Fix $L \in (0,\infty)$ and consider $\mathcal{C}\_L$ defined as follows:
\begin{align\*}
\mathcal{C}\_L := \{ \gamma:[0,1] \rightarrow \mathbb{R}^2 |~ \gamma \text{ is smooth and length($\gamma$)$=L$ }\}. 
\end{align\*}


I am interested in the "blow-up" of $\gamma$, denoted $\gamma\_{+r}$, defined as follows: For any set $S \subseteq \mathbb{R}^2$ and $r>0$
\begin{align\*}
S\_{+r} := \cup\_{z\in S}(z+r\mathbb{D}),
\end{align\*}
where $\mathbb{D}$ is the unit disc in $\mathbb{R}^2$ which is centred at the origin. So $\gamma\_{+r}$ is a bounded open set in $\mathbb{R}^2$. My question is for which $\gamma \in \mathcal{C}\_L$ is $m(\gamma\_{+r})$ maximised? Here $m(\cdot)$ is the Lebesgue measure in $\mathbb{R}^2$. I feel that it should be maximised by the line segment with length $L$.


If this is a version of some well known result, please do indicate it.


The reason for this title is that sometimes the isoperimetric inequality in $\mathbb{R}^2$ is stated as follows: For any Borel subset $A \subseteq \mathbb{R}^2$ with $m(A) < \infty$ and for every $\epsilon >0$, we have $m(A\_{+\epsilon}) \geq m(B\_{+\epsilon})$. Here $B$ is a Euclidean ball with $m(A) = m(B)$.


Thanks!



Isoperimetric type inequality in $\mathbb{R}^2$
The problem of determining the volume of the tubular neighborhood of a submanifold of Euclidean spaces was studied by Hotelling and Weyl (and then others). See <https://www.jstor.org/stable/2371513?seq=1>


The amazing result of Weyl is that, as long as the tubular neighborhood has no "overlaps" (in the sense of non-injective normal exponential map) the volume of the tubular neighborhood is **independent of (isometric) embedding**. (So in the one dimensional case the maximizer is non-unique, since all embeddings are isometric to each other.)


His theorem applies when the submanifold is closed (without boundary); but in your setting of a curve, the end-caps can be explicitly added.


