Critical Points of Functions of Two Variables

A critical point of a multivariable function is a point where the partial derivatives of first order of this function are equal to zero. Examples with detailed solution on how to find the critical points of a function with two variables are presented.
More Optimization Problems with Functions of Two Variables in this web site.

Examples with Detailed Solutions

Example 1

Find the critical point(s) of function f defined by
f(x , y) = x² + y²

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.
f_x(x,y) = 2x = 0
f_y(x,y) = 2y = 0
The solution to the above system of equations is the ordered pair (0,0).
Below is the graph of f(x , y) = x² + y² and it looks 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) = x² - y²

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 following equations f_x(x,y) = 0 and f_y(x,y) = 0 simultaneously.
f_x(x,y) = 2x = 0
f_y(x,y) = - 2y = 0
The solution is the ordered pair (0,0).
The graph of f(x , y) = x² - y² is shown below. f is curving down in the y direction and curving up in the x direction. f is stationary at the point (0,0) but there is no extremum (maximum or minimum). (0,0) is called a saddle point because there is neither a relative maximum nor a relative minimum and the surface close to (0,0) looks like a saddle.

Example 3

Find the critical point(s) of function f defined by
f(x , y) = - x² - y²

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 the following equations f_x(x,y) = 0 and f_y(x,y) = 0 simultaneously.
f_x(x,y) = - 2x = 0
f_y(x,y) = - 2y = 0
The solution to the above system of equations is the ordered pair (0,0).
The graph of f(x , y) = - x² - y² is shown below and it has a relative maximum.

Example 4

Find the critical point(s) of defined by
f(x , y) = x³ + 3x² - 9x + y³ -12y

Solution to Example 4:
The first order partial derivatives are given by
f_x(x,y) = 3x² + 6x - 9
f_y(x,y) = 3y² - 12
We now solve the equations f_x(x,y) = 0 and f_y(x,y) = 0 simultaneously.
3x² + 6x - 9 = 0
3y² - 12 = 0
The solutions, which are the critical points, to the above system of equations are given by
(1,2) , (1,-2) , (-3,2) , (-3,-2)

Exercises

Find, if any, the critical points to the functions below.
1. f(x , y) = 3xy - x³ - y³
2. f(x , y) = e^{(- x² - y² + 2x - 2y - 2)}
3. f(x , y) = (1/2)x² + y³ - 3xy - 4x + 2
4. f(x , y) = x³ + y³ + 2x + 6y

Answers to the Above Exercises

1. (0,0) , (1,1)
2. (1,-1)
3. (16,4) , (1,-1)
4. no critical points.

More on Partial Derivatives and Multivariable Functions

Multivariable Functions