Not quite an answer: it is a result of Dutkovsky that the area of a rectangle containing a curve of perimeter $2\pi$ is at most $4.$

Not quite an answer: it is a result of Dutkovsky that the area of a rectangle containing a curve of perimeter $2\pi$ is at most $4.$

EDIT I have just learned (from Erwin Lutwak) that the result is NOT a result of Dutkovsky, but a result of Lutwak, as in: Lutwak, E. On isoperimetric inequalities related to a problem of Moser. Amer. Math. Monthly 86 (1979), no. 6, 476–477.