Tìm GTLN

L

lp_qt

Tìm x để biểu thức $\dfrac{ax\sqrt{3a^2-x^2}}{12}$ đạt GTLN (x>0, a là hằng số)

$P=\dfrac{ax\sqrt{3a^2-x^2}}{12} \le \dfrac{|a|.x\sqrt{3a^2-x^2}}{12}= \dfrac{\dfrac{\sqrt{6}}{2}|a|.x.\sqrt{3a^2-x^2}}{6\sqrt{6}}$

$\dfrac{\sqrt{6}}{2}|a|.x.\sqrt{3a^2-x^2} \le \left ( \dfrac{\dfrac{3}{2}a^2+x^2+3a^2-x^2}{3} \right )^3=...$

$\Longrightarrow P \le ...$
 
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