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.

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.
fx(x,y) = 2x
fy(x,y) = 2y
We now solve the following equations fx(x,y) = 0 and fy(x,y) = 0 simultaneously.
fx(x,y) = 2x = 0
fy(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.

critical point example 1, minimum point

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.
fx(x,y) = 2x
fy(x,y) = -2y
Solve the following equations fx(x,y) = 0 and fy(x,y) = 0 simultaneously.
fx(x,y) = 2x = 0
fy(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.

critical point example 2, saddle point

Example 3: Find the critical point(s) of defined by

f(x , y) = - x2 - y2

Solution to Example 3:
We first find the first order partial derivatives.
fx(x,y) = - 2x
fy(x,y) = - 2y
We now solve the following equations fx(x,y) = 0 and fy(x,y) = 0 simultaneously.
fx(x,y) = - 2x = 0
fy(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.

critical point example 3, maximum point

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
fx(x,y) = 3x2 + 6x - 9
fy(x,y) = 3y2 - 12
We now solve the equations fx(x,y) = 0 and fy(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 - x
3 - y3
2. f(x , y) = e
(- x2 - y2 + 2x - 2y - 2)
3. f(x , y) = (1/2)x
2 + y3 - 3xy - 4x + 2
4. f(x , y) = x
3 + y3 + 2x + 6y
Answer to 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