:06
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Your first equation seems to be wrong

The Area is given by $$J[y]=\int_0^\ell y dx$$

The length of the differential element: $$(ds)^2=(dx)^2+(dy)^2$$ $$dx=\sqrt{1-(y'(s))^2}ds$$ You get $$J[y]=\int_0^\ell y(s) \sqrt{1-(y'(s))^2}ds$$

Also see Isoperimetric problem in the calculus of variations

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  • $\begingroup$ Ahhh, now I see it. Thank you! $\endgroup$ – Aaron Curran Nov 5 '16 at 18:06

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