10 CHAPTER Optimization 10.1 ANALYSIS OF EXTREMA Introduction Via Examples Many problems we deal with every day involve a search for some “optimal” arrangement. What route must be taken to travel the distance between two cities in the shortest amount of time? When should a farmer harvest his crop in order to maximize his profit? What dimensions will minimize the amount of material required to construct a can of a given volume? Example 6.1 was such an optimization problem; a gardener had a fixed amount of fencing and wanted to find out how to maximize the area of her garden. These and many other questions require the optimization of some quantity. If the quantity we aim to optimize can be expressed as a function of one variable on a particular domain, then knowing about the graph of the function can help us locate the point at which the function achieves its maximum (or minimum, depending on the problem) value. Knowing the rate of change of the function can be very useful in finding this point, regardless of whether or not we actually produce a graph. In this chapter, we will look at some basic types of optimization problems to get an idea of how to set them up and use both calculus-based and non-calculus-based methods to solve them. Throughout the course, as we study new kinds of functions and their derivatives, we will return to the topic of optimization. ◆ EXAMPLE 10.1 The Beta Shuttle flies passengers between Boston and New York City. Currently, it charges $150 for each one-way ticket. At this price, an average of 190 people buy tickets on the Shuttle, so the company receives ($150 per person)·(190 people) =$28,500 for each flight. The owners of the Beta Shuttle hire a consulting firm to help them figure out how the number of people buying tickets would change if the price were changed. The consultants conclude that for each one dollar increase in the price of the ticket, two fewer people would 341 342 CHAPTER 10 Optimization be willing to fly the Beta Shuttle; similarly, for each one dollar decrease in price, two more people would buy tickets. What should the price be to maximize the amount of revenue the Beta Shuttle owners receive? (Assume that the expenses of the flight are independent of the number of people on board because the large expenses like the cost of the plane, the salaries of the pilots, the rental space at the airports, and so forth, will be fixed, regardless of the number of passengers.) For now, let’s assume that the plane’s capacity is not an issue. SOLUTION Revenue, R, can be calculated by multiplying the price of the ticket by the number of people who buy the ticket, R = (price per passenger) · (number of passengers). To express R as a function of one variable we first need formulas for the price and for the number of passengers. Since we know how many tickets will be sold in terms of the change in price from the current price, we let x be the change in the price of a ticket from its current level of $150. 1 If x is positive the price increases; if x is negative it decreases. Let’s express both price and number of passengers as functions of x. The price of a ticket now is $150, so changing the price by x dollars makes the new price 150 + x. price = 150 + x At a price of $150, 190 people are willing to buy tickets; for each $1 increase in price, 2 fewer people will buy tickets. 2 number of passengers = 190 − 2x Taking the product of these two expressions, we obtain R(x) = (150 + x)(190 − 2x). We need to find the value of x that will make R(x) as big as possible. There are several methods available for us to do this. Regardless of the method, we need to figure out the appropriate domain of the function. The relevant questions are: “What is the least possible value of x?” and “What is the greatest possible value of x?” What is the least possible value for x? Because the price of a ticket must be a positive number, the price of the ticket is (150 + x) dollars, and x must be greater than or equal to −150. What is the largest possible value for x?Atfirst it may seem like the sky is the limit, but because the number of passengers is (190 − 2x) and the number of passengers can’tbe negative, the largest possible value for x is the one that makes 190 − 2x = 0; solving gives x = 95. Therefore, the domain is [−150, 95]. One approach to the problem is to get a good graph of R(x). R(x) is a quadratic and the coefficient of the x 2 term is negative, so its graph is a parabola opening downward. (Multiplying out, we get R(x) =−2x 2 −110x + 28,500.) To sketch the graph of the parabola, we’ll first find its x-intercepts. These are the solutions to 1 You can try letting x be the price itself; this makes things a bit more complicated. 2 Our model is based on a completely linear demand curve. In reality this model will not be particularly good near the edges of the demand curve. For example, even if the price of a ticket gets so high that our model predicts that no one will buy a ticket, there is probably some rich fellow out there who will pick up one anyway. Our final answer is not going to be near this point, so this discrepancy with reality doesn’t faze us. 10.1 Analysis of Extrema 343 R(x) = (150 + x)(190 − 2x) = 0 150 + x = 0, or 190 − 2x = 0 x =−150, or x = 95 The zeros of R(x) are at x = 95 and x =−150. (Notice that we already found these zeros when we were finding the domain of R. They are the values of x for which the revenue will be zero. Either the price of a ticket is $0 or there are no passengers.) –150 –27.5 95 x R(x) Figure 10.1 Knowing that a parabola is symmetric about the vertical line through its vertex, we see that the maximum revenue occurs midway between the two zeros. Thus the value of x that makes R(x) largest is the average of −150 and 95; −150 + 95 2 =−27.5. If the Beta Shuttle reduces its fares by $27.50 to a new fare of $122.50, it will achieve its maximum amount of revenue of ($122.50 per person) · (190 − 2(−27.5) people) = ($122.50 per person) · (245 people) = $30,012.50. This represents an increase of $1,512.50 per flight over their current revenue of $28,500 per flight. A second approach to this optimization problem is to use R  to find the maximum. Either we can use R  to identify the x-coordinate of the vertex of the downward-facing parabola or, by looking at the derivative of R(x), we can see where R(x) is increasing and decreasing. (The latter will be useful if we fail to notice that we’re dealing with a parabola.) R(x) =−2x 2 −110x + 28500, therefore R  (x) =−4x−110. Let’s draw a number line to indicate the sign of R  (x) and what that tells us about R(x). R  (x) is continuous, so it can only change sign about a point at which it is zero. R  (x) = 0 when x =−27.5. –150 –27.5 95 + – graph of R sign of R′ R(x) is increasing for values of x less than −27.5 and decreasing for values of x greater than −27.5. Because R(x) is continuous, we know the revenue will be maximized at x =−27.5. 344 CHAPTER 10 Optimization The price of a ticket should be set at $150 − $27.50 = $122.50 in order to maximize the revenue taken in. ◆ ◆ EXAMPLE 10.2 Suppose that we are trying to solve the same problem as in Example 10.1, but now we are told that the Beta Shuttle flight has a maximum seating capacity of 210 people. We can’t simply use our answer to Example 10.1, because we were figuring on 245 people purchasing tickets for the flight. Mathematically speaking, the difference between Example 10.1 and Example 10.2 is a change in the domain. Let’s determine the new domain. Recall that R(x) = (150 + x)(190 − 2x) = (price)(number of passengers). Since the number of passengers must be between 0 and 210 inclusive, 3 we need to find x such that 0 ≤ (190 − 2x) ≤ 210. We solve the following two equations. 190 − 2x = 0 and 190 − 2x = 210 −2x =−190 −2x = 20 x = 95 x =−10 Looking at the number of passengers tells us that −10 ≤ x ≤ 95. None of these x-values pushes the price of the ticket under zero, so this is our domain. Let’s look at the graph of R(x) restricted to the new domain. We’re interested in the portion of the parabola between x =−10 and x = 95. –10 95 x R Figure 10.2 We can see that R(−10) will be our maximum revenue given the plane size constraint in this example. The Shuttle should charge $140 per person to draw 210 passengers and obtain revenue of $29,400, an improvement of $900 per flight from their original revenue of $28,500. Alternatively, looking at the sign of R  (x) allows us to determine the maximum. R  (x) is negative on the interval from −10 to 95, so R(x) is decreasing on this interval. R(−10) must be the highest value attained on this domain. –10 95 – graph of R sign of R′ ◆ 3 The word “inclusive” tells us that the number of passengers can be equal to 0 and can be equal to 210. 10.1 Analysis of Extrema 345 ◆ EXAMPLE 10.3 A machinist is making an open-topped box from a flat sheet of metal, 12 inches by 12 inches in size, by cutting out equal-sized squares from each corner of the sheet and folding up the remaining sides to form a box, as shown. 12" 12" Figure 10.3 To maximize the volume of the box, what size squares should she cut from the corners? SOLUTION The goal is to maximize the volume of the box, so we’d like to get volume as a function of one variable. Let’s begin by using the variable x to label the lengths of the sides of the square pieces she will cut out. Then the box formed will be 12 − 2x inches long, 12 − 2x inches wide, and x inches tall. 12 – 2x 12 – 2x x x xx x x x x x x 12 – 2x 12 – 2x Figure 10.4 The volume, V , of the box is given by V = (length)(width)(height), so we obtain the equation V(x)=(12 − 2x)(12 − 2x)(x). Our goal is to maximize V(x)for 0 ≤ x ≤ 6. If we approach the problem without using derivatives, we find ourselves out of the familiar realm of parabolas; the expression for V is a cubic. We will discuss the graphs of cubics in Section 11.1, but because we have determined the domain it will be simple to graph the function V(x) using a graphing calculator. On the following page is a graph on the appropriate domain. The maximum value of V(x) on [0, 6] appears to be around x = 2. Notice that the beautiful symmetry of the parabola is, alas, lost on the cubic; the maximum value of V(x) does not lie midway between zeros of V . The machinist should cut out squares with sides of length approximately 2 inches each. 346 CHAPTER 10 Optimization 26 x V Figure 10.5 We can nail this down more definitively by using V  to find the maximum of V . We are interested in the points where V  (x) is either zero or undefined, because the sign of V  (x), and hence the direction of V , can only change at these points. V(x)=(12 − 2x)(12 − 2x)(x) = 4x 3 − 48x 2 + 144x In the Exploratory Problems for Chapter 7 you found that if f(x)=a 0 +a 1 x+a 2 x 2 +a 3 x 3 , then f  (x) = a 1 + 2a 2 x + 3a 3 x 2 . Therefore, V  (x) = 12x 2 − 96x + 144 = 12(x 2 − 8x + 12) = 12(x − 6)(x − 2). V  (x) is never undefined; it is zero at x = 2 and x = 6. On the interval [0, 6] the sign of V  can change only around x = 2. We need to check the sign of V  on the intervals [0, 2) and (2, 6). The simplest way to do this is to use a test point from each interval. We draw a number line to show important information about V(x)and V  (x). 02 6 + – graph of V sign of V ′ V(x) is continuous and is increasing from x = 0tox=2and decreasing from x = 2to x=6. So x = 2 gives the maximum volume. The dimensions of the box will be 8 inches × 8 inches × 2 inches with a total volume of 128 cubic inches. ◆ Analysis of Extrema In this section we use information about the derivative of f to identify where f takes on maximum and minimum values. We attempt to make arguments that are intuitively compelling. Our basic premise is, as previously stated, f  > 0 ⇒ f is increasing, f  < 0 ⇒ f is decreasing. This makes sense; where the tangent line to f slopes upward, f is increasing. It actually takes a surprising amount of work to prove this. The proof rests on the Mean Value Theorem and is given in Appendix C. We make a distinction between the absolute highest value of a function on its entire domain and a merely locally high value. This is analogous, for instance, to distinguishing between the point of highest altitude in the state of Massachusetts and the highest point of Winter Hill in Somerville, MA. We do similarly with low values. 10.1 Analysis of Extrema 347 Definitions Let x 0 be a point in the domain of f . x 0 is a local maximum point or relative maximum point of f if f(x 0 )≥f(x) for all x in an open interval around x 0 . 4 The number f(x 0 )is a local maximum value. x 0 is a local minimum point or relative minimum point of f if f(x 0 )≤f(x) for all x in an open interval around x 0 . The number f(x 0 )is a local minimum value. x 0 is a global maximum point or absolute maximum point of f if f(x 0 )≥f(x) for all x in the domain of f . The number f(x 0 ) is the global or absolute maximum value. x 0 is a global minimum point or absolute minimum point of f if f(x 0 )≤f(x) for all x in the domain of f . The number f(x 0 ) is the global or absolute minimum value. The term extrema is used to refer to local and to global maxima and minima. local max at x = – 4 local min at x = –2 absolute min at x = 8 local max and absolute max at x = 4 f x – 6 –5 –4 –3 –2 –112345678 The graph of f with domain [–6, 8]. Figure 10.6 x = 8: absolute minimum point x = 4: absolute maximum point local maximum point x =−2: local minimum point x =−4: local maximum point x =−6: neither absolute nor local extremum Technical Conventions In the language set out above, inputs to the function are called points; outputs are called values. A maximum/minimum point may be just local or both local and global. Because our definitions have “≤” and “≥” as opposed to < and >,iff is a constant function, then every x-value is simultaneously a maximum point and a minimum point. We will use the words “global” and “absolute” interchangeably when describing max- ima and minima. 4 Loosely speaking, “in an open interval around x 0 ” means “right around x 0 , on both sides.” More precisely, the statement above means there is an open interval (a, b), a<x 0 <bsuch that f(x 0 )≥f(x)for all x ∈ (a, b). 348 CHAPTER 10 Optimization We will use the words “local” and “relative” interchangeably when describing maxima and minima. Endpoints of the domain can be absolute extrema but not local maxima and minima. This is because we have said that when at a local extrema you must be able to “look” on both sides of you and endpoints, by their nature, only allow you to “look” on one side. With the exception of the last point, all the definitions given should fit in well with your intuitive notions of local and absolute maximum and minimum. Where Should We Search for the Extrema of f ? Let’s first look at the case in which f is continuous on an open connected domain. The domain could be of the form (a, b), (a, ∞), (−∞, b),or(−∞, ∞). We want to identify points at which the function changes from increasing to decreasing or vice versa; i.e., we are interested in the points where the derivative changes sign. The derivative can change sign only at a point where the derivative is either zero or undefined. Around these points the sign of f  may change or it may not. global max local max global min local min (a) At the extrema f ′ = 0. Not all points at which f ′= 0 are extrema. f global max absolute max global min absolute min f local max local min local min (b) At the extrema f ′ is undefined. Not all points at which f ′is undefined are extrema. Figure 10.7 If f is continuous on an open interval, it may be that f has no extrema. Consider, for example, f(x)=x 3 on the interval (−1, 1). See Figure 10.8. f x f f(x) = x 3 on (–1, 1) f(x) = x 2 on (–2, 2) x –2 –11 2 minimum only f f(x) = x on (–1, 2) x Figure 10.8 If f is continuous on a closed interval, then the Extreme Value Theorem guarantees that f attains both an absolute maximum and an absolute minimum value. The endpoints 10.1 Analysis of Extrema 349 of the domain and points at which f  is zero or undefined are candidates for extrema. See Figure 10.9. f x (i) f(x) = |x| on [–1, 1] f x (ii) f(x) = x on [–1, 2] –112 f x (iii) f(x) = x 2 on [–1, 2] Figure 10.9 If f is not continuous, we also need to look at points of discontinuity. (Keep in mind that f must be defined at x 0 in order for x 0 to be an extreme point.) At points of discontinuity f  is undefined. Notice that f  does not change sign at the extrema in the examples in Figure 10.10. f x local and absolute max local and absolute min local and absolute max no extrema f x f x f on domain [–2, 2] f on domain [–1, 1] Figure 10.10 We’ll refer to the set of points we need to sieve through in order to identify extrema as candidates for extrema or candidates for maxima and minima. We can summarize our observations by identifying all candidates for extrema as follows. Candidates for Maxima and Minima Candidates for extrema are all x 0 in the domain of f for which • f  (x 0 ) = 0, or • f  (x 0 ) is undefined, or • x 0 is an endpoint of the domain.    Critical points of f 350 CHAPTER 10 Optimization Definition The critical points of f are the points x 0 in the domain of f such that either f  (x 0 ) = 0, f  (x 0 ) is undefined, or x 0 is an endpoint of the domain. CAUTION Be aware that many texts define critical points to be only points at which f  is zero or f  is undefined. We are adding in endpoints of the domain so that critical points and candidates for extrema are identical. A point at which the derivative is zero is also sometimes referred to as a stationary point. 5 Question: Suppose f  (a) = 0. Does that mean that x = a is, necessarily, a maximum or a minimum? Answer: No. f  (a) = 0 simply means that the slope of the graph at x = a is zero. f  must change sign around x = a if f has a local extremum there. It is possible that f has neither a maximum nor a minimum at x = a. See Figure 10.11. f x f ′ x f ′(x) = 3x 2 f ′(0) = 0 but x = 0 is not an extremum of f. The sign of f ′ does not change around x = 0. f(x) = x 3 Figure 10.11 Question: If f has a maximum or minimum at x = b, does it follow that f  (b) = 0? Answer: No. It only follows that x = b is a critical point. Think about the absolute value function, f(x)=|x|.f has a minimum at x = 0, and f  (0) is undefined. Think also about the function f(x)=x on the domain [−1, 2]. The absolute maximum occurs at x = 2 and the absolute minimum at x =−1. In neither case is f  zero. (See Figure 10.12.) 5 If f gives an object’s displacement as a function of time, then this terminology makes a lot of sense. f  (t 0 ) = 0 means that at t 0 the object’s velocity is zero. The object is stationary at this moment. . problems to get an idea of how to set them up and use both calculus-based and non-calculus-based methods to solve them. Throughout the course, as we study new kinds of functions and their derivatives,. such an optimization problem; a gardener had a fixed amount of fencing and wanted to find out how to maximize the area of her garden. These and many other questions require the optimization of some. a number line to show important information about V(x )and V  (x). 02 6 + – graph of V sign of V ′ V(x) is continuous and is increasing from x = 0tox= 2and decreasing from x = 2to x=6. So x =