# 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.

 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. 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. 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. 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 multivariable functions. Multivariable Functions