= Maximum-area circular segment with fixed arc length and chord
{title2=$R=a/\sin\theta,\quad l/a=\theta/\sin\theta,\quad A_{\max}=R^2(\theta-\sin\theta\cos\theta)$}
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 $-a\leq x\leq a$, 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.
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