ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 343

\[r - радиус\ основания;\]

\[h - высота:\]

\[2\pi r^{2} + 2\pi rh = S\]

\[2\pi rh = S - 2\pi r^{2}\]

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

\[1)\ V(r) = \pi r^{2}h =\]

\[= \pi r^{2}\left( \frac{S}{2\pi r} - r \right) = \frac{\text{rS}}{2} - \pi r^{3};\]

\[V^{'}(r) = \frac{S}{2} - 3\pi r^{2}.\]

\[2)\ \frac{S}{2} - 3\pi r^{2} \geq 0\]

\[3\pi r^{2} \leq \frac{S}{2}\]

\[r^{2} \leq \frac{S}{6\pi}\]

\[r \leq \sqrt{\frac{S}{6\pi}}.\]

\[3)\ Точка\ максимума:\]

\[r = \sqrt{\frac{S}{6\pi}};\]

\[V(r) = \sqrt{\frac{S}{6\pi}} \bullet \frac{S}{2} - \pi\sqrt{\frac{S^{3}}{6^{3}\pi^{3}}} =\]

\[= \frac{S^{\frac{3}{2}}}{2\sqrt{6\pi}} - \frac{S^{\frac{3}{2}}}{6\sqrt{6\pi}} = \frac{3S^{\frac{3}{2}} - S^{\frac{3}{2}}}{6\sqrt{6\pi}} =\]

\[= \frac{2S^{\frac{3}{2}}}{6\sqrt{6\pi}} = \frac{S^{\frac{3}{2}}}{3\sqrt{6\pi}};\]

\[Ответ:\ \ \frac{S^{\frac{3}{2}}}{3\sqrt{6\pi}}.\]