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.

\[ f(x , y) = x^2 + y^2 \]

__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.

__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.

\[ f(x , y) = - x^2 - y^2 \]

__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.

__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. \( f(x , y) = 3xy - x^3 - y^3 \)

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

2. (1,-1)

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

4. no critical points.