IDZ 10.2 - Option 28. Decisions Ryabushko AP
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- 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)
- Find the second partial derivatives of the functions. Ensure that z "xy = z" yx
2.28 z = arcctg (x - 4y)
- To verify whether the above equation the function u.
- Examine the following function extremum.
4.28 z = xy - 3x2 - 2y2
- 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
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