Toán Lớp 8: tìm max: R= -7x^2 – 4y^2 – 8xy + 18x + 9
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Toán Lớp 8: tìm max: R= -7x^2 – 4y^2 – 8xy + 18x + 9
Comments ( 1 )
R = – \left( {7{x^2} + 4{y^2} + 8xy – 18x – 9} \right)\\
= – \left( {4{x^2} + 8xy + 4{y^2} + 3{x^2} – 2.x\sqrt 3 .\dfrac{9}{{\sqrt 3 }} + \dfrac{{81}}{3} – 36} \right)\\
= – {\left( {2x + 2y} \right)^2} – {\left( {x\sqrt 3 – \dfrac{9}{{\sqrt 3 }}} \right)^2} + 36\\
Do:{\left( {2x + 2y} \right)^2} \ge 0\forall x;y\\
{\left( {x\sqrt 3 – \dfrac{9}{{\sqrt 3 }}} \right)^2} \ge 0\forall x;y\\
\to – {\left( {2x + 2y} \right)^2} – {\left( {x\sqrt 3 – \dfrac{9}{{\sqrt 3 }}} \right)^2} \le 0\\
\to – {\left( {2x + 2y} \right)^2} – {\left( {x\sqrt 3 – \dfrac{9}{{\sqrt 3 }}} \right)^2} + 36 \le 36\\
\to Max = 36\\
\Leftrightarrow \left\{ \begin{array}{l}
2x + 2y = 0\\
x\sqrt 3 – \dfrac{9}{{\sqrt 3 }} = 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 3\\
y = – 3
\end{array} \right.
\end{array}\)