Question

Find the maximum rate of change of f at the given point and the direction in which it occurs. The function f(x, y) = 3y² x, is evaluated at the point (3, 6). What is the maximum rate of change and the direction vector?

Answer 1

The maximum rate of change of the function f(x, y) = 3y² x at the point (3, 6) is 108√2, and it occurs in the direction of the vector (1/√2, 1/√2).

Step-by-step explanation:

The maximum rate of change of a function f(x, y) at a given point can be found using the gradient of f, denoted as ∇f. The direction of the maximum rate of change is the direction of the gradient at that point. For the function f(x, y) = 3y² x, evaluated at the point (3, 6), we first find the partial derivatives with respect to x and y to compute the gradient:

• fx(x, y) = 3y² (derivative of f with respect to x)
• fy(x, y) = 6yx (derivative of f with respect to y)

Then we evaluate the partial derivatives at the point (3, 6):

• fx(3, 6) = 3(6)² = 108
• fy(3, 6) = 6(6)(3) = 108

The gradient vector at (3, 6) is therefore ∇f(3, 6) = (108, 108). The maximum rate of change is the magnitude of the gradient vector, calculated as:

|∇f(3, 6)| = √(108² + 108²) = √(2 * 108²) = 108√2.

The direction vector is the normalized gradient vector, which is the gradient vector divided by its magnitude:

direction vector = ∇f(3, 6) / |∇f(3, 6)| = (108, 108) / 108√2 = (1/√2, 1/√2).

The maximum rate of change of f at the point (3, 6) is 108√2 and occurs in the direction of the vector (1/√2, 1/√2).