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

$f$ defined by

f (x, y) = x^{2} + y^{2}

$f(x , y) = x^2 + y^2$

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.

Example 2

Find the critical point(s) of function

f

$f$ defined by

f (x, y) = x^{2} - y^{2}

$f(x , y) = x^2 - y^2$

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.

Example 3

Find the critical point(s) of function

f

$f$ defined by

f (x, y) = - x^{2} - y^{2}

$f(x , y) = - x^2 - y^2$

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.

Example 4

Find the critical point(s) of defined by

f (x, y) = x^{3} + 3 x^{2} - 9 x + y^{3} - 12 y

$f(x , y) = x^3 + 3x^2 - 9x + y^3 -12y$

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)

Exercises

Find, if any, the critical points to the functions below.
1.

f (x, y) = 3 x y - x^{3} - y^{3}

$f(x , y) = 3xy - x^3 - y^3$
2.

f (x, y) = e^{(- x^{2} - y^{2} + 2 x - 2 y - 2)}

$f(x , y) = e^{(- x^2 - y^2 + 2x - 2y - 2)}$
3.

f (x, y) = \frac{1}{2} x^{2} + y^{3} - 3 x y - 4 x + 2

$f(x , y) = \dfrac{1}{2}x^2 + y^3 - 3xy - 4x + 2$
4.

f (x, y) = x^{3} + y^{3} + 2 x + 6 y

$f(x , y) = x^3 + y^3 + 2x + 6y$

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