Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1529

\[\boxed{\mathbf{1529}\mathbf{.}}\]

\[r - радиус\ основания;\]

\[h - высота\ цилиндра.\]

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

\[h = \frac{V}{\pi r^{2}}.\]

\[S(r) = \left( 2 \bullet \pi r^{2} \right) + (2\pi r \bullet h) =\]

\[= 2\pi r^{2} + 2\pi r \bullet \frac{V}{\pi r^{2}} =\]

\[= 2\pi r^{2} + \frac{2V}{r};\]

\[S^{'}(r) = 2\pi\left( r^{2} \right)^{'} + 2V \bullet \left( \frac{1}{r} \right)^{'} =\]

\[= 2\pi \bullet 2r + 2V \bullet \left( - \frac{1}{r^{2}} \right) =\]

\[= \frac{4\pi r^{3} - 2V}{r^{2}}.\]

\[Промежуток\ возрастания:\]

\[4\pi r^{3} - 2V > 0\]

\[4\pi r^{3} > 2V\]

\[r^{3} > \frac{V}{2\pi}\]

\[r > \sqrt[3]{\frac{V}{2\pi}}.\]

\[r = \sqrt[3]{\frac{V}{2\pi}} - точка\ минимума;\]

\[\frac{h}{r} = \frac{V}{\pi r^{2}}\ :r = \frac{V}{\pi} \bullet \frac{1}{r^{3}} = \frac{V}{\pi} \bullet \frac{2\pi}{V} = 2.\]

\[Ответ:\ \ h = 2r.\]