Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 964

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Год:2020-2021-2022-2023
Тип:учебник

Задание 964

\[1)\ \left\{ \begin{matrix} x + 3y - 3 \geq 0\ \ \ \ \ \\ 2x + 3y - 12 \leq 0 \\ x \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 0 \leq y \leq 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[x + 3y - 3 \geq 0\]

\[3y \geq 3 - x\]

\[y \geq 1 - \frac{1}{3}x\]

\[y(0) = 1 - 0 = 1;\]

\[y(3) = 1 - 1 = 0.\]

\[2x + 3y - 12 \leq 0\]

\[3y \leq 12 - 2x\]

\[y \leq 4 - \frac{2}{3}x\]

\[y(0) = 4 - 0 = 4;\]

\[y(3) = 4 - 2 = 2.\]

\[S = 3 \bullet 1 + \frac{1}{2} \bullet 3 \bullet 1 + \frac{1}{2} \bullet 3 \bullet 2 =\]

\[= 3 + 1,5 + 3 = 7,5.\]

\[Ответ:\ \ 7,5.\]

\[2)\ \left\{ \begin{matrix} x - y + 1 \leq 0\ \ \ \ \ \ \ \\ 5x - 3y + 15 \geq 0 \\ x \leq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 0 \leq y \leq 2,5\ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[x - y + 1 \leq 0\]

\[y \geq x + 1\]

\[y( - 1) = - 1 + 1 = 0;\]

\[y(0) = 0 + 1 = 1.\]

\[5x - 3y + 15 \geq 0\]

\[3y \leq 5x + 15\]

\[y \leq \frac{5}{3}x + 5\]

\[y( - 3) = - 5 + 5 = 0;\]

\[y(0) = 0 + 5 = 5.\]

\[S = 3 \bullet 2,5 - \frac{1}{2} \bullet 1,5 \bullet 2,5 - \frac{1}{2} \bullet 1 \bullet 1 =\]

\[= 7,5 - 1,875 - 0,5 = 5,125.\]

\[Ответ:\ \ 5,125.\]

\[3)\ \left\{ \begin{matrix} (x + y - 1)(x + y - 3) \leq 0 \\ x - y \leq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[(x + y - 1)(x + y - 3) \leq 0\]

\[x + y - 1 \geq 0\]

\[y \geq 1 - x;\]

\[\ x + y - 3 \leq 0\ \ \]

\[y \leq 3 - x;\]

\[1 - x \leq y \leq 3 - x.\]

\[y(0) = 1 - 0 = 1;\ \ \ \]

\[y(0) = 3 - 0 = 3;\]

\[y(1) = 1 - 1 = 0;\]

\[y(1) = 3 - 1 = 2.\]

\[x - y \leq 0\]

\[y \geq x\]

\[y(0) = 0;\]

\[y(1) = 1.\]

\[S =\]

\[= 1 \bullet 1 + \frac{1}{2} \bullet 1 \bullet 1 + \frac{1}{4} \bullet 1 \bullet 1 + \frac{1}{4} \bullet 1 \bullet 1 =\]

\[= 1 + 0,5 + 0,25 + 0,25 = 2.\]

\[Ответ:\ \ 2.\]

\[4)\ \left\{ \begin{matrix} (2x - y + 3)(4x + 3y - 9) \leq 0 \\ x + y \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ - 1 \leq y \leq 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[(2x - y + 3)(4x + 3y - 9) \leq 0\]

\[2x - y + 3 \geq 0\]

\[y \leq 2x + 3\]

\[y \leq 2x + 3.\ \ \]

\[4x + 3y - 9 \leq 0\]

\[3y \leq 9 - 4x\]

\[\ y \leq 3 - \frac{4}{3}\text{x.}\]

\[y(0) = 0 + 3 = 3;\]

\[y(0) = 3 - 0 = 3;\]

\[y(1) = 2 + 3 = 5;\]

\[y(3) = 3 - 4 = - 1.\]

\[x + y \geq 0\]

\[y \geq - x;\]

\[y(0) = 0;\ \ \ \]

\[y(1) = - 1.\]

\[S =\]

\[= 4 \bullet 4 - \frac{1}{2} \bullet 2 \bullet 1 - \frac{1}{2} \bullet 3 \bullet 4 - \frac{1}{2} \bullet 2 \bullet 2 =\]

\[= 16 - 1 - 6 - 2 = 10 - 3 = 7.\]

\[Ответ:\ \ 7.\]

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