Maximum area under a curve by calculus of variations I am asked to find the function that has the maximal area for a given length L when x runs from -a to a. I calculated the integral to be varied as follows:
$$ \int_{-a}^{a}\  y + \lambda \sqrt{1 + (\frac{dy}{dx})^2} dx $$ 
Where lambda is a Lagrange multiplier associated with the length constraint on the curve. After invoking Beltrami's identity I ended up with the following DE but could not figure out how to solve it: 
$$ y + \lambda = C \sqrt{1+(\frac{dy}{dx})^2} $$
I would appreciate your help on this matter.
Edit: The answer given by Mathematica is as follows:
$$ \left\{\left\{y(x)\to \frac{1}{2} \left(C^2 e^{\frac{x}{C}-c_1}+e^{c_1-\frac{x}{C}}-2 p\right)\right\},\left\{y(x)\to \frac{1}{2} \left(C^2 e^{-c_1-\frac{x}{C}}+e^{c_1+\frac{x}{C}}-2 p\right)\right\}\right\} $$
 A: Actually after checking out "Mathematical Methods for Students of Physics and Related Fields" by Sadri Hassani and a website I concluded that the answer is a semicircle whose radius is given by $ \lambda^2 $
I am mirroring the treatment in the book by Hassani
The treatment is as follows,
The function to be minimized is
$$
\mathbf{L}[y] = \int_{-a}^{a}ydx 
$$ 
Under the following conditions and constraints
$$
y(-a) = 0 = y(a)
$$
$$ \mathbf{K}[y]=\int_{-a}^{a}\sqrt{1+y'^2}dx=L $$
Using Euler-Lagrange equation is more convenient in this case and it is given by:
$$
\frac{\partial L}{\partial x}-\frac{d}{dt}\frac{\partial L}{\partial \dot{x}} + \lambda \left(\frac{\partial G}{\partial x} - \frac{d}{dt}\frac{\partial G}{\partial \dot{x}}\right)=0
$$
In our case $L=y$ and $G=\sqrt{1+y'^2}$, substituting thse into above equation gives
$$
1 + \lambda \frac{d}{dx} \frac{y'}{\sqrt{1+y'^2}}=0
$$
Integrating wrt x yields
$$
x + \lambda \frac{y'}{\sqrt{1+y'^2}} = C_1
$$
Solve for $y'$
$$
y'=\pm \frac{C_1-x}{\sqrt{\lambda ^2-(C_1-x)^2}}=\frac{dy}{dx}
$$
Multiply each side by $dx$ and integrate
$$
y = \pm \sqrt{\lambda ^2-(C_1-x)^2}+C_2$$
Or
$$
(x-C_1)^2+(y-C_2)^2=\lambda^2
$$
The unknowns are to be determined from the boundary conditions and the length constraint.
A: Approximate the curve by short line segments, and complete a polygon with the straight line segment from -a to a. Then use the fact that, given the length of the sides, the polygon with largest area can be inscribed in a circle. Not calculus of variations, but gets you quickly to the answer.
