The Dido problem with an arclength constraint It is well known that the solution to the classical Dido problem is a semicircle, and that the solution to the classical isoperimetric problem is a circle. It's also reasonably obvious that the solution to the following variant is a circular arc: > Let $A$ and $B$ be fixed points on a plane, and let $l$ be a length greater than $\overline{AB}$. Which (smooth) curve through $A$ and $B$, of length $l$, maximises the area between itself and the line $AB$? It's a straightforward exercise to verify extremality using the calculus of variations, but are there alternative proofs that do not invoke e.g. the Euler-Lagrange equations? This was originally a homework problem with the isoperimetric inequality given as a hint, and I'm just wondering what the intended solution was...

This diagram might help:

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For any length $l > \overline{AB}$, we can find a circle passing through $AB$ such that the length of a circular arc between $A$ and $B$ is equal to $l$. In the diagram above, suppose that the red, dotted arc $ADB$ and the upper part of the circle both have length $l$. Note that the region bounded by $ADBC$ (i.e.: the pink area) has the same perimeter length as the circle.