shows OPT$(1, a, a, 1,  c, c, s), \, \, 0 \le a, c \le 1.$ The $\le 3-$area is dark. We have a path from $a=c=0$ to $a=c=1$ through this area. Therefore, for all $P$ with $0 \le P \le 1$ we have $0 \le a,  c\le 1$ with $P=ac$ such that OPT$(1, a, a, 1,  c, c, s) \le 3.$ In the notation of Lemma \[LEMOPT\] this corresponds to $\lambda_0 \ge \lambda_1=\lambda_2$ (and visualizes Theorem \[OPT\] for this case.) The following four lemmas are the technical core of our proof.

![OPT$(1,a,a,1,c,c,s)$ over the rectangle $0\le a \le 1, 0 \le c \le 1$ for $s=3$ and $s=14$.](opt_allg_03 "fig:"){width="49.00000%"} ![OPT$(1,a,a,1,c,c,s)$ over the rectangle $0\le a \le 1, 0 \le c \le 1$ for $s=3$ and $s=14$.](opt_allg_14 "fig:"){width="49.00000%"}

\[lem1\] Let $s\ge 8, A(x)\,=\,A(x, s)  := \,(7/10)Q \cdot x.$\
(a) OPT$(y)\,\,:=$ OPT $\left( 1, A(y), \, A(y), \, 1, y, \, y, \, s \right)$ is strictly decreasing for $   0 \le y \le 1/(2Q).$ The start value is OPT$(0)=3.$\
(b) Given $0\le y \le 1/(2Q),$ OPT$(z):= $ OPT$(1, A(y+z), \, A(y-z), \,1,  y+z, \, y-z, \, s)$ is decreasing in $0\le z \le y.$

\[lem2\] Let $ s\, \ge 7 