Ta có : \(2x+2y+z=4\)
\(\Rightarrow z=4-2x-2y\)
Khi đó \(A=2xy+yz+zx\)
\(=2xy+\left(y+x\right)z\)
\(=2xy+\left(y+x\right)\left(4-2x-2y\right)\)
\(=2xy+4y-2xy-2y^2+4x-2x^2-2xy\)
\(=4y+4x-2y^2-2x^2-2xy\)
\(\Rightarrow2A=-4x^2-4xy+8x-4y^2+8y\)
\(=-4x^2-4x\left(y-2\right)-4y^2+8y\)
\(=-4x^2-2.2x\left(y-2\right)-\left(y-2\right)^2+\left(y-2\right)^2-4y^2+8y\)
\(=-\left(2x+y-2\right)^2-3y^2+4y+4\)
\(=-\left(2x+y-2\right)^2-3\left(y^2-\dfrac{4}{3}y-\dfrac{4}{3}\right)\)
\(=-\left(2x+y-2\right)^2-3\left(y^2-2.y.\dfrac{2}{3}+\dfrac{4}{9}-\dfrac{4}{9}-\dfrac{4}{3}\right)\)
\(=-\left(2x+y-2\right)^2-3\left(y-\dfrac{2}{3}\right)^2+\dfrac{16}{3}\le\dfrac{16}{3}\)
\(\Rightarrow A\le\dfrac{8}{3}\)
\(Max_A=\dfrac{8}{3}\Leftrightarrow\left\{{}\begin{matrix}y-\dfrac{2}{3}=0\\2x+y-2=0\\z=4-2x-2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{2}{3}\\x=\dfrac{2}{3}\\z=\dfrac{4}{3}\end{matrix}\right.\)
