Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 350

\[AB = x;\ BC = y;\ AC = z:\]

\[z^{2} = x^{2} + y^{2};\]

\[y^{2} = z^{2} - x^{2}.\]

\[1)\ Прочность\ балки:\]

\[T(x) = xy^{2} = x\left( z^{2} - x^{2} \right) =\]

\[= xz^{2} - x^{3};\]

\[T^{'}(x) = z^{2} - 3x^{2}.\]

\[2)\ z^{2} - 3x^{2} \geq 0\]

\[3x^{2} \leq z^{2}\]

\[x^{2} \leq \frac{z^{2}}{3}\]

\[x \leq \frac{z}{\sqrt{3}}.\]

\[3)\ Точка\ максимума:\]

\[x = \frac{z}{\sqrt{3}};\]

\[y^{2} = z^{2} - \frac{z^{2}}{3} = \frac{2z^{2}}{3}\]

\[y = \frac{z\sqrt{2}}{\sqrt{3}}.\]

\[4)\ Проверим:\]

\[AC = R;AM = \frac{1}{3}R;\ MC = \frac{2}{3}R;\]

\[BM = \sqrt{AM \bullet BM} = \sqrt{\frac{1}{3}R \bullet \frac{2}{3}R} =\]

\[= \frac{R\sqrt{2}}{3};\]

\[AB = \sqrt{AM^{2} + BM^{2}} =\]

\[= \sqrt{\frac{1}{9}R^{2} + \frac{2}{9}R^{2}} = \frac{R\sqrt{3}}{3} = \frac{R}{\sqrt{3}};\]

\[BC = \sqrt{MC^{2} + BM^{2}} =\]

\[= \sqrt{\frac{4}{9}R^{2} + \frac{2}{9}R^{2}} = \frac{R\sqrt{6}}{3} = \frac{R\sqrt{2}}{\sqrt{3}}.\]

\[Ответ:\ \ разметка\ верна.\]