ИДЗ 10.2 – Вариант 18. 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)
1.18 S: x2 - y2 + z2 - 4x + 2y = 14, M0 (3, 1, 4)

2. Find the second partial derivatives of the functions. Ensure that z "xy = z" yx
2.18 z = arctg (5x + 2y)

3. To verify whether the above equation the function u.
4. Examine the following function extremum.
4.18 z = xy (12 - x - y)

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

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