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.

__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) = x^{2} + y^{2} and it looks that at the critical point (0,0) f has a minimum value.

__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) = x^{2} - y^{2} 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.

__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) = - x^{2} - y^{2} is shown below and it has a relative maximum.

__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

3y^{2} - 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)

1. f(x , y) = 3xy - x

2. f(x , y) = e

3. f(x , y) = (1/2)x

4. f(x , y) = x

2. (1,-1)

3. (16,4) , (1,-1)

4. no critical points.