Đáp án:
$\max A= 10 \Leftrightarrow (x;y)=\left(\dfrac53;\dfrac43\right)$
Giải thích các bước giải:
$\quad A = - 2x^2 + 2xy - 5y^2 + 4x +10y$
$\to A = -\dfrac12(4x^2 - 4xy + y^2- 8x + 4y + 4) - \dfrac12(9y^2 - 24y + 16) + 10$
$\to A = - \dfrac12(2x - y - 2)^2 - \dfrac12(3y - 4)^2 +10$
Ta có:
$\quad \begin{cases}(2x - y -2)^2 \geqslant 0\quad \forall x;y\\(3y-4)^2\geqslant 0\quad \forall y\end{cases}$
$\Leftrightarrow \begin{cases}-\dfrac12(2x-y-2)^2 \leqslant 0\\-\dfrac12(3y-4)\leqslant 0\end{cases}$
$\Leftrightarrow - \dfrac12(2x - y - 2)^2 - \dfrac12(3y - 4)^2 +10 \leqslant 10$
Dấu $=$ xảy ra $\Leftrightarrow \begin{cases}2x - y -2 = 0\\3y - 4= 0\end{cases}\Leftrightarrow \begin{cases}x =\dfrac53\\y =\dfrac43\end{cases}$
Vậy $\max A= 10 \Leftrightarrow (x;y)=\left(\dfrac53;\dfrac43\right)$
