\(A=-x^2-4xy-5y^2+6y+1672\)
\(A=-x^2-4xy-4y^2-y^2+6y-9+1681\)
\(A=-\left(x+2y\right)^2-\left(y-3\right)^2+1681\)
\(A=1681-\left[\left(x+2y\right)^2+\left(y-3\right)^2\right]\)
Có: \(\left(x+2y\right)^2+\left(y-3\right)^2\ge0\)
\(\Rightarrow1681-\left[\left(x+2y\right)^2+\left(y-3\right)^2\right]\le1681\)
Dấu = xảy ra khi: \(\left(x+2y\right)^2+\left(y-3\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}\left(x+2y\right)^2=0\\\left(y-3\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x+2y=0\\y-3=0\end{cases}}\Rightarrow\hept{\begin{cases}x+2y=0\\y=3\end{cases}}\Rightarrow\hept{\begin{cases}x=-6\\y=3\end{cases}}\)
Vậy: \(Max_A=1681\) tại \(\hept{\begin{cases}x=-6\\y=3\end{cases}}\)
