Locate relative maxima, minima and saddle points of functions of two variables. Several examples with detailed solutions are presented. 3Dimensional graphs of functions are shown to confirm the existence of these points. More on Optimization Problems with Functions of Two Variables in this web site.
Theorem
Let f be a function with two variables with continuous second order partial derivatives f_{xx}, f_{yy} and f_{xy} at a critical point (a,b). Let
D = f_{xx}(a,b) f_{yy}(a,b)  f_{xy}^{2}(a,b)
 If D > 0 and f_{xx}(a,b) > 0, then f has a relative minimum at (a,b).
 If D > 0 and f_{xx}(a,b) < 0, then f has a relative maximum at (a,b).
 If D < 0, then f has a saddle point at (a,b).
 If D = 0, then no conclusion can be drawn.
We now present several examples with detailed solutions on how to locate relative minima, maxima and saddle points of functions of two variables. When too many critical points are found, the use of a table is very convenient.
Example 1: Determine the critical points and locate any relative minima, maxima and saddle points of function f defined by
f(x , y) = 2x^{2} + 2xy + 2y^{2}  6x.
Solution to Example 1:
Find the first partial derivatives f_{x} and f_{y}.
f_{x}(x,y) = 4x + 2y  6
f_{y}(x,y) = 2x + 4y
The critical points satisfy the equations f_{x}(x,y) = 0 and f_{y}(x,y) = 0 simultaneously. Hence.
4x + 2y  6 = 0
2x + 4y = 0
The above system of equations has one solution at the point (2,1).
We now need to find the second order partial derivatives f_{xx}(x,y), f_{yy}(x,y) and f_{xy}(x,y).
f_{xx}(x,y) = 4
f_{yy}(x,y) = 4
f_{xy}(x,y) = 2
We now need to find D defined above.
D = f_{xx}(2,1) f_{yy}(2,1)  f_{xy}^{2}(2,1) = ( 4 )( 4 )  2^{2} = 12
Since D is positive and f_{xx}(2,1) is also positive, according to the above theorem function f has a local minimum at (2,1).
The 3Dimensional graph of function f given above shows that f has a local minimum at the point (2,1,f(2,1)) = (2,1,6).
Example 2: Determine the critical points and locate any relative minima, maxima and saddle points of function f defined by
f(x , y) = 2x^{2}  4xy + y^{4} + 2.
Solution to Example 2:
Find the first partial derivatives f_{x} and f_{y}.
f_{x}(x,y) = 4x  4y
f_{y}(x,y) =  4x + 4y^{3}
Determine the critical points by solving the equations f_{x}(x,y) = 0 and f_{y}(x,y) = 0 simultaneously. Hence.
4x  4y = 0
 4x + 4y^{3} = 0
The first equation gives x = y. Substitute x by y in the equation  4x + 4y^{3} = 0 to obtain.
 4y + 4y^{3} = 0
Factor and solve for y.
4y(1 + y^{2}) = 0
y = 0 , y = 1 and y = 1
We now use the equation x = y to find the critical points.
(0 , 0) , (1 , 1) and (1 , 1)
We now determine the second order partial derivatives.
f_{xx}(x,y) = 4
f_{yy}(x,y) = 12y^{2}
f_{xy}(x,y) = 4
We now use a table to study the signs of D and f_{xx}(a,b) and use the above theorem to decide on whether a given critical point is a saddle point, relative maximum or minimum.
critical point (a,b)  (0,0)  (1,1)  (1,1) 
f_{xx}(a,b)  4  4  4 
f_{yy}(a,b)  0  12  12 
f_{xy}(a,b)  4  4  4 
D  16  32  32 
 saddle point  relative minimum  relative minimum 
A 3Dimensional graph of function f shows that f has two local minima at (1,1,1) and (1,1,1) and one saddle point at (0,0,2).
Example 3: Determine the critical points and locate any relative minima, maxima and saddle points of function f defined by
f(x , y) =  x^{4}  y^{4} + 4xy .
Solution to Example 3:
First partial derivatives f_{x} and f_{y} are given by.
f_{x}(x,y) =  4x^{3} + 4y
f_{y}(x,y) =  4y^{3} + 4x
We now solve the equations
f_{y}(x,y) = 0 and
f_{x}(x,y) = 0 to find the critical points..
 4x^{3} + 4y = 0
 4y^{3} + 4x = 0
The first equation gives y = x^{3}. Combined with the second equation, we obtain.
 4(x^{3})^{3} + 4x = 0
Which may be written as .
x(x^{4}  1)(x^{4} + 1) = 0
Which has the solutions.
x = 0 , 1 and 1.
We now use the equation y = x^{3} to find the critical points.
(0 , 0) , (1 , 1) and (1 , 1)
We now determine the second order partial derivatives.
f_{xx}(x,y) = 12x^{2}
f_{yy}(x,y) = 12y^{2}
f_{xy}(x,y) = 4
The table below shows the signs of D and f_{xx}(a,b). Then the above theorem is used to decide on what type of critical points it is.
critical point (a,b)  (0,0)  (1,1)  (1,1) 
f_{xx}(a,b)  0  12  12 
f_{yy}(a,b)  0  12  12 
f_{xy}(a,b)  4  4  4 
D  16  128  128 
 saddle point  relative maximum  relative maximum 
A 3Dimensional graph of function f shows that f has two local maxima at (1,1,2) and (1,1,2) and a saddle point at (0,0,0).
Exercise: Determine the critical points of the functions below and find out whether each point corresponds to a relative minimum, maximum, saddle point or no conclusion can be made.
1. f(x , y) = x^{2} + 3y^{2} 2xy  8x
2. f(x , y) = x^{3}  12x + y^{3} + 3y^{2}  9y
Answer to Above Exercise:
1. relative maximum at (1,1) and (1,1) and a saddle point at (0,0)
2. relative maximum at (2,3), relative minimum at (2,1), saddle points at (2,3) and (2,1).
More on partial derivatives and mutlivariable functions.
Multivariable Functions
