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 is available on this website.
In what follows, \( C \) represents the constant of integration where applicable.
Theorem: Second Derivative Test for Functions of Two Variables
Let \( f \) be a function of 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 \( f_{xx}(a, b) < 0 \), then \( f \) has a relative maximum at the point \( (a, b) \).
- c) If \( D < 0 \), 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
Click on each example to view its detailed step-by-step solution.
Example 1
Determine the critical points and locate any relative minima, maxima, and saddle points of function \( f \) defined by:
\[ f(x, y) = 2x^2 + 2xy + 2y^2 - 6x \]Show 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:
\[ 4x + 2y - 6 = 0 \] \[ 2x + 4y = 0 \implies x = -2y \]Substitute \( x = -2y \) into the first equation:
\[ 4(-2y) + 2y - 6 = 0 \implies -6y = 6 \implies y = -1 \]Then \( x = -2(-1) = 2 \). The system has one critical point at \( (2, -1) \).
Find the second-order partial derivatives:
\[ f_{xx}(x, y) = 4, \quad f_{yy}(x, y) = 4, \quad f_{xy}(x, y) = 2 \]Evaluate \( D \) at \( (2, -1) \):
\[ D = f_{xx}(2, -1) f_{yy}(2, -1) - f_{xy}^2(2, -1) = (4)(4) - 2^2 = 16 - 4 = 12 \]Since \( D = 12 > 0 \) and \( f_{xx}(2, -1) = 4 > 0 \), function \( f \) has a local minimum at \( (2, -1) \).
The 3D graph confirms a local minimum at \( (2, -1, f(2, -1)) = (2, -1, -6) \).
Example 2
Determine the critical points and locate any relative minima, maxima, and saddle points of function \( f \) defined by:
\[ f(x, y) = 2x^2 - 4xy + y^4 + 2 \]Show Solution to Example 2
Find the first partial derivatives \( f_x \) and \( f_y \):
\[ f_x(x, y) = 4x - 4y \] \[ f_y(x, y) = -4x + 4y^3 \]Solve \( f_x(x, y) = 0 \) and \( f_y(x, y) = 0 \) simultaneously:
\[ 4x - 4y = 0 \implies x = y \]Substitute \( x = y \) into \( -4x + 4y^3 = 0 \):
\[ -4y + 4y^3 = 0 \implies 4y(y^2 - 1) = 0 \implies y = 0, \; y = 1, \; y = -1 \]Using \( x = y \), the critical points are:
\[ (0, 0), \; (1, 1), \; (-1, -1) \]Find the second-order partial derivatives:
\[ f_{xx}(x, y) = 4, \quad f_{yy}(x, y) = 12y^2, \quad f_{xy}(x, y) = -4 \]Evaluate \( D \) and \( f_{xx} \) at each critical point:
| 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 |
| Conclusion | Saddle Point | Relative Minimum | Relative Minimum |
Example 3
Determine the critical points and locate any relative minima, maxima, and saddle points of function \( f \) defined by:
\[ f(x, y) = -x^4 - y^4 + 4xy \]Show Solution to Example 3
Find the first partial derivatives \( f_x \) and \( f_y \):
\[ f_x(x, y) = -4x^3 + 4y \] \[ f_y(x, y) = -4y^3 + 4x \]Solve \( f_x = 0 \) and \( f_y = 0 \):
\[ -4x^3 + 4y = 0 \implies y = x^3 \]Substitute \( y = x^3 \) into the second equation:
\[ -4(x^3)^3 + 4x = 0 \implies -4x^9 + 4x = 0 \implies -4x(x^8 - 1) = 0 \implies x(x^4 - 1)(x^4 + 1) = 0 \]Real solutions for \( x \) are \( x = 0, 1, -1 \). Using \( y = x^3 \), the critical points are:
\[ (0, 0), \; (1, 1), \; (-1, -1) \]Second-order partial derivatives:
\[ f_{xx}(x, y) = -12x^2, \quad f_{yy}(x, y) = -12y^2, \quad f_{xy}(x, y) = 4 \]| 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 |
| Conclusion | Saddle Point | Relative Maximum | Relative Maximum |
Exercises
Determine the critical points of the functions below and find whether each point corresponds to a relative minimum, maximum, or saddle point. Click each exercise to check your answers.
Exercise 1
\( f(x, y) = x^2 + 3y^2 - 2xy - 8x \)
Show Answer
Critical point at \( (6, 2) \). Since \( f_{xx} = 2 > 0 \) and \( D = 2(6) - (-2)^2 = 8 > 0 \), there is a relative minimum at \( (6, 2) \).
Exercise 2
\( f(x, y) = x^3 - 12x + y^3 + 3y^2 - 9y \)
Show Answer
Critical points are \( (2, 1), (2, -3), (-2, 1), (-2, -3) \):
- Relative minimum at \( (2, 1) \) (\( D = 144 > 0, f_{xx} = 12 > 0 \))
- Relative maximum at \( (-2, -3) \) (\( D = 144 > 0, f_{xx} = -12 < 0 \))
- Saddle points at \( (2, -3) \) and \( (-2, 1) \) (\( D = -144 < 0 \))