How much area in a unit square is not covered by $k$ disjoint disks of maximal area centered at random points within the square? 
1. Paint a $1\times 1$ square in blue.
2. Take $k$ points randomly and uniformly from the square.
3. Paint $k$ disks centered at each point in red. The radius of the disk centered at point $p$ is $d/2$, where $d$ is the distance from $p$ to its closest point.

What is the expected remaining blue area?
Related question:
How much area in a unit square is not covered by $k$ disks of area $1/k$ centered at random points within the square?

Like the related question you linked to, this question is difficult for finite $k$ but can be answered in the limit $k\to\infty$.
It is then akin to Stars in the universe - probability of mutual nearest neighbors, but it doesn't require the volume of a lune and thus has a simpler result.
A point can be painted red only due to the disk centre closest to it, and it's painted red due to that centre if and only if its distance from that centre is at most half the centre's distance to the nearest other centre.
So assume we have a Poisson process of centres with density $\lambda$, and fix some point $P$. The probability that there's no centre within a circle of radius $r$ around $P$ is $\exp(-\lambda\pi r^2)$. Thus the density for the nearest centre $C$ to be at distance $r$ from $P$ is $2\pi\lambda r\exp(-\lambda\pi r^2)$.
Conditional on $C$ being at distance $r$, $P$ is painted due to $C$ if and only if there's no other centre within $2r$ of $C$. We already know that there isn't one in a circle of radius $r$ around $P$, which lies entirely within the circle of radius $2r$ around $C$. Thus, given this condition, the probability that there's no other centre in the bigger circle is $\exp\left(-\lambda\left(4\pi r^2-\pi r^2\right)\right)=\exp(-3\lambda\pi r^2)$.
Thus the unconditional probability for $P$ to be painted is
\begin{align}
\int_0^\infty2\pi\lambda r\exp(-\lambda\pi r^2)\exp(-3\lambda\pi r^2)\,\mathrm dr&=2\pi\lambda\int_0^\infty r\exp(-4\lambda\pi r^2)\,\mathrm dr
\\
&=2\pi\lambda\left[-\frac1{8\pi\lambda}\exp(-4\lambda\pi r^2)\right]_0^\infty
\\
&=
\frac14\;,
\end{align}
so three quarters of the area is expected to remain blue.
Here's code for a simulation to confirm this result.
