ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1511

\[\boxed{\mathbf{1511}\mathbf{.}}\]

\[\text{r\ }см - радиус\ основания;\]

\[\text{h\ }см - высота\ цилиндра.\]

\[S_{пов} = 54\pi\ см^{2}:\]

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

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

\[h = \frac{S - 2\pi r^{2}}{2\pi r} = \frac{54\pi - 2\pi r^{2}}{2\pi r} =\]

\[= \frac{27 - r^{2}}{r}.\]

\[V(r) = \pi r^{2} \bullet h = \pi r^{2} \bullet \frac{27 - r^{2}}{r} =\]

\[= \pi r \bullet \left( 27 - r^{2} \right) = 27\pi r - \pi r^{3};\]

\[V^{'}(r) = 27\pi(r)^{'} - \pi\left( r^{3} \right)^{'} =\]

\[= 27\pi - \pi \bullet 3r^{2} = 3\pi \bullet \left( 9 - r^{2} \right).\]

\[Промежуток\ возрастания:\]

\[9 - r^{2} > 0\]

\[r^{2} < 9\]

\[- 3 < r < 3.\]

\[r = 3 - точка\ максимума;\]

\[V(3) = 27\pi \bullet 3 - \pi \bullet 3^{3} =\]

\[= 81\pi - 27\pi = 54\pi\ \left( см^{3} \right).\]

\[Ответ:\ \ 54\pi\ см^{3}.\]