Решебник по алгебре 11 класс Никольский Параграф 5. Применение производной Задание 100

\[\boxed{\mathbf{100}\mathbf{.}}\]

\[Объем\ цилиндра:\]

\[V = \pi R^{2}h = \pi \cdot \frac{d^{2}}{4}\text{h.}\]

\[Пусть\ x > 0 - диаметр;\]

\[h = \frac{4V}{\pi d^{2}} = \frac{4V}{\pi x^{2}} - высота.\]

\[S_{пов} = 2S_{осн} + S_{бок} = 2\pi кр +\]

\[+ 2\pi r^{2} = 2\pi\left( rh + r^{2} \right) =\]

\[= 2\pi\left( \frac{x}{2} \cdot \frac{4V}{\pi x^{2}} + \frac{x^{2}}{4} \right) =\]

\[= 2\pi\left( \frac{2V}{\text{πx}} + \frac{x^{2}}{4} \right) = \frac{4\pi V}{\text{πx}} +\]

\[+ \frac{2\pi x^{2}}{4} = \frac{4V}{x} + \frac{\pi x^{2}}{2};\]

\[S_{пов} = - \frac{4V}{x^{2}} + \frac{2\text{πx}}{2} = - \frac{4V}{x^{2}} + \pi x.\]

\[- \frac{4V}{x^{2}} + \pi x = 0\]

\[\frac{- 4V + \pi x^{3}}{x^{2}} = 0\]

\[- 4V + \pi x^{3} = 0\]

\[x = \sqrt[3]{\frac{4V}{\pi}} - диаметр\ цилиндра.\]

\[S_{пов}^{''} = \frac{4V}{x^{2}} + \pi > 0:\]

\[x = \sqrt[3]{\frac{4V}{\pi}} - точка\ минимума.\]

\[h = \frac{4V}{\pi x^{2}} = \frac{4V}{\pi\left( \sqrt[3]{\frac{4V}{\pi}} \right)^{2}} = \sqrt[3]{\frac{4V}{\pi}}.\]

\[Ответ:диаметр\ и\ высота\ \]

\[равны\ \sqrt[3]{\frac{4V}{\pi}}.\]