# Area between Smooth Curve and Line is Maximized by Semicircle

## 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 half-plane with an exception of endpoints, which are on the x-axis.

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

Then $y$ is a semicircle.

## Proof

It is described as follows:

if $\displaystyle \frac l \pi \le \lambda < \infty$ then
$y = \sqrt{\lambda^2 - x^2} - \sqrt{\lambda^2 - a^2}$
where
$l = 2 \lambda \, \map \arctan {\dfrac a {\sqrt {\lambda^2 - a^2} } }$;
if $\displaystyle \frac l {2 \pi} \le \lambda < \frac l \pi$ then
$y = \sqrt{\lambda^2 - a^2} - \sqrt{\lambda^2 - x^2}$
where
$l = 2 \lambda \paren{\pi - \map \arctan {\dfrac a {\sqrt {\lambda^2 - a^2} } } }$.

In the first case the area is a definite integral between $x = -a$ and $x = a$:

 $\displaystyle A$ $=$ $\displaystyle \int_{-a}^a \paren {\sqrt{\lambda^2 - x^2} - \sqrt{\lambda^2 - a^2} } \rd x$ $\displaystyle$ $=$ $\displaystyle \sqbrk {\frac 1 2 \paren {x\sqrt{\lambda^2 - a^2} + \lambda^2 \arctan {\dfrac x {\sqrt {\lambda^2 - x^2} } } } - x\sqrt{\lambda^2 - a^2} } \Bigg \vert_{-a}^a$ Primitive of Root of a squared minus x squared $\displaystyle$ $=$ $\displaystyle \frac 1 2 \paren {\lambda^2 \arctan \frac a {\sqrt{\lambda^2 - a^2} } - a \sqrt{\lambda^2 - a^2} }$

Solve the length constraint for $a$.

The solution, satisfying strict positivity of $l$ and the allowed values of $\lambda$, is:

$\displaystyle a = \frac {\lambda \map {\tan}{\frac l {2\lambda} } }{\sqrt{1 + \map {\tan^2}{\frac l {2\lambda} } } }$

Substitution of this into the expression for the area leads to:

$\displaystyle A = \frac \lambda 2 \paren {l - \lambda \sin \frac l \lambda}$

$l$ is a constant, while $\lambda$ is a free variable.

For the allowed values of $\lambda$, the area function is a continuous real function.

By Definition of Local Maximum, the maximum is either in a subinterval of domain of $A$ with respect to $\lambda$ or at one of the boundary values.

By Derivative at Maximum or Minimum, we have to find the extremum of $A$:

$\displaystyle \dfrac {\d A}{\d \lambda} = \frac \lambda 2 \paren {\frac l \lambda + \frac l \lambda \cos \frac l \lambda - 2 \sin \frac l \lambda}$

This vanishes at $\displaystyle\frac l \lambda = (2k + 1)\pi$ and $\displaystyle\frac l \lambda = 0$, where $k \in \Z$.

Then the area at these extremums acquires the following values:

$\displaystyle \map A {\frac l \lambda = 0} = 0$;
$\displaystyle \map A {\frac l \lambda = (2k + 1)\pi} = \frac {l^2} {\paren {1 + 2 k} 2\pi}$

This is maximized for $k = 0$, or $\displaystyle \frac l \lambda = \pi$.

Incidentally, these are also boundary values of the area function.

This concludes the maximization of area when the arc is being varied from a straight line to a semicirle.

The second part considers the variation of the curve from a semicirle to a circle.

In this case the area is that of a semicirle plus a part of the lower semicirle:

 $\displaystyle A$ $=$ $\displaystyle \pi \frac {\lambda^2}{2} + \int_{-\lambda}^{-a} \sqbrk{\sqrt{\lambda^2 - a^2} - \paren{\sqrt{\lambda^2 - a^2} - \sqrt{\lambda^2 - x^2} } } \rd x + \int_{-a}^{a} \sqbrk{\sqrt{\lambda^2 - a^2} } \rd x + \int_{a}^{\lambda} \sqbrk{\sqrt{\lambda^2 - a^2} - \paren{\sqrt{\lambda^2 - a^2} - \sqrt{\lambda^2 - x^2} } } \rd x$ $\displaystyle$ $=$ $\displaystyle \pi \lambda^2 + a \sqrt{\lambda^2 - a^2} - \lambda^2 \arctan \paren{\frac a {\sqrt{\lambda^2 - a^2} } }$ Primitive of Root of a squared minus x squared

Like in the previous case, solve the length constraint for $a$, while satisfying positivity and range conditions:

$\displaystyle a = \frac {\tan{\frac{2\pi\lambda - l}{2 \lambda} } }{\sqrt{1+\tan^2{\frac{2\pi\lambda - l}{2 \lambda} } } }$

Substitution into the area expression leads to:

$A = \frac {\lambda}{2} \paren{l - \lambda \sin \frac l \lambda}$

To find the extremum, compute its derivative with respect to $\lambda$:

$\displaystyle \dfrac {\d A} {\d \lambda} = \frac 1 2 \paren{l + l \cos \frac l \lambda -2 \lambda \sin \frac l \lambda}$

It vanishes if $\displaystyle \frac l \lambda = 0$ or $\frac l \lambda = \pi \paren{1 + 2 k}$, with $k \in \Z$.

From these solutions the one satisfying the range of $\lambda$ is $\frac l \lambda = \pi$.

The area for this value is $\displaystyle \frac {\pi \lambda^2} 2$.

For completeness we have to check the other boundary value in this range, namely, $\displaystyle \lambda = \frac l {2\pi}$.

$\displaystyle \map A {\lambda = \frac l {2\pi} } = \pi \lambda^2$.

Since we have length as an input for this problem, express both areas in terms of length.

$\displaystyle \map A {l = \pi \lambda} = \frac {l^2} {2 \pi}$
$\displaystyle \map A {l = 2 \pi \lambda} = \frac {l^2} {4 \pi}$

Hence, the area is maximized when the curve $y$ is a semicirle.

$\blacksquare$