Toán Lớp 8: Cho x²+y²=52 thì biểu thức M=2x+3y có GTLN là:
A. 52
B. 104
C.26
D.14
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Toán Lớp 8: Cho x²+y²=52 thì biểu thức M=2x+3y có GTLN là:
A. 52
B. 104
C.26
D.14
Comments ( 1 )
Theo\,Bunhia:\\
{\left( {a.x + b.y} \right)^2} \le \left( {{a^2} + {b^2}} \right).\left( {{x^2} + {y^2}} \right)\\
M = 2.x + 3.y\\
\Leftrightarrow {M^2} = {\left( {2.x + 3.y} \right)^2} \le \left( {{2^2} + {3^2}} \right)\left( {{x^2} + {y^2}} \right)\\
\Leftrightarrow {M^2} \le 13.52\\
\Leftrightarrow {M^2} \le {13^2}.4\\
\Leftrightarrow – 13.2 \le M \le 13.2\\
\Leftrightarrow – 26 \le M \le 26\\
\Leftrightarrow GTLN:M = 26\\
Dau\, = :\dfrac{x}{2} = \dfrac{y}{3} \Leftrightarrow x = \dfrac{2}{3}y\\
Do:{x^2} + {y^2} = 52\\
\Leftrightarrow {\left( {\dfrac{2}{3}y} \right)^2} + {y^2} = 52\\
\Leftrightarrow {y^2} = 36\\
\Leftrightarrow \left[ \begin{array}{l}
y = 6 \Leftrightarrow x = \dfrac{2}{3}.6 = 4\\
y = – 6 \Leftrightarrow x = \dfrac{2}{3}.\left( { – 6} \right) = – 4
\end{array} \right.\\
Vậy\,GTLN:M = 26\,khi\,\left( {x;y} \right) = \left\{ {\left( {4;6} \right);\left( { – 4; – 6} \right)} \right\}
\end{array}$