Question

Tìm GTNN của \(A=3x^2+2y^2+2xy-10x-10y+2030\)

Answer 1

Ta có :  

\(A=3x^2+2y^2+2xy-10x-10y+2030\)

\(A=3x^2+2\left(y-5\right)x+2y^2-10y+2030\)

\(\Leftrightarrow3x^2+2\left(y-5\right)x+2y^2-10y+2030+A\ge0\)

\(\Delta'=\left(y-5\right)^2-3\left(2y^2-10y+2030-A\right)\ge0\)

\(\Leftrightarrow-5y^2+20y-6065+3A\ge0\)

\(\Leftrightarrow3A\ge5y^2-20y+6065=5\left(y^2-4y+4\right)+6045\)

\(\Leftrightarrow3A\ge5\left(y-2\right)^2+6045\)

\(\Leftrightarrow A\ge\frac{5}{3}\left(y-2\right)^2+2015\ge2015\)

Vậy  \(MinA=2015\Leftrightarrow\hept{\begin{cases}x=1\\y=2\end{cases}}\)