Question

Cho 3x + y = 3

a) Tìm min M = 3x² + y²

b) Tìm max N = 2xy

Answer 1

a) Áp dụng BĐT Bunhia ta có:

\(\left(3+1\right)\left(3x^2+y^2\right)\ge\left(3x+y\right)^2\)

<=> \(3x^2+y^2\ge3^2:4=\dfrac{9}{4}\)

=> Dấu = xảy ra <=> \(\left\{{}\begin{matrix}3x+y=3\\x=y\end{matrix}\right.\) <=> \(x=y=\dfrac{3}{4}\)

b) Ta có: \(3x+y=3\) => \(y=3-3x\) (1)

Thay (1) vào N ta được:

N = \(2.\left(3-3x\right)x\) = \(6x-6x^2\) = \(-6\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{2}\)

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

=> Dấu = xảy ra <=> \(\left\{{}\begin{matrix}y=3-3x\\x=\dfrac{1}{2}\end{matrix}\right.\) <=> \(x=\dfrac{1}{2};y=\dfrac{3}{2}\)