Optimization

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minimum and maximum

When using functions to analyze situations, we are often interested in the maximum and minimum values that a function takes on. These are often called extreme values. We would like to know the extreme values and where, in the domain, they occur.

There are two views on extreme values.

Global Extreme Values

The global maximum value is the greatest value of the function.
$f(c) \text { is a global maximum if } f(x) \leq f(c) \text { for all } x \text { in the domain of } f$
The global minimum value is the least value of the function.
$f(c) \text { is a global minimum if } f(c) \leq f(x) \text { for all } x \text { in the domain of } f$

A function has at most, one maximum value and one minimum value, but they can occur at multiple domain numbers.

Global extrema are also called absolute extrema.

A global maximum may occur at multiple domain numbers.
A function can also not have a maximum value.

A global minimum may occur at multiple domain numbers.
A function can also not have a minimum value.

In contrast to a single maximum or minimum value, a function may also have values which are the greatest value in their own little neighborhood of the domain, but not the greatest overall value. We see these visually encoded as tops of hills and bottom of valleys on the graph. A function may have many of these local or relative extrema.

Local Extreme Values

$f(c)$ is a local maximum value of the function $f$ , if there exists an $\epsilon > 0$ such that $f(x) \leq f(c)$ for all domain numbers, $x$ , within a distance of $\epsilon$ from $c$ .
$f(c) \text { is a local maximum if } f(x) \leq f(c) \text { for all } \{ x \in domain \, | \, x \in (c - \epsilon , c + \epsilon ) \} \text { for some } \epsilon > 0$
$f(c)$ is a local minimum value of the function $f$ , if there exists an $\epsilon > 0$ such that $f(c) \leq f(x)$ for all domain numbers, $x$ , within a distance of $\epsilon$ from $c$ .
$f(c) \text { is a local minimum if } f(c) \leq f(x) \text { for all } \{ x \in domain \, | \, x \in (c - \epsilon , c + \epsilon ) \} \text { for some } \epsilon > 0$

Local extrema are also called relative extrema.

By default, all global extrema are automatically local extrema.

Extrema

Let $g(k)$ be a function. The graph of $y= g(k)$ is displayed below.

$g$ has no global maximum.
The global minimum of $g$ is $-3.3$ , which occurs at $-8$ .
$g$ has a local minimum of $-3.3$ , which occurs at $-8$ and a local minimum of $-2.3$ , which occurs at $0.9$ .
$g$ has a local maximum of $2.4$ , which occurs at $-4.9$

Let’s run through the idea of a local extrema for the previous example.

Local Explanations

$\blacktriangleright$ $g$ has a local minimum of $-2.3$ , which occurs at $0.9$ .

Let $\epsilon = 0.3$ . Then, $\epsilon > 0$ .
$-2.3 = g(0.9) \leq g(k)$ for all $k$ in the domain within a distance of $0.3$ from $0.9$ , which would be the interval $\left (\answer {0.6}, \answer {1.2}\right )$ .

$\blacktriangleright$ $g$ has a local maximum of $2.4$ , which occurs at $-4.9$ .

Let $\epsilon = 0.4$ . Then, $\epsilon > 0$ .
$2.4 = g(k) \leq g(-4.9)$ for all $k$ in the domain within a distance of $0.4$ from $-4.9$ , which would be the interval $\left (\answer {-5.3}, \answer {-4.5}\right )$ .

$\blacktriangleright$ $g$ has a local minimum of $-3.3$ , which occurs at $-8$ .

Let $\epsilon = 0.5$ . Then, $\epsilon > 0$ .

$-3.3 = g(-8) \leq g(k)$ for all $k$ in the domain within a distance of $0.5$ from $-8$ , which would be the interval $\left [\answer {-8.5}, \answer {-7.5}\right )$ .

Note: As the previous example illustrates, global extrema are also local extrema.
Note: We picked $0.3$ , $0.4$ , and $0.5$ for $\epsilon$ , but we could have selected any small, positive numbers that work.

Extrema

Let $B(w)$ be a function. The graph of $y = B(w)$ is displayed below.

$B$ has no global maximum.
The global minimum of $B$ is $-2.3$ , which occurs at $1.4$ .
$B$ has a local minimum of $-2.3$ , which occurs at $1.4$ and a local minimum of $-0.7$ , which occurs at $-5.9$ .
$B$ has a local maximum of $0.25$ , which occurs at $-3$

Local Explanations for $B(w)$

Let $0 < \epsilon = 0.6$ , then $-2.3 = B(1.4) \leq B(w)$ for all $w$ in the domain within a distance of $\answer {0.6}$ from $\answer {1.4}$ , which would be the interval $(0.8, 2)$ .

Let $0 < \epsilon = 0.25$ , then $-0.7 = B(-5.9) \leq B(w)$ for all $w$ in the domain within a distance of $\answer {0.25}$ from $\answer {-5.9}$ , which would be the interval $(-6.15, -5.65)$ .

Let $0 < \epsilon = 0.5$ , then $0.25 = B(w) \leq B(-3)$ for all $w$ in the domain within a distance of $\answer {0.5}$ from $\answer {-3}$ , which would be the interval $(-3.5, -2.5)$ .

Extrema

Let $T(p)$ be a function. The graph of $y= T(p)$ is displayed below.

$T$ has no global maximum.
$T$ has no global minimum.
$T$ has a local minimum of $-2$ , which occurs at $-8$ .
$T$ has a local minimum of $\answer {-6}$ , which occurs at $3$ .
$T$ has a local maximum of $\answer {6}$ , which occurs at $-1$ .

It appears that $8$ would have been the global maximum of $T$ occurring at $-8$ , however, the point $(-8, 8)$ is missing from the graph and $8$ is not in the range. And, there is no real number “just below $8$ ”. If you choose any real number, $r$ , below $8$ to be the possible candidate for global maximum, there is the number $\frac {8+r}{2}$ . This number is in the range and $r < \frac {8+r}{2} < 8$ . You cannot identify a specific number as the maximum, so there isn’t one.

Local Explanations

Let $0 < \epsilon = \answer {0.5}$ . $T(-8) \leq T(p)$ for all $p$ in the domain within a distance of $0.5$ from $-8$ , which would be the interval $[-8, -7.5)$ .

Let $0 < \epsilon = \answer {0.4}$ . $T(3) \leq T(p)$ for all $p$ in the domain within a distance of $0.4$ from $3$ , which would be the interval $(3.4, 2.6)$ .

Let $0 < \epsilon = \answer {0.3}$ . $T(p) \leq T(-1)$ for all $p$ in the domain within a distance of $0.3$ from $-1$ , which would be the interval $(-1.3, -0.7)$ .

Next

Inside the natural numbers there is the idea of “next”. Each natural number has a next number.

However, the real numbers does not have a “next” property. There is no “next” number for any real number.

Let $r$ be a real number.

Supposed $N$ is the “next” real number. Then we have a problem, because the real number $\frac {r+N}{2}$ is between $r$ and $N$ . Therefore, $N$ cannot be the “next” real number. There is no “next” real number.

A consequance of this is that each open interval contains an infinite number of numbers.

Note: This is not true for closed intervals. For example, the closed interval $[3, 3]$ contains only one number. This is one reason we talk about open intervals.

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