TITLE: Closed curve whose neighborhood is as large as possible
QUESTION [7 upvotes]: Let $C$ be a closed curve in the plane and let $N_\epsilon(C)$ be an $\epsilon$-neighborhood of $C$, like this: 
(ignore the fact that the "curve" is polygonal in this picture, I drew it in MATLAB)
My question is:  given fixed $\epsilon$, if we search among all closed curves $C$ of fixed length $\ell$, what is the optimal curve that makes the area of $N_\epsilon(C)$ as large as possible?  It seems very intuitive to me that the curve should just be a circle, which gives an area of $2\epsilon\ell$ provided that $\epsilon\leq \ell/(2\pi)$.

REPLY [6 votes]: Just to emphasize Thomas Richard's remark about smoothness, unless I've miscalculated, a $\frac{1}{4} L$-square leads to area
$$2 \epsilon L - \epsilon^2 (4-\pi) <  2 \epsilon L \;.$$

      

Added. This is to illustrate my "hypothesis" in the comments: Any smooth curve (not necessarily convex) such that its radius of curvature exceeds $\epsilon$ (at every point) leads to area $2 \epsilon L$.
In the right figure, too-sharp curvature leads to gaps between
the $\epsilon$-disk and the tube boundaries.