\(x+2y=1\Leftrightarrow x=1-2y\Leftrightarrow A=xy=\left(1-2y\right)y=y-2y^2=\frac{1}{8}-\left(2y^2-y+\frac{1}{8}\right)=\frac{1}{8}-2\left(y^2-2.\frac{1}{4}.y+\frac{1}{16}\right)=\frac{1}{8}-2\left(y-\frac{1}{4}\right)^2\)
Vì \(\left(y-\frac{1}{4}\right)^2\ge0\Rightarrow2\left(x-\frac{1}{4}\right)^2\ge0\Rightarrow\frac{1}{8}-2\left(y-\frac{1}{4}\right)^2\ge\frac{1}{8}\)
Dấu "=" xảy ra khi \(\left(y-\frac{1}{4}\right)^2=0\Rightarrow y-\frac{1}{4}=0\Rightarrow y=\frac{1}{4}\Rightarrow x=\frac{1}{2}\)
Vậy Amax=1/8 khi x=1/2 và y=1/4
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