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)f(x)f(c) \ge f(x) for all x in D.
    • Absolute Minimum value of f on D if f(c)f(x)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)f(x)f(c) \ge f(x) when x is near c.
    • Local Minimum value of f if f(c)f(x)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][a, b], then f attains an absolute maximum value f(c)f(c) and an absolute minimum value f(d)f(d) at some numbers c and d in [a,b][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)=1(x21)2f(x) = \frac{1}{(x^2-1)^2} which domain is (1,1)(-1, 1).
    • Step 1: f(x)=4x(x21)3f'(x) = -\frac{4x}{(x^2-1)^{3}}
    • Step 2: List the critical points x=0x = 0, and the endpoints ( limx1+f(x)=\displaystyle\lim_{x \to -1^{+}}f(x) = \infty, limx+1+f(x)=\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,)(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)=xx3f(x) = x-x^3 in domain [3,3][-3, 3]
  • We got:
    • endpoint: f(3)=24f(-3) = 24
    • local minimum:f(1/3)=2930.38f(-\sqrt{1/3}) = -\frac{2}{9}\sqrt{3} \approx -0.38
    • local maximum:f(1/3)=2930.38f(\sqrt{1/3}) = \frac{2}{9}\sqrt{3} \approx 0.38
    • endpoint: f(3)=24f(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)=xx22xf(x) = x - |x^2-2x|, (x0x \ge 0)
    • Rewrite the function: f(x)={x(x22x), if x22x0,x+(x22x), if x22x<0.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)={32x, if x<0 or x>2,1+2x, if 0<x<2.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)
      • x+y2xy\frac{x+y}{2} \ge \sqrt{xy}
      • So 24xy24 \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=acscθ+bsecθl = a \cdot \csc\theta + b \cdot \sec\theta
    • And we want to minimize ll.
      • Don't forget 0<θ<π/20 < \theta < \pi/2
    • To get the critical points, set l(θ)=acscθcotθ+bsecθtanθ=0l'(\theta) = -a \cdot \csc\theta \cot\theta+ b \cdot \sec\theta \tan \theta = 0
    • Then tan3θ=ab\tan^3\theta = \frac{a}{b}, tanθ=ab3\tan\theta = \sqrt[3]{\frac{a}{b}}
      • So, use trigonometry, we got cscθ=1+(a/b)2/3a/b3\csc\theta = \frac{\sqrt{1+(a/b)^{2/3}}}{\sqrt[3]{a/b}}, secθ=1+(q/b)2/3\sec\theta = \sqrt{1+(q/b)^{2/3}}
    • l(θ)=a1+(a/b)2/3a/b3+b1+(q/b)2/3=(a2/3+b2/3)3/2\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 (a2/3+b2/3)3/2(a^{2/3} + b^{2/3})^{3/2}

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