The Largest Rectangle

Let the horizontal side of the rectangle be $a$ , the vertical side of the rectangle be $b$ , the horizontal leg of triangle C be $e$ , and the vertical leg of triangle C be $f$ . We can see that $ef=2C$ , $(a-e)b=2B$ , and $(b-f)a=2A$ . Multiplying the second and third equations together gives us $ab(ab-af-eb+ef)=4AB$ Adding the second and third equations together tells us that $-af-eb=2A+2B-2ab$ . Substituting this and the first equation into our other formula gives us $ab(ab+2A+2B-2ab+2C)=4AB \implies (ab)^2+(-2A-2B-2C)ab +4AB=0 \implies ab=A+B+C \pm \sqrt{(A+B+C)^2-4AB}$ We know that $ab=A+B+C+D$ , so this simplifies to $D^2+4AB=(A+B+C)^2$ . Because $A+B+C$ is fixed, the entire rectangle will be maximized when $D$ is maximized. Thus we need to find a $D$ just under 2018 that satisfies $D^2+4AB=2018^2$ . 2017 and 2015 don't work because those would make the left side odd and the right side even. 2016 doesn't work because $2018^2-2016^2=4*2017$ and 2017 is prime so $A$ and $B$ would have to be 2017 and 1 which can't satisfy the equation $A+B+C=2018$ . The next possibility for $D$ is 2014 which does work, so the answer is $2018+2014=\boxed{4032}$