IDZ 10.2 - Option 18. 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)

1.18 S: x2 - y2 + z2 - 4x + 2y = 14, M0 (3, 1, 4)

• Find the second partial derivatives of the functions. Ensure that z "xy = z" yx

2.18 z = arctg (5x + 2y)

• To verify whether the above equation the function u.
• Examine the following function extremum.

4.18 z = xy (12 - x - y)

• 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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