Set with small internal radius, small perimeter and prescribed area 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.
 A: Not a complete answer, but remark that if the minimizer exists, then it is not unique. Indeed, taking $n$ copies dilated by a factor $1/\sqrt{n}$ of a given set $E$ you do not change the area, the perimeter is increased by a factor of $\sqrt{n}$ and the inradius is decreased by the same factor.
Also, note that squares and discs of the same area have the same product perimeter times inradius; very thin rectangles do twice better than both these shapes. It might be that no minimizer exist, I am not sure at all.
At least, it is clear that this problem is in fact quite different from the isoperimetric problem.
A: Also not a complete answer, but if you do not require that $\partial E$ is connected, the answer is zero, because you can take a unit square, and remove $n^2$ circles centered in a grid with mesh $1/n$ that are of radius $1/n^3$ (so their influence on the perimeter is neglectable).
Finally, the argument of Benoît shows that even if you assume that the boundary is connected, it is \emph{not} the disk that is an optimizer. Indeed, you can take three circles that are tangent to each other. Due to Benoît's argument this shape has the same constant as a disk itself.
Now, this shape has a small concave empty triange inside. Now, one can change the corners of this triangle: "open" one (with an arbitrarily small perturbation) and add to $E$ the small, but noticable zones in two other ones, that does not increase the boundary area, does not increase the incircle radii (there still is sufficient boundary to stop them from growing), and increases the area of $E$.
A: 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 $c=1/2$: 
$$
|E|\le 2Rl= 2 R(E) \cdot |\partial E|.
$$
