Question

Consider the function f(x,y)=x³+y³−3x−12y+20. Determine the four stationary points of this function and enter them in parts a) to d). Note: Enter these in increasing order of their x-coordinates first (if the x coordinates are the same, then enter them in increasing order of their y coordinates).

Answer 1

The four stationary points of the function f(x, y) = x³ + y³ - 3x - 12y + 20 are (-1, -2), (-1, 2), (1, -2), and (1, 2).

Step-by-step explanation:

To find the stationary points of the function f(x, y) = x³ + y³ - 3x - 12y + 20, we need to find the values of x and y where the partial derivatives with respect to x and y are both equal to zero.

First, we find the partial derivative with respect to x:

∂f/∂x = 3x² - 3

To find the stationary points, set this equal to zero and solve for x: 3x² - 3 = 0

Solving this quadratic equation, we find x = 1 and x = -1.

Next, we find the partial derivative with respect to y:

∂f/∂y = 3y² - 12

To find the stationary points, set this equal to zero and solve for y: 3y² - 12 = 0

Solving this quadratic equation, we find y = 2 and y = -2.

Therefore, the four stationary points of the function are:

(-1, -2), (-1, 2), (1, -2), (1, 2)