ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 299

\[a,\ a\ и\ b - стороны\ \]

\[параллелепипеда.\]

\[1)\ d = \sqrt{a^{2} + a^{2}} = \sqrt{2a^{2}} = a\sqrt{2}.\]

\[2)\ 2R = \sqrt{d^{2} + b^{2}} = \sqrt{2a^{2} + b^{2}}\]

\[4R^{2} = 2a^{2} + b^{2}\]

\[b^{2} = 4R^{2} - 2a^{2}\]

\[b = \sqrt{4R^{2} - 2a^{2}}.\]

\[3)\ S(a) = aab = a^{2}\sqrt{4R^{2} - 2a^{2}} =\]

\[= \sqrt{4R^{2}a^{4} - 2a^{6}};\]

\[S^{'}(a) = \frac{4R^{2} \bullet 4a^{3} - 2 \bullet 6a^{5}}{2\sqrt{4R^{2}a^{4} - 2a^{6}}} =\]

\[= \frac{8R^{2}a^{3} - 6a^{5}}{\sqrt{4R^{2}a^{4} - 2a^{6}}}.\]

\[4)\ Промежуток\ возрастания:\]

\[8R^{2}a^{3} - 6a^{5} \geq 0\]

\[2a^{3}\left( 3a^{2} - 4R^{2} \right) \leq 0\]

\[\left( a\sqrt{3} + 2R \right)a\left( a\sqrt{3} - 2R \right) \leq 0\]

\[a \leq - \frac{2R}{\sqrt{3}};\ \]

\[0 \leq a \leq \frac{2R}{\sqrt{3}}.\]

\[5)\ Точка\ максимума:\]

\[a = \frac{2R}{\sqrt{3}},\ \ \ b = \sqrt{4R^{2} - 2 \bullet \frac{4R^{2}}{3}} =\]

\[{= \sqrt{\frac{4R^{2}}{3}} = \frac{2R}{\sqrt{3}}. }{Ответ:\ \ куб\ со\ стороной\ \frac{2R}{\sqrt{3}}.}\]