Consider a unit square that is covered by finitely many (potentially overlapping) rectangles of various sizes, with sides parallel to the sides of the square. We are allowed to choose some of these rectangles, so that the chosen rectangles are disjoint. Our goal is to maximize the total area $A$ of the chosen rectangles. Is there a (positive) value of $c$ such that we can always achieve $A \geq c$?
Because each rectangle contains the center of the square, we can only choose 1 rectangle, which has area a little over $\frac14$. There are also situations where we necessarily end up with an even smaller value of $A$. In the right image, a plus sign is covered with 8 rectangles. Again, we can only choose 1 of 8 rectangles, which gives us $15\%$ of the area of the plus sign. Since many plus signs can almost completely cover a square, this shows that we cannot have $c>0.15$. By tweaking the shape of the plus sign, we can even show that we must have $c<\frac18=0.125$.
My question here is whether such a positive value of $c$ exists at all. The above examples show that if it exists, it must be $<\frac18$.
Is there a $c>0$ such that, given any covering of a unit square with finitely many axis-parallel rectangles, it is possible to choose disjoint rectangles with total area at least $c$?
I do not know the answer to this question.