Extreme Values of Function of Two Variables

Definition 4..1   For all points $(x,y)$ in the $\delta$ neighborhood of $(x_0, y_0)$^4.1
1. If $f(x_{0},y_{0}) \leq f(x,y)$, then $f(x,y)$ takes minimum at $(x_{0},y_{0})$ and $f(x_{0},y_{0})$ is called local minimum of $f(x,y)$.
2. If $f(x_{0},y_{0}) \geq f(x,y)$, then $f(x,y)$ takes maximum at $(x_{0},y_{0})$ and $f(x_{0},y_{0})$ is called local maximum of $f(x,y)$.

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

Figure 4.26: Understanding

If the graph of function is smooth, then $f_{x}(x,y) = 0, f_{y}(x,y) = 0$.

Figure 4.27: Understanding

If the graph has a sharp edge, then the function is not differentiable at the sharp edge.

Theorem 4..5   If $f(x,y)$ 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

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

$2x_0 + y_0 + 1 = 0$ implies $y_0 = -2x_0 -1$. substitute this inot $x_0 + 6y_0 + 1 = 0$ to get $x_0 + 6(-2x_0 -1) + 1 = -11x_0 -5 = 0$. Thus, $x_0 = -\frac{5}{11}$.

SOLUTION If $f(x,y)$ takes the extreme value at $(x_{0},y_{0})$, then

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

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

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

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

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.

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

Figure 4.28: Exercise4-17

$f_{x}(x,y) = 0, f_{y}(x,y) = 0$ is not sufficient condition for the existence of a limit. $f(x,y) = x^2 - y^2$. Find $\Delta$ at $(0,0)$. Then $f_{xx}(0,0) = 2$, $f_{xy}(0,0) = 0$, $f_{yy}(0,0) = -2$. Thus, $\Delta = 2(-2) - 0^2 = -4 < 0$ and $f(0,0)$ is not extreme value.

SOLUTION If $f(x,y)$ takes the extreme value at $(x_{0},y_{0})$, then

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

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

Thus a point $(0,0)$ is a critical point. Now as $(x,y)$ approaches $(0,0)$ along $x$-axis, we have $f(x,y) > f(0,0)$, along $y$-axis we have $f(x,y) < f(0,0)$. Thus $f$ does not take the extreme value at $(0,0)$ $\blacksquare$

Theorem 4..6 (Second Derivative Test)   Let $f(x,y)$ be the class $C^{2}$ at $(x_0, y_0)$ in the region $D$. 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.

Check

For $Q = \frac{A}{2}\left[(h + \frac{Bk}{A})^2 + \frac{(AC - B^2)k^2}{A^2}\right]$, $(h + \frac{Bk}{A})^2$ and $\frac{k^2}{A^2}$ are squares and thus non negative. Therefore, the sign of $Q$ is determined by the sign of $A$ and $AC-B^2$.