Am I solving Dido's isoperimetric (variational) problem correctly? I am trying to solve Dido's isoperimetric problem, more specifically, the version where we have to maximize the area under a curve, given that the two endpoints are on the x-axis, and given a fixed arclength:

That is, we have to maximize
$$J(y)=\int_a^by(x)dx$$
subject to constraint
$$C(y)=\int_a^b\sqrt{1+(y'(x))^2}dx$$
Where we assume $a$ is fixed, but $b$ is allowed to vary.
Am I solving this correctly?
Besides the boundary constraint (because of variable boundary $b$) $L_{y'}(b)=0$, we of course formulate the Euler-Lagrange equation $L_y=\frac d {dx}L_{y'}$, based on $J$ subject to the constraint $C$, so that the Lagrangian becomes:
$$L(y,y')=y(x)+\lambda\sqrt{1+(y'(x))^2}$$
Therefore:
$$L_y=1$$
$$L_{y'}=\lambda \frac {y'(x)} {\sqrt{1+(y'(x))^2}}$$
$$\frac d {dx} L_{y'}=\lambda y''(x)\left(\frac {1} {\sqrt{1+(y'(x))^2}}-\frac {(y'(x))^2}{\left (1+(y'(x))^2\right)^{\frac 3 2}}\right)$$
which is equal to
$$\frac d {dx} L_{y'}=\lambda y''(x)\left(\frac {1}{\left (1+(y'(x))^2\right)^{\frac 3 2}}\right)$$

This gives the Euler-Lagrange equation $$\lambda
y''(x)=\left (1+(y'(x))^2\right)^{\frac 3 2}$$

I have no idea how to solve this ODE, and moreover, it doesn't seem like this is what I should be getting.
Did I derive this result correctly? If so, How do I solve it?
note: I know that it is also possible to solve by parameterizing $x=x(t), y(x)=y(x(t))$. I want to do this as well, later, but I would first like to understand the approach I'm taking here.
 A: You have derived the radius of circle correctly. Maybe you did not realize that no need to anything further than seeing how the particular situation constant$~ \lambda $ applies everywhere.
If point $b$ is moving w.r.t. point $a$ on x-axis we have to understand each segment situation in all the three cases with differing radii $\lambda, d $ symbolically given by a single relation in the last line of answer.

The Lagrange multiplier $\lambda$  is nothing but the associated geometrical invariant radius in this case.
If  distance $ (a-b) =d $ and arc length $L$ are given then we have three cases

*

*$ L= d$ gives optimal semi-circular area segment


*$ L> d$ gives optimal area of major segment


*$ L< d$ gives optimal area of minor segment
In the last two cases we need to iterate /numerically and find radius $\lambda$ from:
$$L=\lambda \cdot  2 \sin^{-1} \frac{d}{2\lambda}$$
which relates $\lambda$ with the given constant length quantities $d,l$.
A: Since the maximum configuration to Dido's isoperimetric problem is a semicircle, it is natural to use polar coordinates, which we will try in this answer.

*

*method: Polar coordinates $(r,\theta)$ with Lagrange multiplier $\lambda$. Infinitesimal arclength:
$$ (ds)^2~=~(dr)^2 + (rd\theta)^2.\tag{1a}$$
Area functional:
$$\begin{align}\widetilde{A} ~=~&A+\lambda (\ell-\ell_0)
~=~\int_0^{\pi} \! d\theta~L-\lambda\ell_0, \cr 
L~=~&\frac{r^2}{2}+\lambda\sqrt{r^2+\dot{r}^2}, \qquad 
\dot{r}~=~\frac{dr}{d\theta}. \end{align}\tag{1b}$$
Momentum:
$$ p~=~\frac{\partial L}{\partial\dot{r}}
~=~\frac{\lambda\dot{r}}{\sqrt{r^2+\dot{r}^2}}. \tag{1c}$$
Infinitesimal variation:
$$ \delta\widetilde{A} ~=~\int_0^{\pi}\! d\theta~ EL~\delta r + [p \delta r]^{\theta=\pi}_{\theta=0} +(\ell-\ell_0) \delta\lambda. \tag{1d}$$
This leads to natural/Neumann BC:
$$p(0)~=~0~=~p(\pi)\quad\Leftrightarrow\quad \dot{r}(0)~=~0~=~\dot{r}(\pi),\tag{1e}$$
and the constraint
$$ \ell~=~\int_0^{\pi} \! d\theta\sqrt{r^2+\dot{r}^2}~=~\ell_0\tag{1f}.$$
Energy is a constant:
$$\begin{align} E~=~p\dot{r}-L~=~&-\frac{\lambda r^2}{\sqrt{r^2+\dot{r}^2}}-\frac{r^2}{2}\cr
\Leftrightarrow\qquad \sqrt{r^2+\dot{r}^2}~=~& \frac{-\lambda}{\frac{E}{r^2}+\frac{1}{2}}.\end{align} \tag{1g}$$
The stationary points $\dot{r}=0$ of the profile $\theta\mapsto r(\theta)$ apparently satisfy the 2nd order equation
$$ 0~=~\frac{r^2}{2}+\lambda r+E, \tag{1h}$$
with roots
$$ r_{\pm}~=~-\lambda\pm\sqrt{D}, \qquad D~=~\lambda^2-2E. \tag{1i}$$
We conclude that
$$ r_-~\leq~r~\leq~r_+. \tag{1j}$$
Case $r_- < 0$. Note that $r=r_+$ whenever $\dot{r}=0$, i.e. at stationary points and endpoints. It follows that
$$ r~=~r_+ \tag{1k} $$
is a constant. $\Box$
Case $r_- \geq 0$. Then $E\geq 0$ and $\lambda\leq 0$. Hm. Eq. (1g) seems fairly complicated to solve. Let's try another approach...
Perturbative variation around constant profile. Let
$$ r(\theta)~=~r_0+\eta(\theta), \qquad r_0~=~\frac{\ell_0}{\pi}, \qquad \eta~\ll~\ell_0. \tag{1l} $$
Lagrangian to quadratic order:
$$ L_2~=~\frac{r_0^2}{2} +\lambda r_0 + (r_0+\lambda)\eta + \frac{\eta^2}{2} + \frac{\lambda}{2r_0}\dot{\eta}^2. \tag{1m} $$
Euler-Lagrange (EL) equation
$$\frac{\lambda}{r_0}\ddot{\eta}~=~\eta+(r_0+\lambda).\tag{1n}$$
BC:
$$ \dot{\eta}(0)~=~0~=~\dot{\eta}(\pi).\tag{1o}$$
Solution:
$$ \eta+(r_0+\lambda)~=~a\cos\left(\theta\sqrt{\frac{-\lambda}{r_0}} \right).\tag{1p}$$
In order for $r_0$ to be a stationary profile, apparently
$$ \lambda~=~-r_0. \tag{1q}$$
Then the fluctuation (1p) is a zero-mode at quadratic order.
More generally, we can consider a Fourier series
$$ \eta(\theta)~=~\sum_{n\in\mathbb{N}}a_n\cos(n\theta). \tag{1r}$$
It is a straightforward exercise to see that the higher modes $n>0$ lower the value of (the quadratic approximation to) the area functional (1b), while the first mode $n=1$ is a zero-mode.
The first mode $n=1$ presumably stops being a zero-mode at higher order. Hm. Let's try another approach...


*method: Radial coordinate $r$ as a function of arclength $s$. Area functional:
$$A ~=~\int_0^{\ell_0} \! ds~L, \qquad 
L~=~\frac{r}{2}\sqrt{1-\dot{r}^2}, \qquad 
\dot{r}~=~\frac{dr}{ds}. \tag{2a}$$
Momentum:
$$ p~=~\frac{\partial L}{\partial\dot{r}}
~=~-\frac{r\dot{r}}{2\sqrt{1-\dot{r}^2}}. \tag{2b}$$
Infinitesimal variation:
$$ \delta A ~=~\int_0^{\ell_0}\! ds~ EL~\delta r + [p \delta r]^{s=\ell_0}_{s=0} . \tag{2c}$$
This leads to natural/Neumann BC:
$$p(0)~=~0~=~p(\ell_0)\quad\Leftrightarrow\quad \dot{r}(0)~=~0~=~\dot{r}(\ell_0).\tag{2d}$$
Energy is a negative constant:
$$ E~=~p\dot{r}-L~=~-\frac{r}{2\sqrt{1-\dot{r}^2}}~<0. \tag{2e}$$
Define positive constant
$$ R~:=~-2E~>~0. \tag{2f}$$
Then the first integral reads
$$ \left(\frac{r}{R}\right)^2~=~1-\dot{r}^2.   \tag{2g}$$
So
$$0~<~r~\leq~R.\tag{2h}$$
Note that $r=R$ whenever $\dot{r}=0$, i.e. at stationary points and endpoints. It follows that
$$ r~=~R \tag{2i}$$
is constant. It is not hard to see that a semicircle (2i) is a maximum configuration to Dido's isoperimetric problem. $\Box$
A: First, integrate both sides with regards to $x$. Next, we can get an explicit function $x=f(y')+C_1$, $C_1$ is a constant. Here we have a fact : 
                        $y'(x)=0$ if and only if $x=C_1$
Geometrically, curve $y$ reaches the maximum value when $x=C_1$. Hence, $x$ increases for $x<C_1$ and x decrease for $x>C_1$. So, $y'(x)$ will never be zero when $x$ in $(0,C_1) \cup (C_1,1)$. Finally, with some algebraic manipulations, we can get a circle equation (with radius $\lambda$).
