2.5:Optimization          2.6:Further optimization problems          2.7: Applications of calculus to business and economics
                                     Math 16A (Autumn 2005)
                                                   Thomas Scanlon
                                            University of California, Berkeley
                                                          Week 6
   Thomas Scanlon                                                                             University of California, Berkeley
   Math 16A (Autumn 2005)
   2.5:Optimization          2.6:Further optimization problems          2.7: Applications of calculus to business and economics
          In many cases, one is interested in optimizing some quantity. In
          most cases, this optimization corresponds to maximizing or
          minimizing some function.
   Thomas Scanlon                                                                             University of California, Berkeley
   Math 16A (Autumn 2005)
   2.5:Optimization          2.6:Further optimization problems          2.7: Applications of calculus to business and economics
   A mathematical example of minimization
          Example
          Find the value(s) of x minimizing the function
          f (x) = x3 − x2 − x + 1 for x ≥ 0.
   Thomas Scanlon                                                                             University of California, Berkeley
   Math 16A (Autumn 2005)
   2.5:Optimization          2.6:Further optimization problems          2.7: Applications of calculus to business and economics
   A solution
          The minima (if they exist at all) are among the endpoints of the
          domain and the solutions to f ′(x) = 0.
          Wecompute the value of the function f(x) at the only endpoint,
          x = 0 finding that f(0) = 1.
          The first derivative is f ′(x) = 3x2 − 2x − 1 = (3x + 1)(x − 1)
          which has roots                                   −1
                                                     x = 3 and 1
          As we are only interested in nonnegative values of x, we only
          consider x = 1. Here, f (1) = 0.
          As lim f(x) = ∞, we conclude that 1 is the absolute minimum.
                x→∞
   Thomas Scanlon                                                                             University of California, Berkeley
   Math 16A (Autumn 2005)