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.

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.

\[ f(x , y) = 2x^2 + 2xy + 2y^2 - 6x \]

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).

__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).

__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).

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

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

2. relative maximum at (2,-3), relative minimum at (2,1), saddle points at (-2,-3) and (-2,1).

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