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