Question

Найдите наибольшее значение функции y=22√2 sinx-22x+5,5π+21

Answer 1

Ответ:

$\boxed{43}$

Объяснение:

$y = 22\sqrt{2} \sin x - 22x + 5,5\pi + 21$

$T(\sin x,\cos x) = 2\pi \Longtightarrow \Longrightarrow x \in [0;2\pi]$

$y' = (22\sqrt{2} \sin x - 22x + 5,5\pi + 21)' = (22\sqrt{2} \sin x)' - (22x)' + (5,5\pi)' + (21)'=$

$= 22\sqrt{2}( \sin x)' - 22(x)' + 0 + 0= 22\sqrt{2}\cos x - 22$

$y'' = (22\sqrt{2}\cos x - 22)' = (22\sqrt{2}\cos x)' - (22)' = 22\sqrt{2}(\cos x)' - 0= -22\sqrt{2}\sin x$

$y' = 0$

$22\sqrt{2}\cos x - 22 = 0$

$22\sqrt{2}\cos x = 22|:22$

$\sqrt{2}\cos x = 1|:\sqrt{2}$

$\cos x = \dfrac{1}{\sqrt{2} }$

$x = \pm \arccos \bigg (\dfrac{1}{\sqrt{2} } \bigg ) + 2 \pi n, n \in\mathbb Z$

$x = \pm \dfrac{\pi}{4} + 2 \pi n, n \in\mathbb Z$

$n = 0: x = \dfrac{\pi}{4}$

$n = 1:x = -\dfrac{\pi}{4} + 2 \pi = \dfrac{-\pi + 8\pi}{4} = \dfrac{7\pi}{4}$

Критические точки: $x = \dfrac{\pi}{4}$ или $x = \dfrac{7\pi}{4}$

$y''\bigg(\dfrac{\pi}{4} \bigg) = -22\sqrt{2}\sin \bigg (\dfrac{\pi}{4} \bigg) = -22\sqrt{2} \cdot \dfrac{1}{\sqrt{2} } =-22 <0$

$y''\bigg(\dfrac{\pi}{4} \bigg) < 0 \Longrightarrow x = \dfrac{\pi}{4}$ - максимум функции

$y''\bigg(\dfrac{7\pi}{4} \bigg) = -22\sqrt{2}\sin \bigg (\dfrac{7\pi}{4} \bigg) = -22\sqrt{2}\sin \bigg ( -\dfrac{\pi}{4} + 2 \pi \bigg) =$

$= -22\sqrt{2}\sin \bigg ( -\dfrac{\pi}{4} \bigg) = -1 \cdot (-22) \cdot\sqrt{2}\sin \bigg ( \dfrac{\pi}{4} \bigg) = 22\sqrt{2} \cdot \dfrac{1}{\sqrt{2} } = 22 > 0$

$y''\bigg(\dfrac{7\pi}{4} \bigg) < 0 \Longrightarrow x = \dfrac{7\pi}{4}$ - минимум функции

$y \bigg(\dfrac{\pi}{4} \bigg) = 22\sqrt{2} \sin \bigg(\dfrac{\pi}{4} \bigg) - 22\bigg(\dfrac{\pi}{4} \bigg) + 5,5\pi + 21 = 22\sqrt{2} \cdot \dfrac{1}{\sqrt{2} } - 5,5\pi + 5,5\pi + 21=$

$=22+ 21 = 43$ - наибольшее значение функции

$y \bigg(\dfrac{7\pi}{4} \bigg) = 22\sqrt{2} \sin \bigg(\dfrac{\pi}{4} \bigg) - 22\bigg(\dfrac{7\pi}{4} \bigg) + 5,5\pi + 21 =$

$= -22\sqrt{2} \cdot \dfrac{1}{\sqrt{2} } + 5,5\pi + 5,5\pi + 21= 11\pi - 1$ - наименьшее значение функции