Variation of Dido's problem

Question

Find curve of given length $l<2 R$ that maximizes area enclosed between it and arc of circle radius $R$ while passing through two circle points $A,B$ not lying on the x-axis.

Answer 1

Too long for a comment, I would like to present my "line of attack". My best guess is that the supremum is attained for an arc of circumference from $A$ to $B$, long $2R$.

First note that the area between the segment $AB$ and the arc of circumference of radius $R$ does not depend on our choice of the candidate maximiser if the trial curve is not contained in the region between the segment $AB$ and the arc of radius $R$ : if we could satisfy this condition, the problem would boil then down to maximizing the area between the trial curve (of length $<2R$) and the segment $AB$.

Let us then consider a circumference going through $A$ and $B$, such that the length of one arc from $A$ to $B$ (named in the sequel “good arc”, to distinguish it from the other arc from $A$ to $B$, the “bad” arc) equals $2R$. By symmetry we can always place the "good" arc on the opposite side, with respect to the segment $AB$ and the arc from $A$ to $B$ of radius $R$ you draw on your diagram.

This circumference will have a radius $\hat{R}$, and it maximizes the contained areas among all curves of length $2\pi \hat{R}$, by Dido’s argument.

Let us now consider a trial curve, such that the “bad” arc is unaffected, and the “good” arc is deformed such that its length remains constant. Clearly the total area contained within the varied curve decreases. As the area between the segment $AB$ and the “bad” arc is unvaried by construction, we conclude that the area between the segment $AB$ and the deformed “good” arc decreased. Should be straightforward to prove that the "good arc" does not intersect the region between the segment $AB$ and the arc of radius $R$.

This proves the “good” arc is a maximiser. Your constraints $l <2R$ excludes it, so that it is just a supremum.