Question

Cho \(3x+y=1\)

a) Tìm min \(M=3x^2+y^2\)

b) Tìm max \(N=xy\)

Answer 1

Ta có: \(y=1-3x\)

a/ \(M=3x^2+y^2=3x^2+\left(1-3x\right)^2\)

\(\Leftrightarrow12x^2-6x+1=\left(12x^2-\frac{2.2.3x}{2}+\frac{3}{4}\right)+\frac{1}{4}\)

\(=\left(2\sqrt{3}x-\frac{\sqrt{3}}{2}\right)^2+\frac{1}{4}\ge\frac{1}{4}\)

Vậy GTNN là 0,25 đạt được khi x = 0,25

b/ \(N=xy=x\left(1-3x\right)=-3x^2+x\)

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

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

Vậy max là \(\frac{1}{12}\) đạt được khi \(x=\frac{1}{6}\)