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}