Given a regular set $E\subset \mathbb R^2$ define
$$
 R(E) = \sup\{r\colon \exists x,\ B(x,r)\subseteq E\}
$$
to be the radius of the largest circle contained in $E$ and let $|\partial E|$ be the length of the perimeter of $E$.


Among all planar sets $E$ with fixed area, what is the one which minimizes the product $|\partial E| \cdot R(E)$?


Do you have any keyword to look for?


**edit:** Actually I'm looking for an estimate of the kind:
$$
R(E) \ge c \frac{|E|}{|\partial E|}
$$
for $E$ *simply connected*. And even a non optimal constant $c$ would be useful to me... I didn't realized that simply connectedness played a role. Sorry for that.



Set with small internal radius, small perimeter and prescribed area
For the simply connected version of your question: yes, such an estimate exists. Namely: take a radius $R=R(E)$ (closed) "coloring" disk, and let us move its center along the boundary $\partial E$. Note that while the center of the disk makes a path of length $l=|\partial E|$, the disk "colors" the area that is at most $2\pi Rl + \pi R^2$. Indeed, you count the initial area $\pi R^2$ plus estimate the increase in area per $\Delta l$ displacement of the disk as its boundary length $2\pi R$ times the displacement $\Delta l$.


(In fact, this upper bound can be improved to $2 R l$, but it requires a bit more words to be pronounced. Namely, you say that there are two disjoint discs, so you can count only increase terms -- area of difference between displaced and not displaced discs, -- and you estimate such an increase term more carefully as $2R\Delta l$. See also: analogous formulae in the theory of Minkowski sums/convex polyhedra, like the one for the area or the volume of $\varepsilon$-neighborhood.)


On the other hand, by definition of $R$ any point of $E$ is at distance at most $R$ from the boundary $\partial E$. Thus, all the set $E$ will be colored. Hence,
$$
|E|\le 2\pi Rl + \pi R^2
$$
and as $2\pi R\le l$, the $|E|\le (2\pi + \frac{1}{2}) R l$. So you can take $c=(2\pi + \frac{1}{2})^{-1}$.


In fact, the improved estimate mentioned earlier tells us that you can even take $c=1/2$: 
$$
|E|\le 2Rl= 2 R(E) \cdot |\partial E|.
$$


