Question

cho các số thực x, y thỏa mã : x\(\ge\) 1 , x+y \(\le\) 4

tìm min : A= x^2+3xy+4y^2

Answer 1

\(A=x^2+3xy+4y^2\)

\(=\dfrac{9}{16}x^2+3xy+4y^2+\dfrac{7}{16}x^2\)

\(=4\left(\dfrac{9}{64}x^2+\dfrac{3}{4}xy+y^2\right)+\dfrac{7}{16}x^2\)

\(=4\left(\dfrac{3}{8}x+y^2\right)+\dfrac{7}{16}x^2\ge4\cdot0+\dfrac{9}{17}\cdot1^2=\dfrac{7}{16}\)

Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}x=1\\y=-\dfrac{3}{8}\end{matrix}\right.\)

Vậy với \(\left\{{}\begin{matrix}x=1\\y=-\dfrac{3}{8}\end{matrix}\right.\) khi \(A_{Min}=\dfrac{7}{16}\)