Question

Знайти екстремуми функції
Терміново!!!!!

Answer 1

Ответ:

$z = 2 {x}^{3} + x {y}^{2} + 5 {x}^{2} + {y}^{2}$

$z'_x = 6 {x}^{2} + {y}^{2} + 10x$

$z'_y = 2xy + 2y$

$\left \{ {{z'_x = 0} \atop {z'_y = 0} } \right. \\ \\ \left \{ {{ 6 {x}^{2} + {y}^{2} + 10 x = 0} \atop {2xy + 2y = 0} } \right. \\ \\ \left \{ {{x = - 1} \atop {y {}^{2} = - 10x - 6 {x}^{2} } } \right. \\ \\ {y}^{2} = 10 - 6 = 4 \\ y = \pm2 \\ \\ y_1 = 2 \\ 6 {x}^{2} + 10x + 4 = 0 \\ 3 {x}^{2} + 5x + 2 = 0 \\ D = 25 - 24 = 1 \\ x_1 = \frac{ - 5 + 1}{6} = - \frac{2}{3} \\ x_2 = - 1 \\ \\ y_2 = - 2 \\ 6 {x}^{2} + 10x + 4 = 0 \\ x_1 = -\frac{2}{3} \\ x_2 = - 1$

Ответ:

(-2/3; 2), (-1; 2), (-2/3; -2), (-1; -2) - экстремумы