= Circular arc maximizes area below a fixed chord
For a single-valued <curve> below the chord joining $(-a,0)$ to $(a,0)$, with length $2a<L\leq\pi a$, the maximum enclosed area is attained by the lower minor <circular arc>. Equivalently it minimizes $\int_{-a}^a y\,dx$ at fixed length. Its equation and length are
$$
y=\sqrt{R^2-a^2}-\sqrt{R^2-x^2},\qquad
R\geq a,\qquad L=2R\arcsin(a/R).
$$
To prove global minimality for $R>a$, augment the objective by $R\int\sqrt{1+y'^2}\,dx$. The augmented integrand is linear in $y$ and strictly convex in $y'$. The displayed arc satisfies its <Euler-Lagrange equation>, so the supporting-line inequality for the convex slope dependence, followed by <integration by parts>, proves it minimizes the augmented <functional> among <curves> with the prescribed endpoints. On the fixed-length class the multiplier term is constant. The <semicircle> endpoint follows by a limit; its endpoint tangents are vertical but its length is finite.
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