IDZ 10.2 - Option 27. 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)
1.27 S: z = 2x2 - 3y2 + 4x - 2y + 10, M0 (-1, 1, 3)

2. Find the second partial derivatives of the functions. Ensure that z "xy = z" yx
2.27 z = ln (5x2 - 3y4)

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

5. Find the maximum and minimum values of the function z = z (x, y) in D, given the limited lines.
5.27 z = 4 - 2x2 - y2, D: y = 0, y = √1-x2