# 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 $\dfrac l \pi \le \lambda < \infty$ 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 $\dfrac l {2 \pi} \le \lambda < \dfrac l \pi$ 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 $x = a$:

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

Solve the length constraint for $a$.

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

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

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

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

$l$ 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 $A$:

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

This vanishes at $\dfrac l \lambda = \paren {2 k + 1} \pi$ and $\dfrac l \lambda = 0$, where $k \in \Z$.

Then the area at these extremums acquires the following values:

$\map A {\dfrac l \lambda = 0} = 0$;
$\map A {\dfrac l \lambda = \paren {2 k + 1} \pi} = \dfrac {l^2} {\paren {1 + 2 k} 2 \pi}$

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

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 semicircle.

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

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

 $\ds A$ $=$ $\ds \pi \frac {\lambda^2} 2 + \int_{-\lambda}^{-a} \paren {\sqrt {\lambda^2 - a^2} - \paren {\sqrt {\lambda^2 - a^2} - \sqrt {\lambda^2 - x^2} } } \rd x + \int_{-a}^a \paren {\sqrt {\lambda^2 - a^2} } \rd x + \int_a^\lambda \paren {\sqrt {\lambda^2 - a^2} - \paren {\sqrt {\lambda^2 - a^2} - \sqrt {\lambda^2 - x^2} } } \rd x$ $\ds$ $=$ $\ds \pi \lambda^2 + a \sqrt {\lambda^2 - a^2} - \lambda^2 \arctan \paren {\frac a {\sqrt {\lambda^2 - a^2} } }$ 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:

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

Substitution into the area expression leads to:

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

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

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

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

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

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

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

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

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

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

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

$\blacksquare$