__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) = 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 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^{(- x2 - y2 + 2x - 2y - 2)}
3. f(x , y) = (1/2)x^{2} + y^{3} - 3xy - 4x + 2
4. f(x , y) = x^{3} + y^{3} + 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 |