An isoperimetric problem with the "wrong" perimeter There is a scarcity on the web of examples of solving constrained problems using the calculus of variations.  Yes, everybody shows how to prove the shortest distance between two points is a straight line, and you can find the classical isoperimetric problem and the brachistrone, but that seems to be about it.
As a practice problem, say you were asked to find the curve $y(x)$ of length $\pi$ going from $(x=-1,y=0)$ to $(x=+1,y=0)$ that maximizes the area between the curve and the $X$ axis:
$$A = \int_{-1}^{+1} y(x)\,dx \\
\int_{-1}^{+1} \sqrt{1+\left(\frac{dy}{dx} \right)^2}\,dx = \pi \\ 
\mbox{maximize}_{y(x)} A
$$
Because $\pi$ is precisely the length of a semicircle of radius $1$ centered on the origin, it is intuitively clear that the maximizing curve is that semicircle.  In fact, if some other curve of the same length  yielded a larger area, then the familiar unrestricted isoperimetric problem would be maximized by that curve attached to its own reflection about the $X$ axis, and the answer would not be a circle. No calculus of variations needed.
Now the real problem:  

For $2 < L < \pi$, find the curve $y(x)$ of length $L$ going from $(x=-1,y=0)$ to $(x=+1,y=0)$ that maximizes the area between the curve and the $X$ axis:
  $$A = \int_{-1}^{+1} y(x)\,dx \\
\int_{-1}^{+1} \sqrt{1+\left(\frac{dy}{dx} \right)^2}\,dx = L \\ 
\mbox{maximize}_{y(x)} A
$$

My intuition tells me that the answer will be a circular arc of the appropriate length, entered about some point on the negative $Y$ axis.  [Disclaimer -- my intuition could be wrong!] There may be a clever proof of this without using the calculus of variations,  but I'd like to see this problem tackled using the calculus of variations with the constraint realized via a Lagrange multiplier. 
(The reason I specified that $L < \pi$ is to avoid the potential of messiness of double-valued functions going outside the $-1 \leq x \leq 1$ strip.)
 A: In general, an extremum of $I(y) = \int_a^b F(x,y,y') dx$ amongst all $C^2$ functions $y$ with $y(a)=A, y(b)=B$, $J(y) = \int_a^b G(x,y,y') dx = L$ is an extremal of $$K(y)  = \int_a^b F + \lambda G dx$$ for some constant $\lambda$ - a Lagrange multiplier, as you say. You can find the details of the Lagrange multiplier argument worked out in Gelfand and Fomin's book on Calculus of Variations (iirc), or on p.14 here, or here
In your case the new functional $K$ is $\int_{-1}^1 y + \lambda \sqrt{1+y'^2} dx$. You can look for extremals in the usual way: Beltrami's equation is
$$ y + \lambda \sqrt{1+y'^2} - \lambda y'^2 / \sqrt{1+y'^2} = C$$ for some constant C, and if you tidy up and solve it you get the equation of a circle with centre on the $y$-axis at height $\sqrt{\lambda^2-1}$ and radius $\lambda$, where
$$ L = |2 \lambda \sin^{-1}(1/\lambda)|$$
You can avoid the arc length issue by working in polar coordinates - for bigger $L$ the solution will still be a circle, but $y$ won't be expressible as a function of $x$.
