Week 9 - Optimization #

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.

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 $[a, b]$.

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.
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.

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.

Five Steps:
- Draw a picture of the situation;
- Label everything with variables;
- Write down the thing you’re 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.

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 $144 \ge xy$, which is maximum value of xy.