 Maximum-area circular segment with fixed arc length and chord

Fix two attachments a distance $2a$ apart on a straight line and a simple fence of length $2l$ in one half-plane, with $a<l<\pi a/2$. There is a unique $\theta \in (0,\pi /2)$ satisfying $l/a=\theta /\sin \theta$: its derivative has numerator $\sin \theta -\theta \cos \theta >0$, since that numerator starts at zero and has derivative $\theta \sin \theta >0$. A radius-$R$ minor circular arc with half-angle $\theta$ then has the required chord and length. Its graph is $y=\sqrt{R^2-x^2}-\sqrt{R^2-a^2}$, for $-a\le x\le a$, and sector-minus-triangle geometry gives the displayed area.

To prove global maximality, attach the complementary major arc of this circle, lying in the opposite half-plane, to any competing fence. The resulting closed curve has perimeter $2l+(2\pi R-2l)=2\pi R$. Its enclosed area is the competitor's area plus a fixed major-segment area. The planar isoperimetric inequality bounds the total by $\pi R^2$, with equality only for a circle. Subtracting that fixed added area proves the minor arc is the unique maximizing shape on the chosen side. The same arc follows from the Beltrami identity for the area-minus-length functional.