# Week 9 - Optimization

## 1. Maximum And Minimum Values

• Let c be a number in the domain D of a function f. Then f(c) is the
• Absolute Maximum value of f on D if $f(c) \ge f(x)$ for all x in D.
• Absolute Minimum value of f on D if $f(c) \le f(x)$ for all x in D.
• An absolute maximum or minimum is sometimes called a global maximum or minimum. The maximum and minimum values of f are called extreme values of f.
• The number f(c) is a
• Local Maximum value of f if $f(c) \ge f(x)$ when x is near c.
• Local Minimum value of f if $f(c) \le f(x)$ when x is near c.

### 1.1. The Extreme Value Theorem

• Definition: If f is continuous on a closed interval $[a, b]$, then f attains an absolute maximum value $f(c)$ and an absolute minimum value $f(d)$ at some numbers c and d in $[a, b]$.

### 1.2. Find The Maximum And Minimum Values

• Four Steps:
• Differentiate the function.
• List the critical points and endpoints.
• A critical number of a function f is a number c in the domain of f such that either f'(c) = 0 or f(c) does not exist.
• Check all of the points.
• Check the limiting behavior, if you're working on an open interval.
• For Example: find the maximum and minimum values of function $f(x) = \frac{1}{(x^2-1)^2}$ which domain is $(-1, 1)$.
• Step 1: $f'(x) = -\frac{4x}{(x^2-1)^{3}}$
• Step 2: List the critical points $x = 0$, and the endpoints ( $\displaystyle\lim_{x \to -1^{+}}f(x) = \infty$, $\displaystyle\lim_{x \to +1^{+}}f(x) = \infty$)
• Step 3: We got the minimum point (x = 0) and no maximum point.
• Step 4: the domain is an open interval, and when x = -1/1, f(x) goes infinity. So it's fine.
• Let try another domain: $(1, \infty)$
• We know that it doesn't have critical points, so there is no maximum and minimum values in this domain.

#### Don't forgot to check the endpoints

• Example: $f(x) = x-x^3$ in domain $[-3, 3]$
• We got:
• endpoint: $f(-3) = 24$
• local minimum:$f(-\sqrt{1/3}) = -\frac{2}{9}\sqrt{3} \approx -0.38$
• local maximum:$f(\sqrt{1/3}) = \frac{2}{9}\sqrt{3} \approx 0.38$
• endpoint: $f(3) = -24$
• So this time the maximum and minimum values are the endpoints.

#### Consider the points where the function is not differentiable

• Example: $f(x) = x - |x^2-2x|$, ($x \ge 0$)
• Rewrite the function: $f(x) = \left\{ \begin{array}{rl} x-(x^2-2x) \text{, if } x^2-2x \ge 0,\\ x+(x^2-2x) \text{, if } x^2-2x < 0. \end{array} \right.$
• Differentiate it: $f'(x) = \left\{ \begin{array}{rl} 3-2x &\text{, if } x < 0 \text{ or } x > 2,\\ -1+2x &\text{, if } 0 < x < 2. \end{array} \right.$
• Now List the critical points:
• x = 2 is the point that the function can't be differentiable.
• x = 1/2 is the local extreme value
• x = 0 and x = 3 are the endpoints.
• Then we got our conclusion.

#### Handle a real problem

• Five Steps:

• Draw a picture of the situation;
• Label everything with variables;
• Write down the thing I'm tying to optimize;
• Solve my goal for a single variable;
• Apply calculus.
• Example: build the best fence for your sheep

• We've got 52 meters of fencing, and to build a biggest rectangular pen with three sides.
• Since it's a rectangle we got 2x + y = 52 and we are trying to get maximum value of xy
• xy = x(52-2x) so we got a function f(x) = x(52-2x)
• Now to use calculus, then we got x = 13.

## 2. Optimize Functions

• Sometime we don't need calculus.
• Example: two numbers(x, y) sum to 24. How large can their product be?
• x + y = 24, calculate max(xy).
• We can use calculus, but let take another way, use AM-GM theorem(inequality of arithmetic and geometric means)
• $\frac{x+y}{2} \ge \sqrt{xy}$
• So $24 \ge xy$, which is maximum value of xy.

## 3. Optimization in Action

• Example: How large of an object can you carry around a corner?
• I want to move this red stick around this corner without it getting stuck.
• The real problem is to figure out the longest length, the maximum length of a stick that can be navigated around that corner. which end up solving a minimization problem.
• Base on the diagram, we got: $l = a \cdot \csc\theta + b \cdot \sec\theta$
• And we want to minimize $l$.
• Don't forget $0 < \theta < \pi/2$
• To get the critical points, set $l'(\theta) = -a \cdot \csc\theta \cot\theta+ b \cdot \sec\theta \tan \theta = 0$
• Then $\tan^3\theta = \frac{a}{b}$, $\tan\theta = \sqrt[3]{\frac{a}{b}}$
• So, use trigonometry, we got $\csc\theta = \frac{\sqrt{1+(a/b)^{2/3}}}{\sqrt[3]{a/b}}$, $\sec\theta = \sqrt{1+(q/b)^{2/3}}$
• \begin{aligned}l(\theta) &= \frac{a \cdot \sqrt{1+(a/b)^{2/3}}}{\sqrt[3]{a/b}} + b \cdot \sqrt{1+(q/b)^{2/3}} \\ &= (a^{2/3} + b^{2/3})^{3/2}\end{aligned}
• So the longest length is $(a^{2/3} + b^{2/3})^{3/2}$