ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1522

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\[r\ - радиус\ основания;\]

\[h - высота\ цилиндра:\]

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

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

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

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

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

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

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

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

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

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

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

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

\[S(r) = 2\pi \bullet \sqrt[3]{\frac{V^{2}}{4\pi^{2}}} + 2V \bullet \sqrt[3]{\frac{2\pi}{V}} =\]

\[= \sqrt[3]{\frac{8\pi^{3}V^{2}}{4\pi^{2}}} + 2\sqrt[3]{\frac{V^{3} \bullet 2\pi}{V}} =\]

\[= \sqrt[3]{2\pi V^{2}} + 2\sqrt[3]{2\pi V^{2}} = 3\sqrt[3]{2\pi V^{2}}.\]

\[Ответ:\ \ 3\sqrt[3]{2\pi V^{2}}.\]