Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 1094

\[a - сторона\ основания;\]

\[h - высота\ призмы.\]

\[1)\ Расстояние\ до\ центра\ призмы:\]

\[\left( \frac{a}{\sqrt{3}} \right)^{2} + \left( \frac{h}{2} \right)^{2} = R^{2}\]

\[\frac{a^{2}}{3} + \frac{h^{2}}{4} = R^{2}\]

\[4a^{2} + 3h^{2} = 12R^{2}\]

\[4a^{2} = 12R^{2} - 3h^{2}\]

\[a^{2} = 3R^{2} - \frac{3h^{2}}{4} = \frac{3}{4}\left( 4R^{2} - h^{2} \right).\]

\[2)\ V(h) = S_{осн} \bullet h = \frac{a^{2}\sqrt{3}}{4} \bullet h =\]

\[= \frac{\sqrt{3}h}{4} \bullet \frac{3}{4}\left( 4R^{2} - h^{2} \right) =\]

\[= \frac{3\sqrt{3}}{16} \bullet \left( 4R^{2}h - h^{3} \right);\]

\[V^{'}(h) = \frac{3\sqrt{3}}{16}\left( 4R^{2}(h)^{'} - \left( h^{3} \right)^{'} \right) =\]

\[= \frac{3\sqrt{3}}{16}\left( 4R^{2} - 3h^{2} \right).\]

\[3)\ Промежуток\ возрастания:\]

\[4R^{2} - 3h^{2} \geq 0\]

\[3h^{2} \leq 4R^{2}\]

\[h^{2} \leq \frac{4R^{2}}{3}\]

\[- \frac{2R}{\sqrt{3}} \leq h \leq \frac{2R}{\sqrt{3}}.\]

\[4)\ Точка\ максимума:\]

\[h = \frac{2R}{\sqrt{3}}.\]

\[Ответ:\ \ \frac{2R}{\sqrt{3}}.\]