Maxima and Minima of Functions of Two Variables

Locate relative maxima, minima and saddle points of functions of two variables. Several examples with detailed solutions are presented. 3-Dimensional 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 \[ D = f_{xx}(a,b) f_{yy}(a,b) - f_{xy}^2(a,b) \] a) If \( D > 0 \) and \( f_{xx}(a,b) > 0 \), then \( f \) has a relative minimum at the point \((a,b)\).
b) If \( D > 0 \) and \( f_{xx}(a,b) \lt 0 \), then \( f \) has a relative maximum at the point \((a,b)\).
c) If \( D \lt 0 \), then \( f \) has a saddle point at the point \((a,b)\).
d) If \( D = 0 \), then no conclusion can be drawn.

Examples with Detailed Solutions

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 3-Dimensional graph of function \( f \) given above shows that \( f \) has a local minimum at the point (2,-1,\( f(2,-1) \)) = (2,-1,-6).
critical points example 1, local minimum

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 3-Dimensional 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).
critical points example 2, 2 local minima and one saddle point

critical points example 2, 2 local minima and one saddle point

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 3-Dimensional 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).

critical points example 3, 2 local maxima and one saddle point

critical points example 3, 2 local maxima and one saddle point

Exercises

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. \( f(x , y) = x^2 + 3 y^2 - 2 xy - 8x \)
2. \( f(x , y) = x^3 - 12 x + y^3 + 3 y^2 - 9y \)

Answers to the Above Exercises

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).