Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1512

\[\boxed{\mathbf{1512}\mathbf{.}}\]

\[AB = BC = AC = x;\]

\[\text{SO} = h:\]

\[x + h = SO + AC = 9\]

\[h = 9 - x\]

\[1 \leq x \leq 8.\]

\[V(x) = \frac{1}{3}Sh = \frac{1}{3} \bullet \frac{x^{2}\sqrt{3}}{4} \bullet (9 - x) =\]

\[= \frac{\sqrt{3}}{12} \bullet \left( 9x^{2} - x^{3} \right);\]

\[V^{'}(x) = \frac{\sqrt{3}}{12} \bullet \left( 9\left( x^{2} \right)^{'} - \left( x^{3} \right)^{'} \right) =\]

\[= \frac{\sqrt{3}}{12} \bullet \left( 9 \bullet 2x - 3x^{2} \right) =\]

\[= \frac{3\sqrt{3}}{12} \bullet \left( 6x - x^{2} \right).\]

\[Промежуток\ возрастания:\]

\[6x - x^{2} > 0\]

\[x \bullet (6 - x) > 0\]

\[x \bullet (x - 6) < 0\]

\[0 < x < 6.\]

\[x = 6 - точка\ максимума.\]

\[Ответ:\ \ 6.\]