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 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
, 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:


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


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

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 mapping $y$.

Corresponding Euler's Equation reads:


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

Integrating $x$ once yields:


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

Solve this for $y'$:


 * $\displaystyle y' = \pm \frac {c_1 - x} {\sqrt {\lambda^2 - c_1^2 + 2 c_1 x - x^2}}$.

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:

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 $\dfrac {\d l} {\d \lambda}$:


 * $\displaystyle \dfrac {\d l} {\d \lambda} = 2 \paren {\map {\arctan} {\frac 1 {\sqrt {\lambda^2 - 1} } } - \frac 1 {\sqrt {\lambda^2 - 1} } } < 0$.

Hence the domain of $l$ is determined by boundary values.

At the boundary of $\lambda = a$ we have:

To calculate the limit of such composite function denote:

It follows that


 * $\displaystyle \lim_{\lambda \to a^+} \map g {\lambda} = + \infty$


 * $\displaystyle \lim_{y \to \infty} \map f y = \frac \pi 2$

Arctangent is continuous for all $x \in \R$.

Then, by Limit of Composite Function:


 * $\displaystyle 2 a \lim_{\lambda \to a^+} \map {\arctan} {\frac a {\sqrt{\lambda^2 - a^2} } } = \pi a$.

At the boundary of $\lambda = + \infty$ we have:

In the second case it is:


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

Similarly to the previous case:


 * $\displaystyle \dfrac {\d l} {\d \lambda} = 2 \pi - 2 \map {\arctan} {\frac 1 {\sqrt {\lambda^2 - 1} } } + \frac 1 {\sqrt {\lambda^2 - 1} } > 0$

Therefore:


 * $\forall l \ge 2a ~\exists \lambda \ge a $.

Hence, within these constraints the real solution maximizing the area with fixed endpoints is an arc of a circle.