Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 979

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\[h - высота\ кузова;\ \]

\[\text{a\ }и\ b - его\ длина\ и\ ширина:\]

\[\frac{a}{b} = \frac{5}{2} = k \rightarrow \ a = 5k;\ \ \ b = 2k.\]

\[S_{пов} = ab + 2ah + 2bh =\]

\[= 10k^{2} + 10kh + 4kh =\]

\[= 10k^{2} + 14kh = 2S;\]

\[14kh = 2S - 10k^{2}\]

\[h = \frac{2S - 10k^{2}}{14k}.\]

\[V(k) = abh =\]

\[= 5k \bullet 2k \bullet \frac{2S - 10k^{2}}{14k} =\]

\[= \frac{20Sk^{2} - 100k^{4}}{14k} =\]

\[= \frac{5}{7}\left( 2Sk - 10k^{3} \right).\]

\[V^{'}(k) =\]

\[= \frac{5}{7} \bullet \left( 2S \bullet (k)^{'} - 10 \bullet \left( k^{3} \right)^{'} \right);\]

\[V^{'}(k) = \frac{5}{7} \bullet \left( 2S - 10 \bullet 3k^{2} \right) =\]

\[= 10 \bullet \frac{S - 15k^{2}}{7}.\]

\[Промежуток\ возрастания:\]

\[S - 15k^{2} > 0\]

\[15k^{2} < S\]

\[k^{2} < \frac{S}{15}\]

\[- \sqrt{\frac{S}{15}} < k < \sqrt{\frac{S}{15}}.\]

\[k = \sqrt{\frac{S}{15}} - точка\ максимума;\]

\[a = 5\sqrt{\frac{S}{15}};\ \ \ \ \text{\ \ }b = 2\sqrt{\frac{S}{15}}.\]

\[Ответ:\ \ 5\sqrt{\frac{S}{15}}\ и\ \ 2\sqrt{\frac{S}{15}}.\]