Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
*)\\
C = - {x^2} - 2{y^2} - 2xy + 2x - 2y - 15\\
= - \left( {{x^2} + 2xy + {y^2}} \right) + 2.\left( {x + y} \right) - {y^2} - 4y - 15\\
= - {\left( {x + y} \right)^2} + 2.\left( {x + y} \right) - \left( {{y^2} + 4y + 4} \right) - 11\\
= - \left[ {{{\left( {x + y} \right)}^2} - 2\left( {x + y} \right) + 1} \right] - {\left( {y + 2} \right)^2} - 10\\
= - {\left( {x + y - 1} \right)^2} - {\left( {y + 2} \right)^2} - 10 \le - 10,\,\,\,\forall x,y\\
\Rightarrow {C_{\max }} = - 10 \Leftrightarrow \left\{ \begin{array}{l}
{\left( {x + y - 1} \right)^2} = 0\\
{\left( {y + 2} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x + y - 1 = 0\\
y = - 2
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 3\\
y = - 2
\end{array} \right.\\
*)\\
D = 15 - 10x - 10{x^2} + 24xy - 16{y^2}\\
= - \left( {{x^2} + 10x + 25} \right) - \left( {9{x^2} - 24xy + 16{y^2}} \right) + 40\\
= 40 - {\left( {x + 5} \right)^2} - {\left( {3x - 4y} \right)^2} \le 40,\,\,\,\forall x,y\\
\Rightarrow {D_{\max }} = 40 \Leftrightarrow \left\{ \begin{array}{l}
{\left( {x + 5} \right)^2} = 0\\
{\left( {3x - 4y} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = - 5\\
y = \frac{{15}}{4}
\end{array} \right.
\end{array}\)
