smc

Let us first draw a unit square. We will now pick arbitrary points $E$ and $F$ on $\overline{AB}$ and $\overline{AD}$ respectively. We shall say that $AE=AF=x$ Thus, our problem has been simplified to maximizing the area of the blue quadrilateral. If we drop an altitude from $E$ to $\overline{DC}$, and call the foot of the altitude $G$, we can find the area of $EFDC$ by noting that $[EFDC]=[EGDF]+[EGC]$. We can now finish the problem. Since $EG=1$ and $FD=1-x$, we have: $$[EFDC]=[EGDF]+[EGC]=DG\cdot \frac{EG+FD}{2}+\frac{(CG)(EG)}{2}=\frac{x(2-x)}{2}+\frac{1-x}{2}=\frac{1+x-x^2}{2}$$ To maximize this, we compete the square in the numerator to have: $$[EFDC]=\frac{-(x-\frac{1}{2}{)}^2+\frac{5}{4}}{2}=\frac{5}{8}-\frac{(x-\frac{1}{2}{)}^2}{2}$$ Finally, we see that $\frac{(x-\frac{1}{2}{)}^2}{2}\ge 0$, as: $${\left(x-\frac{1}{2}\right)}^2\ge 0$$ So, $$\frac{(x-\frac{1}{2}{)}^2}{2}\ge 0,$$ where the first inequality was from the Trivial Inequality, and the second came from dividing both sides by $2$, which does not change the inequality sign. Thus, the maximum area is $\boxed{\frac{5}{8}}$ or $\boxed{\frac{5}{8}},$ when $\frac{(x-\frac{1}{2}{)}^2}{2}=0.$