Extreme Values of Function of Two Variables

Definition 4..1 For all points $(x,y)$

$(x,y)$

in the $\delta$ neighborhood of $(x_0, y_0)$

$(x_0, y_0)$

^4.1
1. If $f(x_{0},y_{0}) \leq f(x,y)$ , then $f(x,y)$

$f(x,y)$

takes minimum at $(x_{0},y_{0})$ and $f(x_{0},y_{0})$ is called local minimum of $f(x,y)$

$f(x,y)$

.
2. If $f(x_{0},y_{0}) \geq f(x,y)$ , then $f(x,y)$

$f(x,y)$

takes maximum at $(x_{0},y_{0})$ and $f(x_{0},y_{0})$ is called local maximum of $f(x,y)$

$f(x,y)$

.

NOTE A locam minimum and a local maximum togrther are called extrema.

Theorem 4..5 If $f(x,y)$

$f(x,y)$

has a extremum at $(x_{0},y_{0})$ , then . $(f_{x}(x_{0},y_{0}), f_{y}(x_0, y_0))$ exists and $(f_{x}(x_{0},y_{0}), f_{y}(x_0, y_0)) = (0,0)$ , or $(f_{x}(x_{0},y_{0}), f_{y}(x_0, y_0))$ does not exist.

Proof Since $f(x,y_{0})$ takes a extreme value at $x = x_{0}$ , $f_{x}(x_{0},y_{0}) = 0$ or $f_{x}(x_{0},y_{0})$ does not exist. Similarly for $f(x_0,y)$ . $f_{y}(x_{0},y_{0}) = 0$ or $f_{y}(x_{0},y_{0})$ does not exist $\blacksquare$

Example 4..16 Find the extrema of the following function

$\displaystyle f(x,y) = x^2 + xy + 3y^2 + x +y$

SOLUTION If $f(x,y)$

$f(x,y)$

takes the extreme value at $(x_{0},y_{0})$ , then

$\displaystyle f_{x}(x_{0},y_{0}) = 0$ implies $\displaystyle 2x_{0} + y_{0} + 1 = 0$

$\displaystyle f_{y}(x_{0},y_{0}) = 0$ implies $\displaystyle x_{0} + 6y_{0} + 1 = 0$

Now solve for $x_{0},y_{0}$ . Then

$\displaystyle x_{0} = -\frac{5}{11}, y_{0} = -\frac{1}{11}.$

Thus

$f(x,y)$

may takes the extreme value at $(-\frac{5}{11}, -\frac{1}{11})$ . Now we have to check to see if this is a local maximum or minimum.owari

Exercise 4..16 Find the extrema of the following function.

$\displaystyle f(x,y) = x^2 - y^2$

SOLUTION If $f(x,y)$

$f(x,y)$

takes the extreme value at $(x_{0},y_{0})$ , then

$\displaystyle f_{x}(x_{0},y_{0}) = 0 \Rightarrow 2x_{0} = 0$

$\displaystyle f_{y}(x_{0},y_{0}) = 0 \Rightarrow 2y_{0} = 0$

Thus a point $(0,0)$

$(0,0)$

is a critical point. Now as $(x,y)$

$(x,y)$

approaches $(0,0)$

$(0,0)$

along

$x$

-axis, we have $f(x,y) > f(0,0)$

$f(x,y) > f(0,0)$

, along

$y$

-axis we have $f(x,y) < f(0,0)$

$f(x,y) < f(0,0)$

. Thus

$f$

does not take the extreme value at $(0,0)$

$(0,0)$

$\blacksquare$

Theorem 4..6 (Second Derivative Test) Let $f(x,y)$

$f(x,y)$

be the class $C^{2}$ at $(x_0, y_0)$

$(x_0, y_0)$

in the region $D$

$D$

. If $f_{x}(x_{0},y_{0}) = f_{y}(x_0,y_0) = (0,0)$ , then denote $f_{xx}(x_0,y_0) = A, f_{xy}(x_0,y_0) = B, f_{yy}(x_0,y_0) = C, \Delta = AC - B^2$ .

1. If $\Delta > 0, A > 0$ , then $f(x_0,y_0)$ is a local minimum.

2. If $\Delta > 0, A < 0$ , then $f(x_0,y_0)$ is a local maximum.

3. If $\Delta < 0$ , then $f(x_0,y_0)$ is a saddle point

4. If $\Delta = 0$ , then test is no conclusive.

Subsections

Taylor Theorem for Two Variables