Critical Points of Functions of Two Variables

Definition

A critical point of a function with two variables is a point where the partial derivatives of first order are equal to zero. 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 - 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

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 mutlivariable functions. Multivariable Functions



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Updated: 2 April 2013

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