Area between Smooth Curve and Line with Fixed Endpoints is Maximized by Arc of Circle

Theorem
Let $y$ be a smooth curve, embedded in 2-dimensional Euclidean space.

Let $y$ have a total length of $l$.

Let it be contained in the upper halfplane with exception of endpoints, which are on the x-axis and are given.

Suppose, $y$, together with a line segment connecting $y$'s endpoints, maximizes the enclosed area.

Then $y$ is an arc of a circle.

Proof
Without loss of generality, we choose our point of reference such that $y$ intersect x-axis at points $\tuple {-a, 0}$ and $\tuple {a, 0}$ for some $a > 0$.

The area below the curve $y$ is a functional of the following form:


 * $\displaystyle A \sqbrk y = \int_{-a}^a y \rd x$

Furthermore, $y$ has to satisfy the following conditions:


 * $\displaystyle \map y {-a} = \map y a = 0$


 * $\displaystyle L \sqbrk y = \int_{-a}^a \sqrt{1 + y'^2} \rd x = l$

By Simplest Variational Problem with Subsidiary Conditions, there exists a constant $\lambda$ such that the functional:


 * $\displaystyle A \sqbrk y + \lambda L \sqbrk y = \int_{-a}^a \paren {y + \lambda \sqrt {1 + y'^2} }\rd x$

is extremized by the function $y$.

Corresponding Euler's Equation reads:


 * $\displaystyle 1 + \lambda \dfrac \d {\d x} \frac {y'} {\sqrt{1 + y'^2} } = 0$

Integrating $x$ once yields:


 * $\displaystyle x + \lambda \frac {y'} {\sqrt{1 + y'^2} } = C_1$

Integration yields:


 * $\paren {x - C_1}^2 + \paren {y - C_2}^2 = \lambda^2$

This is an equation for a circle with radius $\lambda$ and center $\tuple {C_1, C_2}$.

To find $C_1, C_2, \lambda$, apply boundary conditions and the length constraint.

From the boundary conditions we have that:


 * $\displaystyle \paren {- a - C_1}^2 = \tuple {a - C_1}^2 = \lambda^2$.

Take the difference of these two equations:


 * $4 a C_1 = 0 \implies C_1 = 0$

because $a>0$.

Apply one of the boundary conditions again, i.e. $\tuple{a,0}$:


 * $a^2 + C_2^2 = \lambda^2$

Then:


 * $C_2 = \pm \sqrt {\lambda^2 - a^2}$.

which can be used to get rid of $C_2$.

The last parameter to find is $\lambda$.

We have two cases:


 * the curve is an arc of the upper semicirle


 * the curve is a union of upper semicircle with two arcs of lower semicircle

In the first case the length constraint is:

$\displaystyle l = 2 \lambda \map {\arctan} {\frac a {\sqrt{\lambda^2 - a^2} } }$

It has solutions for $2 \le l \le \pi$.

In the second case it is:


 * $\displaystyle l = 2 \lambda \paren {\pi - \arctan \frac{a}{\sqrt{\lambda^2 - a^2} } }$

It has solutions for $\pi \le l < \infty$.