Question

[Trần Gia Lâm]

cho 3x+y=1
a> tìm GTNN M= 3x^2 + y^2
b> tìm GTLN N= xy

Answer 1

[Đinh Đức Hùng]

\(3x+y=1\Rightarrow y=1-3x\) (1)

a ) Thay (1) vào M ta được :

\(M=3x^2+\left(1-3x\right)^2=3x^2+9x^2-6x+1=12x^2-6x+1\)

\(=\left(\sqrt{12}x\right)^2-2\sqrt{12}x.\frac{3}{\sqrt{12}}+\frac{9}{12}+\frac{1}{4}=\left(\sqrt{12}x-\frac{3}{\sqrt{12}}\right)^2+\frac{1}{4}\ge\frac{1}{4}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=\frac{1}{4}\)

Vậy \(M_{min}=\frac{1}{4}\) tại \(x=y=\frac{1}{4}\)

b ) Thay (1) vào N ta được :

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

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

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

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