IDZ 10.2 - Option 26. Decisions Ryabushko AP

1. Find the equation of the tangent plane and the normal to the given surface S at the point M0 (x0, y0, z0)
1.26 S: x2 + y2 - z2 + 6xy - z = 8, M0 (1, 1, 0)

2. Find the second partial derivatives of the functions. Ensure that z "xy = z" yx
2.26 z = arctg (3x + 2y)

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

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