Technical Optimisation

One of the most practical uses of differential calculus is to optimise desired quantities, which amounts to either minimising a function $f(x)$ on some subset $I \subseteq \mathbb R$ , or maximising it. How do we do that?

Let’s first formulate the notion of a global minimum and local minimum respectively.

Definition 1. Let $I \subseteq \mathbb R$ and $f : I \to \mathbb R$ be a function.

We say that $(c, f(c))$ is a global minimum point of $f$ on $I$ if for any $x \in I$ , $f(x) \geq f(c)$ .
We say that $(c, f(c))$ is a local minimum point of $f$ if there exists $\delta > 0$ such that $(c, f(c))$ is a global minimum point on $(c - \delta , c + \delta) \cap I$ .

To abbreviate, we will say that $c$ is a global (resp. local) minimum point of $f$ .

In a symmetric manner, we can define the global maximum and local maximum respectively. In fact, we have the following symmetrical result.

Lemma 1. Suppose that $c$ is a global (resp. local) maximum for a function $f$ . Then $c$ is a global (reap. local) minimum for $-f$ .

Proof. Employ the equivalence of $f(x) \geq f(c) \iff -f(x) \leq -f(c)$ .

We say that $c$ is a global (reap. local) extremum if it is either a global (resp. local) minimum or a global (resp. local) maximum. By the extreme value theorem, we will always obtain a global extremum. What about a local extremum?

Theorem 1. Let $f$ be a function that is differentiable at $c$ . Suppose $c$ is a local extremum at $f$ . Then $f'(c) = 0$ .

Proof. The proof of this result is essentially the same as that of Rolle’s theorem.

Since global and local extrema constitute all possible extremum points, we have the critical value test.

Theorem 2. Let $f: [a, b] \to \mathbb R$ be a continuous function that is differentiable on $(a, b)$ . Then $f$ obtains global extrema either at $a, b$ or interior points $c$ such that $f'(c)$ is undefined or $f'(c) = 0$ . These possible points are known as critical points.

Does the converse hold? That is, does $f'(c) = 0$ imply that $c$ is a local extremum? Not if we consider $c = 0$ and the function $f(x) = x^3$ . How then do we determine the nature of this stationary point?

Suppose $f'(c) = 0$ . Denote $f'(c^-) \leq 0$ (resp. $f'(c^+) \geq 0$ ). to mean that there exists $\delta > 0$ such that for any $x \in (c-\delta, c)$ (resp. $x \in (c, c + \delta)$ ), $f'(x) \leq 0$ (resp. $f'(x) \geq 0$ ). We can use the first derivative test to answer our aforementioned question.

Theorem 3 (First Derivative Test). Let $f : [a, b] \to \mathbb R$ be differentiable at $c \in (a, b)$ . Suppose $f'(c) = 0$ .

If $f'(c^-) \leq 0$ and $f'(c^+) \geq 0$ , then $c$ is a local minimum of $f$ .
If $f'(c^-) \geq 0$ and $f'(c^+) \leq 0$ , then $c$ is a local maximum of $f$ .
If $f'(c^-), f'(c^+) \geq 0$ or $f'(c^-),f'(c^+) \geq 0$ , then $c$ is not a local extremum. In this case, we say that $c$ is a stationary point of inflection.

Proof. We will prove the first result.

Since $f'(c^-) \leq 0$ , find $\delta_1 > 0$ such that for $x \in (c - \delta_1, c)$ , $f'(x) \leq 0$ .
Since $f'(c^+) \geq 0$ , find $\delta_2 > 0$ such that for $x \in (c, c + \delta_2)$ , $f'(x) \geq 0$ .

Define $\delta := \min\{\delta_1,\delta_2\}$ . We claim that for any $x \in (c-\delta, c+\delta)$ , $f(x) \geq f(c)$ .

Suppose $x < c$ . By the mean value theorem, there exists

$c_1 \in (x,c) \subseteq (c-\delta, c) \subseteq (c-\delta_1,c),$

implying $f'(c_1) \leq 0$ , such that

$\displaystyle f(c) = f(x) + f'(c_1) ( c - x) \leq f(x) + 0 \cdot 0 = f(x).$

A similar argument for $x > c$ yields

$f(x) = f(c) + \underbrace{(\text{stuff})}_{\geq 0} \underbrace{(x-c)}_{> 0} \geq 0.$

Can we have a simpler computation for this process? Yes, but this does not help us with inflection points. Furthermore, we can make no conclusion if $f''(c) = 0$ .

Theorem 4 (Second Derivative Test). Let $f : [a, b] \to \mathbb R$ be twice-differentiable at $c \in (a, b)$ . Suppose $f'(c) = 0$ .

If $f''(c) > 0$ , then $c$ is a local minimum of $f$ .
If $f''(c) < 0$ , then $c$ is a local maximum of $f$ .
If $f''(c) = 0$ , the test is inconclusive.

Proof. We will prove the first result since the second result follows by symmetry.

By the definition of differentiation, for sufficiently small $h > 0$ ,

$\displaystyle \frac{f'(c+h)}{h}=\frac{f'(c+h) - f'(c)}{h} = f''(c) + o(h) > 0.$

Hence, $f'(c+h) > (f''(c) + o(h)) \cdot h > 0$ , which after taking $h \to 0^+$ , implies that $f'(c^+) \geq 0$ . Likewise, from the left side,

$\displaystyle \frac{-f'(c-h)}{h}=\frac{f'(c) - f'(c-h)}{h} = f''(c) + o(h) > 0,$

which yields $-f'(c^-) \leq 0 \iff f'(c^-) \geq 0$ .

For the last result, consider the functions $x^4$ for $0$ being a local minimum, $-x^4$ for $0$ being a local maximum, and $x^3$ for $0$ being a stationary point of inflection since in all cases, $f''(0) = 0$ .

In the next post, we will use this test to maximise profit.

—Joel Kindiak, 31 Oct 24, 1316H

KindiakMath

Technical Optimisation

Response

Leave a comment Cancel reply

Technical Optimisation

Share this:

Response

Leave a comment Cancel reply