\(A\left(-2;1\right);B\left(4;4\right);C\left(-2;0\right);D\left(4;0\right)\)
Gọi \(M\left(a;\frac{a^2}{4}\right);N\left(a;0\right)\) với \(-2< a< 4\)
\(S_{ABM}=S_{ACDB}-\left(S_{ACNM}+S_{BDNM}\right)\)
Do \(S_{ACDB}\) cố định nên \(S_{ABM}\) lớn nhất khi \(P=S_{ACNM}+S_{BDNM}\) nhỏ nhất
\(P=\frac{1}{2}\left(y_M+y_A\right)\left(x_M-x_A\right)+\frac{1}{2}\left(y_M+y_B\right)\left(x_B-x_M\right)\)
\(\Rightarrow2P=\left(\frac{a^2}{4}+1\right)\left(a+2\right)+\left(\frac{a^2}{4}+4\right)\left(4-a\right)\)
\(\Rightarrow2P=\frac{a^3}{4}+\frac{1}{2}a^2+a+2+a^2-\frac{a^3}{4}+16-4a\)
\(\Rightarrow2P=\frac{3a^2}{2}-3a+18=\frac{3}{2}\left(a-1\right)^2+\frac{33}{2}\ge\frac{33}{2}\)
\(\Rightarrow P_{min}\) khi \(a=1\Rightarrow M\left(1;\frac{1}{4}\right)\)