Toán Lớp 8: Tìm GTLN của
A = 2x – x^2 + 2
B = – y^2 – 3y + 1/3
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Toán Lớp 8: Tìm GTLN của
A = 2x – x^2 + 2
B = – y^2 – 3y + 1/3
Comments ( 2 )
{A_{\max }} = 3 \Leftrightarrow x = 1\\
{B_{\max }} = \dfrac{{31}}{{12}} \Leftrightarrow y = – \dfrac{3}{2}
\end{array}\)
A = 2x – {x^2} + 2\\
= 3 + \left( { – {x^2} + 2x – 1} \right)\\
= 3 – \left( {{x^2} – 2x + 1} \right)\\
= 3 – \left( {{x^2} – 2.x.1 + {1^2}} \right)\\
= 3 – {\left( {x – 1} \right)^2}\\
{\left( {x – 1} \right)^2} \ge 0,\,\,\,\,\forall x\\
\Rightarrow A = 3 – {\left( {x – 1} \right)^2} \le 3,\,\,\,\,\forall x\\
\Rightarrow {A_{\max }} = 3 \Leftrightarrow {\left( {x – 1} \right)^2} = 0 \Leftrightarrow x – 1 = 0 \Leftrightarrow x = 1\\
\Rightarrow {A_{\max }} = 3 \Leftrightarrow x = 1\\
B = – {y^2} – 3y + \dfrac{1}{3}\\
= \dfrac{{31}}{{12}} + \left( { – {y^2} – 3y – \dfrac{9}{4}} \right)\\
= \dfrac{{31}}{{12}} – \left( {{y^2} + 3y + \dfrac{9}{4}} \right)\\
= \dfrac{{31}}{{12}} – \left[ {{y^2} + 2.y.\dfrac{3}{2} + {{\left( {\dfrac{3}{2}} \right)}^2}} \right]\\
= \dfrac{{31}}{{12}} – {\left( {y + \dfrac{3}{2}} \right)^2}\\
{\left( {y + \dfrac{3}{2}} \right)^2} \ge 0,\,\,\,\forall y\\
\Rightarrow B = \dfrac{{31}}{{12}} – {\left( {y + \dfrac{3}{2}} \right)^2} \le \dfrac{{31}}{{12}},\,\,\,\forall y\\
\Rightarrow {B_{\max }} = \dfrac{{31}}{{12}} \Leftrightarrow {\left( {y + \dfrac{3}{2}} \right)^2} = 0 \Leftrightarrow y + \dfrac{3}{2} = 0 \Leftrightarrow y = – \dfrac{3}{2}\\
\Rightarrow {B_{\max }} = \dfrac{{31}}{{12}} \Leftrightarrow y = – \dfrac{3}{2}
\end{array}\)