Toán Tìm Max C=-x^2-2y^2-2xy+2x-2y-15 D=15-10x-10x^2+24xy-16y^2 11/08/2021 By Bella Tìm Max C=-x^2-2y^2-2xy+2x-2y-15 D=15-10x-10x^2+24xy-16y^2
Giải thích các bước giải: Ta có: \(\begin{array}{l}*)\\C = – {x^2} – 2{y^2} – 2xy + 2x – 2y – 15\\ = – \left( {{x^2} + 2xy + {y^2}} \right) + 2.\left( {x + y} \right) – {y^2} – 4y – 15\\ = – {\left( {x + y} \right)^2} + 2.\left( {x + y} \right) – \left( {{y^2} + 4y + 4} \right) – 11\\ = – \left[ {{{\left( {x + y} \right)}^2} – 2\left( {x + y} \right) + 1} \right] – {\left( {y + 2} \right)^2} – 10\\ = – {\left( {x + y – 1} \right)^2} – {\left( {y + 2} \right)^2} – 10 \le – 10,\,\,\,\forall x,y\\ \Rightarrow {C_{\max }} = – 10 \Leftrightarrow \left\{ \begin{array}{l}{\left( {x + y – 1} \right)^2} = 0\\{\left( {y + 2} \right)^2} = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x + y – 1 = 0\\y = – 2\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 3\\y = – 2\end{array} \right.\\*)\\D = 15 – 10x – 10{x^2} + 24xy – 16{y^2}\\ = – \left( {{x^2} + 10x + 25} \right) – \left( {9{x^2} – 24xy + 16{y^2}} \right) + 40\\ = 40 – {\left( {x + 5} \right)^2} – {\left( {3x – 4y} \right)^2} \le 40,\,\,\,\forall x,y\\ \Rightarrow {D_{\max }} = 40 \Leftrightarrow \left\{ \begin{array}{l}{\left( {x + 5} \right)^2} = 0\\{\left( {3x – 4y} \right)^2} = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = – 5\\y = \frac{{15}}{4}\end{array} \right.\end{array}\) Trả lời
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
*)\\
C = – {x^2} – 2{y^2} – 2xy + 2x – 2y – 15\\
= – \left( {{x^2} + 2xy + {y^2}} \right) + 2.\left( {x + y} \right) – {y^2} – 4y – 15\\
= – {\left( {x + y} \right)^2} + 2.\left( {x + y} \right) – \left( {{y^2} + 4y + 4} \right) – 11\\
= – \left[ {{{\left( {x + y} \right)}^2} – 2\left( {x + y} \right) + 1} \right] – {\left( {y + 2} \right)^2} – 10\\
= – {\left( {x + y – 1} \right)^2} – {\left( {y + 2} \right)^2} – 10 \le – 10,\,\,\,\forall x,y\\
\Rightarrow {C_{\max }} = – 10 \Leftrightarrow \left\{ \begin{array}{l}
{\left( {x + y – 1} \right)^2} = 0\\
{\left( {y + 2} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x + y – 1 = 0\\
y = – 2
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 3\\
y = – 2
\end{array} \right.\\
*)\\
D = 15 – 10x – 10{x^2} + 24xy – 16{y^2}\\
= – \left( {{x^2} + 10x + 25} \right) – \left( {9{x^2} – 24xy + 16{y^2}} \right) + 40\\
= 40 – {\left( {x + 5} \right)^2} – {\left( {3x – 4y} \right)^2} \le 40,\,\,\,\forall x,y\\
\Rightarrow {D_{\max }} = 40 \Leftrightarrow \left\{ \begin{array}{l}
{\left( {x + 5} \right)^2} = 0\\
{\left( {3x – 4y} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = – 5\\
y = \frac{{15}}{4}
\end{array} \right.
\end{array}\)