A critical point of a multivariable function is a point where the partial derivatives of the first order of this function are equal to zero. Examples with detailed solutions on how to find the critical points of a function with two variables are presented.
Explore more Optimization Problems with Functions of Two Variables on this website.
In what follows, \( C \) represents the constant of integration where applicable.
Examples with Detailed Solutions
Click on each example to view its detailed step-by-step solution.
Example 1
Find the critical point(s) of function \( f \) defined by:
\[ f(x, y) = x^2 + y^2 \]Show Solution to Example 1
We first find the first-order partial derivatives:
\[ f_x(x, y) = 2x \] \[ f_y(x, y) = 2y \]We now solve the following equations \( f_x(x, y) = 0 \) and \( f_y(x, y) = 0 \) simultaneously:
\[ 2x = 0 \implies x = 0 \] \[ 2y = 0 \implies y = 0 \]The solution to the above system of equations is the ordered pair \( (0, 0) \).
Below is the graph of \( f(x, y) = x^2 + y^2 \), showing that at the critical point \( (0, 0) \), \( f \) has a minimum value.
Example 2
Find the critical point(s) of function \( f \) defined by:
\[ f(x, y) = x^2 - y^2 \]Show Solution to Example 2
Find the first-order partial derivatives of function \( f \):
\[ f_x(x, y) = 2x \] \[ f_y(x, y) = -2y \]Solve the equations \( f_x(x, y) = 0 \) and \( f_y(x, y) = 0 \) simultaneously:
\[ 2x = 0 \implies x = 0 \] \[ -2y = 0 \implies y = 0 \]The solution is the ordered pair \( (0, 0) \).
The graph of \( f(x, y) = x^2 - y^2 \) is shown below. The surface curves down in the \( y \)-direction and up in the \( x \)-direction. \( f \) is stationary at \( (0, 0) \), but there is no extremum (maximum or minimum). \( (0, 0) \) is called a saddle point because the surface close to \( (0, 0) \) resembles a riding saddle.
Example 3
Find the critical point(s) of function \( f \) defined by:
\[ f(x, y) = -x^2 - y^2 \]Show Solution to Example 3
We first find the first-order partial derivatives:
\[ f_x(x, y) = -2x \] \[ f_y(x, y) = -2y \]We now solve \( f_x(x, y) = 0 \) and \( f_y(x, y) = 0 \) simultaneously:
\[ -2x = 0 \implies x = 0 \] \[ -2y = 0 \implies y = 0 \]The solution is the ordered pair \( (0, 0) \).
The graph of \( f(x, y) = -x^2 - y^2 \) is shown below, illustrating a relative maximum.
Example 4
Find the critical point(s) of function \( f \) defined by:
\[ f(x, y) = x^3 + 3x^2 - 9x + y^3 - 12y \]Show Solution to Example 4
The first-order partial derivatives are given by:
\[ f_x(x, y) = 3x^2 + 6x - 9 \] \[ f_y(x, y) = 3y^2 - 12 \]We now solve the equations \( f_x(x, y) = 0 \) and \( f_y(x, y) = 0 \) simultaneously:
\[ 3x^2 + 6x - 9 = 0 \implies 3(x + 3)(x - 1) = 0 \implies x = -3 \text{ or } x = 1 \] \[ 3y^2 - 12 = 0 \implies 3(y^2 - 4) = 0 \implies y = -2 \text{ or } y = 2 \]Combining all possible combinations of \( x \) and \( y \), the critical points are:
\[ (1, 2), \; (1, -2), \; (-3, 2), \; (-3, -2) \]Exercises
Find, if any, the critical points for the functions below. Click each exercise to check your answers.
Exercise 1
\( f(x, y) = 3xy - x^3 - y^3 \)
Show Answer
Exercise 2
\( f(x, y) = e^{-x^2 - y^2 + 2x - 2y - 2} \)
Show Answer
Exercise 3
\( f(x, y) = \dfrac{1}{2}x^2 + y^3 - 3xy - 4x + 2 \)
Show Answer
Exercise 4
\( f(x, y) = x^3 + y^3 + 2x + 6y \)