Cái này mình tìm đc GTNN thôy!\(2x+y=6\Rightarrow2x=6-y\Rightarrow x=\dfrac{6-y}{2}=3-\dfrac{y}{2}\)Ta có:
\(K=xy=\left(3-\dfrac{y}{2}\right)y=3y-\dfrac{y^2}{2}\)
\(=\left(\dfrac{y^2}{2}-3y+\dfrac{9}{2}\right)-\dfrac{9}{2}=\left(\dfrac{y}{\sqrt{2}}-\dfrac{3\sqrt{2}}{2}\right)^2-\dfrac{9}{2}\)Với mọi giá trị của y ta có:
\(\left(\dfrac{y}{\sqrt{2}}-\dfrac{3\sqrt{2}}{2}\right)^2\ge0\Rightarrow\left(\dfrac{y}{\sqrt{2}}-\dfrac{3\sqrt{2}}{2}\right)^2-\dfrac{9}{2}\ge\dfrac{-9}{2}\)Vậy \(Min_K=-\dfrac{9}{2}\)
Để \(K=-\dfrac{9}{2}\) thì \(\dfrac{y}{\sqrt{2}}-\dfrac{3\sqrt{2}}{2}=0\)
\(\Rightarrow\dfrac{y}{\sqrt{2}}=\dfrac{3\sqrt{2}}{2}\)
\(\Leftrightarrow2y=\sqrt{2}.3\sqrt{2}\)
\(\Leftrightarrow2y=6\Rightarrow y=3\)
\(\Rightarrow x=\dfrac{6-3}{2}=1,5\)
