Find the minimum possible value of the largest of $xy$, $1-x-y+xy$, and $x+y-2xy$ if $0\leq x \leq y \leq 1$.
We claim that the minimum is $\frac{4}{9}.$  When $x = y = \frac{1}{3},$
\begin{align*}
xy &= \frac{1}{9}, \\
(1 - x)(1 - y) &= \frac{4}{9}, \\
x + y - 2xy &= \frac{4}{9}.
\end{align*}The rest is showing that one of $xy,$ $(1 - x)(1 - y),$ $x + y - 2xy$ is always at least $\frac{4}{9}.$

Note that
\[xy + (1 - x - y + xy) + (x + y - 2xy) = 1.\]This means if any of these three expressions is at most $\frac{1}{9},$ then the other two add up to at least $\frac{8}{9},$ so one of them must be at least $\frac{4}{9}.$

Let $s = x + y$ and $p = xy.$  Then
\[s^2 - 4p = (x + y)^2 - 4xy = (x - y)^2 \ge 0.\]Assume $x + y - 2xy = s - 2p < \frac{4}{9}.$  Then
\[0 \le s^2 - 4p < \left( 2p + \frac{4}{9} \right)^2 - 4p.\]This simplifies to $81p^2 - 45p + 4 > 0,$ which factors as $(9p - 1)(9p - 4) > 0.$  This means either $p < \frac{1}{9}$ or $p > \frac{4}{9}$; either way, we are done.

Therefore, the maximum value is $\boxed{\frac{4}{9}}.$