Final answer:
The maximum value of the function f(x, y, z) = x²yz + 1 on the intersection of the plane z=1 with the sphere x² + y² + z² = 26 is 26.5, which occurs at the coordinates (-5/√2, 5/√2, 1).
Step-by-step explanation:
The student is tasked with finding the extreme values of the function f(x, y, z) = x²yz + 1 when it is constrained by the intersection of a plane and a sphere. Specifically, the constraints are given by z=1 (the equation of a plane) and x² + y² + z² = 26 (the equation of a sphere). To find the maximum value of the function f on this intersection, we first substitute the plane's equation z=1 into both the function and the sphere's equation. This gives us a modified function f(x, y) = x²y + 1 and a new constraint in the form of a circle x² + y² + 1² = 26, which simplifies to x² + y² = 25. The circle has a radius of 5, and the maximum of f occurs when the product x²y is maximized. Taking into account the constraint x² + y² = 25 and the symmetry of the circular constraint, we can infer that the maximum product occurs when x and y are equal or when x and y are -5/√2 and 5/√2 respectively (since the product of two numbers with a fixed sum is maximized when the numbers are equal). Substituting these values into the modified function f(x, y), we obtain f(-5/√2, 5/√2) = 25/2 + 1 = 26.5 as the maximum value of f.