Lời giải:Ta có: \(2x + 2y + z = 4 \Leftrightarrow z = 4 - 2x - 2y.\)
\(\begin{array}{l} \Rightarrow A = 2xy + yz + zx = 2xy + y\left( {4-2x-2y} \right) + x\left( {4-2x-2y} \right)\\ = {\rm{ }}2xy + 4y-2xy-2{y^2} + 4x-2{x^2}-2xy\\ = -2{x^2}-2xy + 4x-2{y^2} + 4y\\ = - \left[ {\left( {{x^2} + 2xy + {y^2}} \right) - \frac{8}{3}\left( {x + y} \right) + \frac{{16}}{9}} \right] - \left( {{x^2} - \frac{4}{3}x + \frac{4}{9}} \right) - \left( {{y^2} - \frac{4}{3}y + \frac{4}{9}} \right) + \frac{8}{3}\\ = - {\left( {x + y - \frac{4}{3}} \right)^2} - {\left( {x - \frac{2}{3}} \right)^2} - {\left( {y - \frac{2}{3}} \right)^2} + \frac{8}{3}\end{array}\)
Mà: \({\left( {x + y - \frac{4}{3}} \right)^2} \ge 0;\;{\left( {x - \frac{2}{3}} \right)^2} \ge 0;\;{\left( {y - \frac{2}{3}} \right)^2} \ge 0\;\forall x,\;y\)
\( \Rightarrow A = - {\left( {x + y - \frac{4}{3}} \right)^2} - {\left( {x - \frac{2}{3}} \right)^2} - {\left( {y - \frac{2}{3}} \right)^2} + \frac{8}{3} \ge \frac{8}{3}\)\(\forall x,\;y\)
Dấu “=” xảy ra \( \Leftrightarrow \left\{ \begin{array}{l}x + y - \frac{4}{3} = 0\\x - \frac{2}{3} = 0\\y - \frac{2}{3} = 0\end{array} \right. \Leftrightarrow x = y = \frac{2}{3}.\)
Vậy Amax = \(\frac{8}{3}\) tại \(\left\{ \begin{array}{l}x = y = \frac{2}{3}\\z = \frac{4}{3}\end{array} \right.\).
Chọn D.
