Answer: Therefore at (x, y, z) = (3/8, 1/4, 1/6), the function f(x, y, z) assumes its minimum values.
Explanation:
First, let's find the partial derivatives of f(x, y, z) with respect to x, y, and z:
∂f/∂x = 8x - 3
∂f/∂y = 8y - 2
∂f/∂z = 6z - 1
To find the critical points, we set these partial derivatives equal to zero and solve for x, y, and z:
8x - 3 = 0 --> x = 3/8
8y - 2 = 0 --> y = 1/4
6z - 1 = 0 --> z = 1/6
Therefore, the critical point is (x, y, z) = (3/8, 1/4, 1/6).
To determine if this point is a minimum, we can use the second derivative test.
∂^2f/∂x^2 = 8
∂^2f/∂y^2 = 8
∂^2f/∂z^2 = 6
Since the second partial derivatives are all positive, the critical point (3/8, 1/4, 1/6) is a local minimum.