CK-12-Calculus

(Marvins-Underground-K-12) #1

http://www.ck12.org Chapter 3. Applications of Derivatives


3.7 Optimization


Learning Objectives


A student will be able to:



  • Use the First and Second Derivative Tests to find absolute maximum and minimum values of a function.

  • Use the First and Second Derivative Tests to solve optimization applications.


Introduction


In this lesson we wish to extend our discussion of extrema and look at the absolute maximum and minimum values
of functions. We will then solve some applications using these methods to maximize and minimize functions.
Absolute Maximum and Minimum
We begin with an observation about finding absolute maximum and minimum values of functions that are continuous
on a closed interval. Suppose thatfis continuous on a closed interval[a,b].Recall that we can find relative minima
and maxima by identifying the critical numbers off in(a,b)and then applying the Second Derivative Test. The
absolute maximum and minimum must come from either the relative extrema off in(a,b)or the value of the
function at the endpoints, f(a)orf(b).Hence the absolute maximum or minimum values of a function fthat is
continuous on a closed interval[a,b]can be found as follows:



  1. Find the values offfor each critical value in(a,b);

  2. Find the values of the functionfat the endpoints of[a,b];

  3. The absolute maximum will be the largest value of the numbers found in 1 and 2; the absolute minimum will
    be the smallest number.


The optimization problems we will solve will involve a process of maximizing and minimizing functions. Since
most problems will involve real applications that one finds in everyday life, we need to discuss how the properties
of everyday applications will affect the more theoretical methods we have developed in our analysis. Let’s start with
the following example.
Example 1:
A company makes high-quality bicycle tires for both recreational and racing riders. The number of tires that the
company sells is a function of the price charged and can be modeled by the formulaT(x) =−x^3 + 36. 5 x^2 + 50 x+ 250 ,
wherexis the priced charged for each tire in dollars. At what price is the maximum number of tires sold? How
many tires will be sold at that maximum price?
Solution:
Let’s first look at a graph and make some observations. Set the viewing window ranges on your graphing calculator
to[− 10 , 50 ]forxand[− 500 , 10000 ]fory.The graph should appear as follows:

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