\(S=\left(3x+1\right)\left(6-x\right)=18-3x^2+6-x=-3x^2+17x+6\)
Đặt \(A=-3x^2+17x+6\)
\(S_{max}\Leftrightarrow A_{max}=-3\left(x^2-\dfrac{17}{3}x-2\right)=-3\left[x^2-2.x.\dfrac{17}{6}+\left(\dfrac{17}{6}\right)^2-\left(-\dfrac{17}{6}\right)^2-2\right]\)
\(=-3\left[\left(x-\dfrac{17}{6}\right)^2-\dfrac{361}{36}\right]\)
\(=-3\left(x-\dfrac{17}{6}\right)^2+\dfrac{361}{12}\)
Vì \(-3\left(x-\dfrac{17}{6}\right)^2\le0\forall x\in R\)
\(\Rightarrow-3\left(x-\dfrac{17}{6}\right)^2+\dfrac{361}{12}\le\dfrac{361}{12}\forall x\in R\)
\(\Rightarrow A_{max}=\dfrac{361}{12}\Leftrightarrow\left(x-\dfrac{17}{6}\right)^2=0\Leftrightarrow x=\dfrac{17}{6}\)
\(\Rightarrow S_{max}\Leftrightarrow x=\dfrac{17}{6}\)
