Determining Extreme Values of Functions of Several Variables
We recently looked at the definitions of local and absolute extrema on the Maximum and Minimum Values of Functions of Several Variables and we saw that if we have an extreme value at a point $P_0$ on a function of several variables, then the partial derivatives at that point are all equal to zero.
We are now about to look at ways to find extreme values (maximum and minimum values) of functions of several variables. However, before we do so though, we must first look at some more important extended definitions.
| Definition: Let $z = f(x, y)$ be a two variable real-valued function. A point $(a, b) \in D(f)$ is called a Critical Point of $f$ if both of the partial derivatives of $f$ at $(a,b)$ equal zero, that is: $\frac{\partial}{\partial x} f(a, b) = 0$ and $\frac{\partial}{\partial y} f(a,b) = 0$. |
Critical points can also be defined in a similar manner for functions of three or more variables.
| Definition: Let $z = f(x, y)$ be a two variable real-valued function. Then a point $(a, b) \in D(f)$ is called a Singular Point of $f$ if either $\frac{\partial}{\partial x} f(a, b)$ or $\frac{\partial}{\partial y} f(a, b)$ does not exist, or neither exists, in other words, the gradient of $f$ at $(a, b)$, $\nabla f(a, b)$ does not exist. |
Singular points can also be defined for functions $z = f(x_1, x_2, ..., x_n)$ of three or more variables in which one or more of the partial derivatives of $f$ does not exist and so the gradient $\nabla f(x_1, x_2, ..., x_n)$ does not exist. Furthermore, some people also classify what we call "singular points" as critical points too.
Also, recall from the Sets of Points in One, Two, and Three Dimensions the definition of a boundary point. We will now formalize this definition in terms of a two variable function.
| Definition: The point $(a, b) \in D(f)$ is called a Boundary Point of $f$ if for all real numbers $r > 0$ there exists a point $P \in D(f)$ and $Q \not \in D(f)$ such that $P, Q \in B_r ((a,b))$. |
From the definition above, we see that $(a, b) \in D(f)$ is a boundary point if for any disk $\mathcal D$ with center $(a,b)$ and radius $r > 0$, there exists points both in the domain of $f$ and not in the domain of $f$.
Now that we have gotten those definitions out of the way, we will introduce a theorem that will tell us that if we want to find a the maximum and minimum values of a function (provided that they exist), then all we need to do is check the three types of points (critical, singular, and boundary) specified above.
| Theorem 1: Let $z = f(x, y)$ be a two variable real-valued function. Then if $f$ has an absolute/local maximum/minimum value at $(a,b) \in D(f)$ only if one of the following is true: a) $(a, b) \in D(f)$ is a critical point of $f$. b) $(a, b) \in D(f)$ is a singular point of $f$. c) $(a, b) \in D(f)$ is a boundary point of $D(f)$. |
- Proof: We will show that if $(a, b)$ is not a critical point, singular point, or boundary point, then $f$ cannot be an absolute or local maximum/minimum. Suppose that $(a, b)$ is not a boundary point of $D(f)$. Then $(a, b)$ is contained within the interior of $f$. Furthermore, if $(a, b)$ is not a singular point of $f$ then $\nabla f(a, b)$ exists. If $\nabla f(a, b) \neq (0, 0)$ then either $\nabla f(a, b) > 0$ and $- \nabla f(a, b) < 0$ or $\nabla f(a, b) < 0$ and $-\nabla f(a, b) > 0$. In either case, $f$ is increasing in one direction and decreasing in another direction and so $f$ cannot have an absolute/local maximum/minimum at $(a, b)$. $\blacksquare$
Thus if $f$ attains maximum and minimum values, and if we can determine all of the critical, singular, and boundary points of a function and compare their values, then we can determine the extreme values of $f$. We will subsequently look at some approaches for obtaining these extrema.
Example 1
Show that the function $f(x, y) = x^2 - y^2$ does not have an extreme values.
We first note that $f$ has no singular or boundary points, and so $f$ can obtain extreme values only at critical points. Let's first determine the gradient of $f$. We have that:
(1)Now $\nabla f(x, y) = 0$ at the point $(0, 0) \in D(f)$ . So $(0, 0)$ is our only critical point. Now note that $f(x, 0) > 0$ for $(x, 0) \in D(f)$ such that $x \neq 0$. Similarly, note that $f(0, y) < 0$ for $(0, y) \in D(f)$ such that $y \neq 0$. Thus $f$ does not have an extreme value at $(0, 0) \in D(f)$. The following image is a portion of the graph of $f$. As you can see, the origin is not an extreme value of $f$.
