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

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

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

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

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. $$f(x , y) = 3xy - x^3 - y^3$$
2. $$f(x , y) = e^{(- x^2 - y^2 + 2x - 2y - 2)}$$
3. $$f(x , y) = \dfrac{1}{2}x^2 + y^3 - 3xy - 4x + 2$$
4. $$f(x , y) = x^3 + y^3 + 2x + 6y$$

### 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