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Find the extreme values of f(x, y, z) = x²yz + 5 on the intersection of the plane z = 1 with the sphere x² + y² + z² = 17.

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Final answer:

The extreme values of f(x, y, z) = x²yz + 5 on the intersection of the plane z = 1 with the sphere x² + y² + z² = 17 occur at ( √2, √2, 1) and (- √2, - √2, 1), where f takes the value of 7.

Step-by-step explanation:

To find the extreme values of the given function subject to the given constraints, we use the method of Lagrange multipliers. First, we express the constraints as the system of equations z = 1 and x² + y² + z² = 17. By substituting z = 1 into the second equation, we obtain x² + y² = 16, which represents a circle in the xy-plane with a radius of 4.

Next, we set up the Lagrangian function L(x, y, z, λ) = x²yz + 5 - λ(x² + y² + z² - 17). Solving the system of equations formed by setting the partial derivatives of L with respect to x, y, z, λ equal to zero, we find the critical points. These points are then evaluated on the constraint z = 1, leading to the solutions (√2, √2, 1) and (- √2, - √2, 1), where the function f takes the extreme value of 7.

In summary, the extreme values of the given function on the intersection of the specified plane and sphere occur at the points (√2, √2, 1) and (- √2, - √2, 1), where the function takes the value of 7. These points represent critical values on the constrained surface, and the Lagrange multipliers method allows us to identify these extreme points systematically.

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