Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1528

\[\boxed{\mathbf{1528}\mathbf{.}}\]

\[r - радиус\ основания;\]

\[h - высота\ цилиндра.\]

\[Диагональ\ осевого\ сечения\ \]

\[цилиндра\ совпадает\ с\ \]

\[диаметром\ шара:\]

\[(2r)^{2} + h^{2} = d^{2}\]

\[4r^{2} + h^{2} = (2R)^{2}\]

\[4r^{2} = 4R^{2} - h^{2}\]

\[r^{2} = R^{2} - \frac{h^{2}}{4}.\]

\[V(h) = \pi r^{2} \bullet h =\]

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

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

\[= \pi R^{2} - \frac{3\pi}{4}h^{2}.\]

\[Промежуток\ возрастания:\]

\[\pi R^{2} - \frac{3\pi}{4}h^{2} > 0\]

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

\[3\pi h^{2} < 4\pi R^{2}\]

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

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

\[h = \frac{2R}{\sqrt{3}} - точка\ максимума;\]

\[r = \sqrt{R^{2} - \frac{h^{2}}{4}} = \sqrt{R^{2} - \frac{4R^{2}}{3 \bullet 4}} =\]

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

\[Ответ:\ \ h = \frac{2R}{\sqrt{3}};\ \ r = R\sqrt{\frac{2}{3}}.\]