Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 337

\[a,b,c - стороны\ треугольника:\]

\[a + c = l\]

\[c = l - a.\]

\[1)\ b = \sqrt{c^{2} - a^{2}} =\]

\[= \sqrt{(l - a)^{2} - a^{2}} =\]

\[= \sqrt{l^{2} - 2al + a^{2} - a^{2}} =\]

\[= \sqrt{l^{2} - 2al}.\]

\[2)\ S(a) = \frac{1}{2}ab = \frac{1}{2}a\sqrt{l^{2} - 2al} =\]

\[= \frac{\sqrt{a^{2}l^{2} - 2a^{3}l}}{2};\]

\[S^{'}(a) = \frac{1}{2} \bullet \frac{2al^{2} - 2 \bullet 3a^{2}l}{2\sqrt{a^{2}l^{2} - 2a^{3}l}} =\]

\[= \frac{al^{2} - 3a^{2}l}{2\sqrt{a^{2}l^{2} - 2a^{3}l}}.\]

\[3)\ al^{2} - 3a^{2}l \geq 0\]

\[l - 3a \geq 0\]

\[3a \leq l\]

\[a \leq \frac{l}{3}.\]

\[4)\ Точка\ максимума:\]

\[a = \frac{l}{3};\text{\ \ \ }\]

\[c = l - \frac{l}{3} = \frac{2l}{3};\]

\[b = \sqrt{\frac{4l^{2}}{9} - \frac{l^{2}}{9}} = \sqrt{\frac{l^{2}}{3}} = \frac{l}{\sqrt{3}}.\]

\[Ответ:\ \ катеты\ \frac{l}{3}\ и\ \frac{l}{\sqrt{3}};\ \]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }гипотенуза\ \frac{2l}{3}.\]