ИДЗ 10.2 – Вариант 13. Решения Рябушко А.П.

1. Find the equation of the tangent plane and the normal to the given surface S at the point M0 (x0, y0, z0)
1.13 S: z = x2 - y2 - 2xy - x - 2y, M0 (-1, 1, 1)

2. Find the second partial derivatives of the functions. Ensure that z "xy = z" yx
2.13 z = tg√xy

3. To verify whether the above equation the function u.
4. Examine the following function extremum.
4.13 z = (x - 5) 2 + y2 + 1

5. Find the maximum and minimum values of the function z = z (x, y) in D, given the limited lines.
5.13 z = 3x2 + 3y2 - 2x - 2y + 2, D: x = 0, y = 0, x + y - 1 = 0

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