# Homework Help: Calculus of Variations; Maximum enclosed area problem.

1. Oct 26, 2016

### Aaron Curran

The problem reads: "You are given a string of fixed length l with one end fastened at the origin O, and you are to place the string in the (x, y) 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
$$J(y) = \frac{1}{c}\int_{0}^{l} y(s)\sqrt{1 + (y'(s))^2}ds$$
Where s is the arc-length along the string, and then use the following first integral of the Euler-Lagrange equation
$$F - y'\frac{\partial F}{\partial y'} = C$$
Where $y' = dy/ds$ and $F=y\sqrt{1 + (y')^2}$

My attempt at a solution:
I know that the rules of this forum 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!

2. Oct 26, 2016

### stevendaryl

Staff Emeritus
Your expression for $ds$ is confusing you. The relationship between $ds$ and $dx$ is:

$ds = \sqrt{1 + (\frac{dy}{dx})^2} dx$

What you wrote is incorrect, because $y' \equiv \frac{dy}{ds}$, not $\frac{dy}{dx}$. And what you want is $dx$ in terms of $ds$, not $ds$ in terms of $dx$. So how do you find $dx$ in terms of $ds$ and $dy$?

Now, you plug in your answer for $dx$ in the expression for the area under the curve, which is just $\int y dx$.

Last edited: Oct 26, 2016
3. Oct 26, 2016

### Ray Vickson

I don't understand the formulation. If I were doing it I would regard it as a maximization problem with a constraint. For a string of length $c$ we have the problem
$$\begin{array}{rl} \max &J = \int_0^{x_1} y \, dx \\ \text{subject to} & \int_0^{x_1} \sqrt{1 + y'^2} \, dx = c\\ \text{and}& y(0) = y(x_1) = 0 \end{array}$$
We could convert this to an unconstrained problem using a Lagrange multiplier method;
$$\max K = \int_0^{x_1} L(y,y') \, dx \equiv \int_0^{x_1} y \, dx + u \int_0^{x_1} \sqrt{1+y'^2} \, dx$$
with boundary conditions $y(0) = y(x_1) = 0$.
Here, the real constant $u$ is the Lagrange multiplier and $L(y,y') = y + u \sqrt{1 + y'^2}.$

4. Oct 26, 2016

### stevendaryl

Staff Emeritus
But you can convert $\int y dx = \int y \frac{dx}{ds} ds = \int y \sqrt{1-(\frac{dy}{ds})^2} ds$. Presumably, that is equivalent to doing the Lagrange multiplier thing.

5. Oct 29, 2016

### Ray Vickson

Yes, but that is not the formulation originally given in the question, which is to maximize $c^{-1} \int_0^{l} y \sqrt{1+y'(s)^2} \, ds$. When I said I did not understand the formulation, that is the one I was referring to.

6. Oct 29, 2016

### stevendaryl

Staff Emeritus
Oh, I didn't notice the sign. I think the sign of $y'$. I think the sign is wrong in the original post.