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

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 an 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:
{{WLOG}}, 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:
:$\ds A \sqbrk y = \int_{-a}^a y \rd x$ 
Furthermore, $y$ has to satisfy the following conditions:
:$\map y {-a} = \map y a = 0$
:$\ds L \sqbrk y = \int_{-a}^a \sqrt {1 + y'^2} \rd x = l$
By Simplest Variational Problem with Subsidiary Conditions, there exists a constant $\lambda$ such that the functional:
:$\ds A \sqbrk y + \lambda L \sqbrk y = \int_{-a}^a \paren {y + \lambda \sqrt {1 + y'^2} } \rd x$
is extremized by the mapping $y$.
The corresponding Euler's Equation reads:
:$1 + \lambda \dfrac \d {\d x} \dfrac {y'} {\sqrt {1 + y'^2} } = 0$
Integrating {{WRT|Integration}} $x$ once yields:
:$x + \lambda \dfrac {y'} {\sqrt {1 + y'^2} } = C_1$
Solve this for $y'$:
:$\ds y' = \pm \frac {c_1 - x} {\sqrt {\lambda^2 - c_1^2 + 2 c_1 x - x^2} }$
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:
{{begin-eqn}}
{{eqn | l = \paren {-a - C_1}^2 
      | r = \lambda^2
}}
{{eqn | l = \paren {a - C_1}^2 
      | r = \lambda^2
}}
{{end-eqn}}
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, that is, at $\tuple {a, 0}$:
:$a^2 + C_2^2 = \lambda^2$
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 semicirle with two arcs of lower semicirle.
In the first case the length constraint is:
:$l = 2 \lambda \map \arctan {\dfrac a {\sqrt {\lambda^2 - a^2} } }$
For real $\lambda$, if $\lambda \ge a$, then $l \in \R$.
To find extremal values, consider the derivate $\dfrac {\d l} {\d \lambda}$:
{{begin-eqn}}
{{eqn | l = \dfrac {\d l} {\d \lambda} 
      | r = 2 \paren {\map \arctan {\frac 1 {\sqrt {\lambda^2 - 1} } } - \frac 1 {\sqrt {\lambda^2 - 1} } } 
}}
{{eqn | o = < 
      | r = 0
      | c = Tangent Inequality
}}
{{end-eqn}}
Hence the domain of $l$ is determined by boundary values.
At the boundary of $\lambda = a$ we have:
{{begin-eqn}}
{{eqn | l = \lim_{\lambda \mathop \to a^+} l 
      | r = \lim_{\lambda \mathop \to a^+} 2 \lambda \map \arctan {\frac a {\sqrt {\lambda^2 - a^2} } } 
}}
{{eqn | r = 2 a \lim_{\lambda \mathop \to a^+} \map \arctan {\frac a {\sqrt {\lambda^2 - a^2} } } 
      | c = Product Rule for Limits of Real Functions
}}
{{end-eqn}}
To calculate the limit of such composite function denote:
{{begin-eqn}}
{{eqn | l = \map f y
      | r = \map \arctan y
}}
{{eqn | l = \map g {\lambda}
      | r = \frac a {\sqrt {\lambda^2 - a^2} }
}}
{{end-eqn}}
It follows that:
:$\ds \lim_{\lambda \mathop \to a^+} \map g \lambda = + \infty$
:$\ds \lim_{y \mathop \to \infty} \map f y = \frac \pi 2$
Arctangent is continuous for all $x \in \R$.
Then, by Limit of Composite Function:
:$\ds 2 a \lim_{\lambda \mathop \to a^+} \map \arctan {\frac a {\sqrt {\lambda^2 - a^2} } } = \pi a$
At the boundary of $\lambda = + \infty$ we have:
{{begin-eqn}}
{{eqn | l = \lim_{\lambda \mathop \to \infty} l
      | r = \lim_{\lambda \mathop \to \infty} 2 \lambda \map \arctan {\frac a {\sqrt {\lambda^2 - a^2} } } 
}}
{{eqn | r = \lim_{\lambda \mathop \to \infty} 2 \frac {\map \arctan {\frac a {\sqrt {\lambda^2 - a^2} } } } {\frac 1 \lambda} 
      | c = Indeterminate limit $\infty \cdot 0$
}}
{{eqn | r = \lim_{\lambda \mathop \to \infty} 2 \frac {-\frac a {\lambda \sqrt {\lambda^2 - a^2} } } {-\frac 1 {\lambda^2} }
      | c = L'Hôpital's Rule
}}
{{eqn | r = \lim_{\lambda \mathop \to \infty} 2 a \frac \lambda {\sqrt {\lambda^2 - a^2} } 
}}
{{eqn | r = 2 a
}}
{{end-eqn}}
In the second case the length constraint is:
:$l = 2 \lambda \paren {\pi - \arctan \dfrac a {\sqrt {\lambda^2 - a^2} } }$
Similarly to the previous case:
{{begin-eqn}}
{{eqn | l = \dfrac {\d l} {\d \lambda} 
      | r = 2 \pi - 2 \map \arctan {\frac 1 {\sqrt {\lambda^2 - 1} } } + \frac 1 {\sqrt {\lambda^2 - 1} } 
}}
{{eqn | o = >
      | r = 0
      | c = Tangent Inequality
}}
{{end-eqn}}
Hence the domain of $l$ is determined by boundary values.
At the boundary of $\lambda = a$ we have:
{{begin-eqn}}
{{eqn | l = \lim_{\lambda \mathop \to a^+} l
      | r = 2 \lambda \paren {\pi - \arctan \dfrac a {\sqrt {\lambda^2 - a^2} } }
}}
{{eqn | r = 2 \pi a - \pi a
}}
{{eqn | r = \pi a
}}
{{end-eqn}}
As $\lambda$ approaches the infinity we have:
{{begin-eqn}}
{{eqn | l = \lim_{\lambda \mathop \to \infty} l
      | r = \lim_{\lambda \mathop \to \infty} 2 \lambda \paren {\pi - \arctan \dfrac a {\sqrt {\lambda^2 - a^2} } }
}}
{{eqn | r = \lim_{\lambda \mathop \to \infty} 2 \lambda \pi - 2 a
}}
{{eqn | r = \infty
}}
{{end-eqn}}
Therefore:
:$\forall l \ge 2 a: \exists \lambda \ge a$
Hence, within these constraints the real solution maximizing the area with fixed endpoints is an arc of a circle.
{{qed}}