Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1527

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\[r - радиус\ основания;\]

\[h - высота\ цилиндра.\]

\[Периметр\ сечения\ цилиндра = \text{\ p}:\]

\[2 \bullet (2r + h) = p\]

\[4r + 2h = p\]

\[2h = p - 4r\]

\[h = \frac{p}{2} - 2r.\]

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

\[= \frac{1}{2}\text{pπ}r^{2} - 2\pi r^{3};\]

\[V^{'}(r) = \frac{1}{2}\text{pπ}\left( r^{2} \right)^{'} - 2\pi\left( r^{3} \right)^{'} =\]

\[= \frac{1}{2}p\pi \bullet 2r - 2\pi \bullet 3r^{2} =\]

\[= p\pi r - 6\pi r^{2}.\]

\[Промежуток\ возрастания:\]

\[p\pi r - 6\pi r^{2} > 0\]

\[pr - 6r^{2} > 0\]

\[r \bullet (p - 6r) > 0\]

\[r \bullet (6r - p) < 0\]

\[0 < r < \frac{p}{6}.\]

\[r = \frac{p}{6} - точка\ максимума;\]

\[V\left( \frac{p}{6} \right) = \pi\left( \frac{p}{6} \right)^{2} \bullet \left( \frac{p}{2} - 2 \bullet \frac{p}{6} \right) =\]

\[= \frac{\pi p^{2}}{36} \bullet \left( \frac{3p}{6} - \frac{2p}{6} \right) =\]

\[= \frac{\pi p^{2}}{36} \bullet \frac{p}{6} = \frac{\pi p^{3}}{216}.\]

\[Ответ:\ \ \frac{\pi p^{3}}{216}.\]