ИДЗ 10.2 – Вариант 28. 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)
1.28 S: z = x2 + y2 - 4xy + 3x - 15, M0 (-1, 3, 4)

2. Find the second partial derivatives of the functions. Ensure that z "xy = z" yx
2.28 z = arcctg (x - 4y)

3. To verify whether the above equation the function u.
4. Examine the following function extremum.
4.28 z = xy - 3x2 - 2y2

5. Find the maximum and minimum values of the function z = z (x, y) in D, given the limited lines.
5.28 z = 5x2 - 3xy + y2 + 4, D: x = -1, x = 1, y = -1, y = 1

Detailed solution. Designed in PDF format for easy viewing of IDZ solutions on smartphones and PCs. In MS Word (doc format) sent additionally.