Question

1,Cho 3x+y=1

Tìm GTLN của N=xy

Answer 1

Từ \(3x+y=1\Rightarrow y=1-3x\)

\(\Rightarrow N=xy=x\left(1-3x\right)=x-3x^2=-3\left(x^2-\frac{1}{3}x\right)\)

\(=-3\left(x^2-2.\frac{1}{6}.x+\frac{1}{36}-\frac{1}{36}\right)\)

\(=-3\left(x-\frac{1}{6}\right)^2+\frac{1}{12}\)

Vì \(-3\left(x-\frac{1}{6}\right)^2\le0\forall x\) nên \(N=-3\left(x-\frac{1}{6}\right)^2+\frac{1}{12}\le\frac{1}{12}\forall x\) có GTNN là \(\frac{1}{12}\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}y=1-3x\\x-\frac{1}{6}=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{1}{6}\\y=\frac{1}{2}\end{cases}}}\)

Vậy \(N_{min}=\frac{1}{12}\) tại \(x=\frac{1}{6};y=\frac{1}{2}\)