Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 952

\[\boxed{\mathbf{952}\mathbf{.}}\]

\[Пусть\ x - ширина\ досок\ и\ \]

\[a - угол\ наклона\ к\ основанию:\]

\[h = AM = DN =\]

\[= AB \bullet \sin{\angle BAM} = x \bullet \sin a;\]

\[BM = CN = AB \bullet \cos{\angle BAM} =\]

\[= x \bullet \cos a;\]

\[BC = AD + BM + CN =\]

\[= x + 2x \bullet \cos a.\]

\[Сечение\ желоба\ имеет\ форму\ \]

\[равнобокой\ трапеции:\]

\[S(x) = \frac{1}{2} \bullet h \bullet (AD + BC) =\]

\[= \frac{1}{2} \bullet x \bullet \sin a \bullet \left( x + x + 2x \bullet \cos a \right);\]

\[S(x) =\]

\[= \frac{x \bullet \sin a \bullet 2x \bullet \left( 1 + \cos a \right)}{2} =\]

\[= x^{2} \bullet \sin a \bullet \left( 1 + \cos a \right);\]

\[S(x) =\]

\[= x^{2} \bullet \left( \sin a + \sin a \bullet \cos a \right) =\]

\[= x^{2} \bullet \left( \sin a + \frac{1}{2}\sin{2a} \right).\]

\[S^{'}(a) =\]

\[= x^{2} \bullet \left( \left( \sin a \right)^{'} + \frac{1}{2} \bullet \left( \sin{2x} \right)^{'} \right);\]

\[S^{'}(a) =\]

\[= x^{2} \bullet \left( \cos a + \frac{1}{2} \bullet 2\cos{2a} \right);\]

\[S^{'}(a) = x^{2} \bullet \left( \cos a + \cos{2a} \right).\]

\[Промежуток\ возрастания:\]

\[\cos a + \cos{2a} = 0\]

\[\cos a + \cos^{2}a - \sin^{2}a = 0\]

\[\cos a + \cos^{2}a - 1 + \cos^{2}a = 0\]

\[2\cos^{2}a + \cos a - 1 = 0.\]

\[Пусть\ y = \cos a:\]

\[2y^{2} + y - 1 = 0\]

\[D = 1^{2} + 4 \bullet 2 = 1 + 8 = 9\]

\[y_{1} = \frac{- 1 - 3}{2 \bullet 2} = - 1;\text{\ \ }\]

\[y_{2} = \frac{- 1 + 3}{2 \bullet 2} = \frac{1}{2}.\]

\[(y + 1)(y - 0,5) > 0\]

\[y < - 1\ или\ y > 0,5.\]

\[1)\ \cos x < - 1 - нет\ корней.\]

\[2)\ \cos x > \frac{1}{2}\]

\[- \frac{\pi}{3} + 2\pi n < x < \frac{\pi}{3} + 2\pi n.\]

\[3)\ a = \frac{\pi}{3} - точка\ максимума.\]

\[\angle BAM = 90{^\circ} - a = \frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6}.\]

\[\angle BAD = \angle BAM + 90{^\circ} =\]

\[= \frac{\pi}{6} + \frac{\pi}{2} = \frac{2\pi}{3}.\]

\[Ответ:\ \ \frac{2\pi}{3}.\]