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daniil7213 @daniil7213

August 2022 1 6 Report

Найдите наибольшее значение функции
[tex]y = 12 \sin(x) - 6 \sqrt{3} x + \sqrt{3} \pi + 22[/tex]

На отрезке [ 0; pi/2]

Answers & Comments

mathkot

Ответ:

28

Объяснение:

$x \in \bigg [ 0 ; \dfrac{\pi}{2} \bigg]$

$y = 12 \sin x - 6\sqrt{3}x + \sqrt{3}\pi + 22$

$y' = (12 \sin x - 6\sqrt{3}x + \sqrt{3}\pi + 22)' = (12 \sin x)' - (6\sqrt{3}x)' + (\sqrt{3}\pi)' + (22)' =$

$=12 ( \sin x)' - 6\sqrt{3}(x)' +0 + 0 = 12 \cos x - 6\sqrt{3}$

$y' = 0$

$12 \cos x - 6\sqrt{3} = 0$

$12 \cos x= 6\sqrt{3}|:12$

$\cos x = \dfrac{\sqrt{3} }{2}$

$x = \pm \arccos \bigg (\dfrac{\sqrt{3} }{2} \bigg) + 2 \pi n, n \in \mathbb Z$

$x =\pm \dfrac{\pi}{6} +2 \pi n, n \in \mathbb Z$

$n = 0: \left [ \begin{array}{ccc} \boxed{ x = \dfrac{\pi}{6} } \\ \\x = -\dfrac{\pi}{6} \end{array}\right$

$n = 0: \left [ \begin{array}{ccc} x = \dfrac{\pi}{6} + 2\pi \\ \\ x = -\dfrac{\pi}{6} + 2\pi = \dfrac{12\pi - \pi}{6} = \dfrac{11\pi}{6} \end{array}\right$

$\bigg \{ \dfrac{\pi}{6} \bigg \} \subset \bigg [ 0 ; \dfrac{\pi}{2} \bigg]$

$y(0) = 12\sin 0 - 6\sqrt{3} \cdot 0 + \sqrt{3}\pi + 22 = \sqrt{3}\pi + 22$

$y \bigg(\dfrac{\pi}{2} \bigg) = 12 \sin \bigg(\dfrac{\pi}{2} \bigg) - 6\sqrt{3} \cdot \dfrac{\pi}{2} + \sqrt{3}\pi + 22 = 12 + 22 - 3\sqrt{3}\pi + \sqrt{3}\pi =$

$= 34 + \pi \sqrt{3}(1 - 3) = 34 - 2\pi \sqrt{3}$

$y(0) > y\bigg(\dfrac{\pi}{2} \bigg)$

$y \bigg(\dfrac{\pi }{6} \bigg) = 12 \sin \bigg(\dfrac{\pi}{6} \bigg) - 6\sqrt{3} \cdot \dfrac{\pi}{6} + \sqrt{3}\pi + 22 = 12 \cdot 0,5 - \sqrt{3}\pi + \sqrt{3}\pi + 22 =$