Q = - x2 - 2y2 - 2xy + 8x + 6y + 13
= - x2 - y2 - 42 - 2xy + 8y + 8x - y2 - 2y - 1 + 30
= 30 - (x2 + y2 + 42 + 2xy - 8y - 8x) - (y2 + 2y + 1)
= 78 - (x + y - 4)2 - (y + 1 )2 \(\le\) 30
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x+y-4=0\\y+1=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=-1\end{matrix}\right.\)
\(Q=-x^2-2y^2-2xy+8x+6y+13\)
\(=-\left(x^2+2xy-8x\right)-2y^2+6y+13\)
\(=-\left[x^2+2x\left(y-8\right)+\left(y-8\right)^2\right]-2y^2+6y+13-\left(y-8\right)^2\)\(=-\left(x+y-8\right)^2-2y^2+6y+13-y^2+16y-64\)\(=-\left(x+y-8\right)^2-3y^2+22y-51\)
\(=-\left(x+y-8\right)^2-3\left(y^2-\dfrac{22}{3}y+\dfrac{121}{9}\right)-\dfrac{32}{3}\)\(=-\left(x+y-8\right)^2-3\left(y-\dfrac{11}{9}\right)^2-\dfrac{32}{3}\le-\dfrac{32}{2}\forall x\)Vậy Max Q = \(-\dfrac{32}{3}\) khi \(\left\{{}\begin{matrix}x+y-8=0\\y-\dfrac{11}{9}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-\dfrac{61}{9}=0\\y=\dfrac{11}{9}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{61}{9}\\y=\dfrac{11}{9}\end{matrix}\right.\)
