Đặt $AM=AP=CN=CD=x$
$S_{MNPQ}=S_{ABCD}-(S_{BMN}+S_{NCP}+S_{PDQ}+S_{QAM})=6-\bigg [2. \dfrac{1}{2}x^2 +2.\dfrac{1}{2}(2-x)(3-x)\bigg]$
$=6-x^2-(2-x)(3-x) =\dfrac{25}{8}-\dfrac{1}{8}(4x-5)^2 \le \dfrac{25}{8}$
Vậy $\max S_{MNPQ} =\dfrac{25}{8}$ khi chỉ khi $4x-5=0 \Leftrightarrow x=\dfrac{5}{4}$