Let's define a stair case polygon to be a polygon look like the following.
Let the lower left most point to be the origin, all edges are axis-aligned.
One can see if we remove a stair case polygon from a rectangle, we get another stair case polygon.
Let $S_k$ be some $k$ vertices stair case polygon. Let $A(x)$ be the area of the polygon x. Let $R_k$ be the largest axis-aligned rectangle that is contained in $S_k$
What can we say about the lower bound of $\frac{A(R_k)}{A(S_k)}$?
Currently I know $\frac{A(R_k)}{A(S_k)} \geq \frac{1}{(k-4)/2 + 1}$
Define the maximal rectangle as a rectangle that can't be stretched in any direction without going outside the polygon. There are only $(k-4)/2 + 1$ maximal rectangles, the union covers the entire polygon. Then at least one have area $\geq \frac{1}{(k-4)/2 + 1}$
I wonder does there exist a better lower bound? Or can we prove that's the best possible?
Motivation, I originally had a conjecture, that for any bounded continuous monotonic decreasing convex function defined in $[0,a]$ as $f(x)$. The maximal $xf(x)$ over the area under $f(x)$ is a constant. It turns out to be false, counterexample $f(x) = 1/x$. So now I'm doing this for a discrete version, the maximal $xf_k(x)$ over the area under $f_k(x)$ is a function of $k$, where $f_k(x)$ is the average value of $f(x)$ in each of the $k$ partitions of $[0,a]$.