Find the local maximum and minimum values and saddle point(s) of the function f(x, y) = x³ + y³ - 3x² - 3y² - 9x

Solution:

Given function f(x, y) = x³ + y³ - 3x² - 3y² - 9x

A function, an expression, rule, or law that defines a relationship between one variable (the independent variable) and another variable (the dependent variable)

To find the local maxima or minima, we need to equate the first derivative of the function to zero.

Let’s first differentiate f(x, y) w.r.t ‘x’

f’(x, y) = 3x² - 6x - 9 = 0

⇒ 3(x² - 2x - 3) = 0

⇒ (x - 3)(x + 1) = 0

⇒ x = 3, -1

Now, differentiate w.r.t ‘y’

f’(x, y) = 3y² - 6y = 0

⇒ 3y(y - 2) = 0

⇒ y = 0, 1

Put (3, 0) in f(x, y), we get f(3, 0) = 3³ + 0³ - 3(2)² - 3(0)² - 9(3)

= 27 - 12 - 27 = -12

Put (-1, 1) in f(x, y), we get f(-1, 1) = (-1)³ + 1³ - 3(-1)² - 3(1)² - 9(-1) = -1 + 1 - 3 - 3 + 9 = 3

Put (3, 1) in f(x, y), we get f(3, 1) = 3³ + 1³ - 3(3)² - 3(1)² - 9(3) = 27 + 1 - 27 - 3 - 27 = -29

Put ( -1, 0) in f(x, y), we get f(-1, 0) = (-1)³ + 0³ - 3(-1)² - 3(0)² - 9(-1) = -1 - 3 + 9 = 5

In all possible values, the minimum value is -29 and the maximum value is 5

---

Find the local maximum and minimum values and saddle point(s) of the function f(x, y) = x³ + y³ - 3x² - 3y² - 9x

Summary:

The local maximum and minimum values and saddle point(s) of the function f(x, y) = x³ + y³ - 3x² - 3y² - 9x are 5 and -29 respectively.