CHAPTER 7 OPTIMIZATION: METHODOLOGY 7.1 DERIVATIVES: ADDITIONAL INTERPRETATIONS 7 2 IDENTIFICATION OF MAXIMA AND MINIMA 7 3 CURVE SKETCHING 7 4 RESTRICTED-DOMAIN CONSIDERATIONS ADDITIONAL EXERCISES CHAPTER TESTCHAPTER OBJECTIVES G Enhance understanding of the meaning of first and second derivatives Q Reinforce understanding of the nature of concavity G_ Provide a methodology for determining optimization conditions for mathematical functions lustrate methods for sketching the ge..cral shape of mathematical functions 2 Mlustrate a wide variety of applications of optimization proceduresOEE eg Differential calculus offers considerable insight regarding the betas or mathematical functions. It im particulurly usefull in eetimating the srw)! a) representation of a funetionin twodimensians, This contrasts wit) the brute force” mathods of sketching functions, which we discussed js ( hu A. We want to illist rate thin attribute of ditferentinl calculus hy skeichiny the function . fix) aie (Example 17) In this chapter the tools developed in Chap. 15 will be extended. We witl further our understanding of the first and second derivatives. We will see how these derivatives can be useful in describing the behavior of mathematical functions, A major objec. tive of the chapter isto develop a method for determining where a function achieves maximum or minimum values. We will show how these calculus-based optimiz tion procedures facilitate the sketching of functions DERIVATIVES: ADDITIONAL INTERPRETATIONS In this section we will continue to expand our understanding of derivatives, The First Derivative As mentioned in the previous chapter, the first derivative represents the instanta. neous rate of change in f(x) with respect to a change in x. DEFINITION: INCREASING FUNCTION The function fis said to be an increasing function on an interval Jil for any x, and x, within the interval, x, < x, implies that flx,)< fx.) Increasing functions can also be identified by slope conditions. If the firet derivative of f is positive throughout an interval, then the slope ia posi- tive and f is an increasing function on the interval. That is, at any point within the interval, a alight increase in the value of x will be accompanied by an increase in the value of f(x). The curves in Fig. 7.10 and 7.1b are the graphs of increasing functions of x because the tangent slope at any point in positive. * Technically speaking, these are detinitions for strictly inereasing (dacreasing) RinctionsDERIVATIVES: ADDITIONAL INTERPRETATIONS 335 fe) Decrenung 4) Dberaang fur Figure 7.1 The relationship between /’(x) and increasing /decreasing functions. DEFINITION: DECREASING FUNCTION The f interval Lif tor b ction fis said to be 4 decreasing function o) any x, and x, within the x, < x, anphies that fly) >A As with increasing functions, decreasing functions can be identified by tangent slope conditions. If the first derivative of fia negative throughout an in- terval, then the lope ta negative and f ia a decreaning function on the interval. That is, at any point within the interval a alight inerease in ths value of x will be accompanied by a decrease in the value of f(x). The curves in Fig. 16.1cand are the graphs of decreasing functions of z. reasing) on an snlerval, the function 1s increasing, NOTE If afunction is increasing 'echnically speaking, these are dafinicions for strictly Increasing (dsereeaing) functionsCHAPTER 7 OPTIMIZATION: METHODOLOGY Given f(x) = 5x" — 20r + 3, determine the intervals over which fean be described as (a) en increasing function, (3) a decreasing function, and (c) neither increasing nor decreasing. SOLUTION ‘To determine whether / is increasing ar decreasing, we should first find /” P(x) = 10x = 20 f will be an increasing function when {°(s) > 0, or when 10a = 20 > 0 or 10x > 20 J will be @ decreasing function when f'(x) <0, o when 10x -20<0 or lor < 20 or ree { will be neither increasing or decreasing when {'(x) = 0, or when 10x—20= or 10z= 20 or x2 Summarizing, fin a decreasing function when x <2, neither increasing nor decreasing at x= 2, and an increasing function when x > 2. Sketch the graph of / to see whether these conclusions seem reasonable a ‘The necond derivative f(x) is a menaure of the instantaneous rate of change in {'(x) with respect to a change in x, In other words, it indicates the rate at which the slope of the function is changing with respect to a change in x —whether the slope of the function is increasing or decreasing at « particular instant If {"(x) is negative on an interval I of f, the first derivative is de- creasing on I. Graphically, the slope is decreaning in value on the inter- val. If f(x) ia positive on an interval I of [, the first derivative is increas- ing on I. Graphically, the slope is increasing on the interval. Examine Fig. 7 2, Bither mentally construct tangent lines or lay a straight ecige on the curve to represent the tangent line at various points. Along the curve from A to B the slope lightly negative near A and becomes more and more negative as we get closer to B. In fact, the slope of the tangent line goes from a valueDERIVATIVES, ADDITIONAL INTERPRETATIONS. 337 rio ‘Slope inereing Figur, 7-2 The relationship between /(x) and increasing /decreasing slope |/"(x)] of 0 at A to its most negative value at point B. Thus the slope is decreasing in value cover the interval between A and B, and we would expect /”(x) to be negative on this, interval [ive (“(2) < 0} Having reached its most negative value at point B, the slope continues to be negative on the interval between B and C; however, the slope becomes less and less negative, eventually equaling 0 at C. If the slope assumes values which are becom: ing less negative (e.g., —5, — 4, —3, —2, —1, 0), the slope is increasing in value. As such, we would expect /”(x) to be positive on this interval [i.e., /"(x) > 0]. Between C and D the slope becomes more and more positive, assuming ite largest positive value at D. Since the slope is increasing in value on this interval, we would expect /"(z) to be positive. Between D and B the slope continues to be positive, but it is becoming lees and less positive, eventually equaling 0 at E, If the slope is positive but becoming amaller (ex. 5, 4, 3, 2, 1, 0), we would expect /"(x) to be negative on the interval Figury 7 3 summarizes the first- and second-derivative conditions for the four regions of the function, These relationships can be difficult to understand. Take your time studying these figures and retrace the logic if necessary. é fx} oecrmasing Ns) ner fla) incrensing Gnupn of Graoh of Graph ot f oneave op ‘ncaa up ‘Concave down Piguee 7.3 Joint characteristies for {’(x) and f(x)338 CHAPTER 7 OPTIMIZATION: METHODOLOGY Concavity and Inflection Points In Chap. 6 the concept of concavity was briefly introduced. A more formal defini- tion of concavity follows. DEFINITION: CONCAVITY The graph of a function fis concave up (down) on an interval if /” increases (decreases) on the entire interval. ‘This definition suggests that the graph of a function is concave up on an interval ifthe slope increases over the entire interval. For any point within auch an interval the curve representing f will lie above the tangent line drawn at the point. Similarly, the graph of a function is concave down on an interval if the slope decreases over the entire interval, For any point within such an interval the curve representing { will lie below the tangent line drawn at the point. InFig. 7.4 the graph of fis concave down between A and B, and it is concave up between H and C. Note that between A and B the curve lies below its tangent lines ‘and between and C the curve lies above its tangent lines. Point B ia where the concavity changes from concave down to concave up. A point at which the concay- ity changes is called an inflection point. Thus, point B is an inflection point. fe) ‘ | Conenmuo (carve limabave pe . ie, rot AG Give le below tangent tne | re, 7.4 Representation of coneavity conditions ‘There are relationships between the wecond derivative and the concavity of the graph of a function which are going to be of considerable value later in thie chapter. ‘These relationships are as follows. RELATIONSHIPS BETWEEN THE SECOND DERIVATIVE AND CONCAVITY 1 IFf*t2) <0 0n an interval 0 x b, the graph of fis concave down over that interval. For any point x= © within the interval, fis said to be concave down at jc, fc.eee DERIVATIVES: ADDITIONAL INTERPRETATIONS. 339 UW JE" (3) > O.0n ang interval a = x= ‘over that interval Far any point x= c within the interval, (43 vaie’ 1» he concave up at [c, (cl). IMF (a) © 0 at any point x ¢ in the domain of f no c be draum abou the concavity at fc. {ci} b, the graph of fs comeave up clusion ean Be very careful not to reverse the logic of thees relationships! Because of relation. ship III we cannot make etatementa about the sign of the second derivative knowing the concavity of the graph of a function. For example, we cannot state that if the graph of a function is concave down at x= a, f(a) < 0. ‘To determine the concavity of the graph of the generalized quadratic function f(z) = ax? + bx + ¢, let's find the first and second darivativ ["z)—2ax+b and fs) = 20 Ifa > 0, then /*(z) ~ 20> 0. From relationship 11, the graph of / is concave wp whenever a >0. Ifa<0, then f%(z) = 2a <0, Prom relationship 1, the graph of / ia concave down ‘whenever a < 0. Thia is entirely consistent with our discussione in Chap. 6 which concluded that if > 0, /graphe as a parabola which ix concave up, And ifa< 0, graphses a parabola whieh ia concave down, For fisp=x* ~ 1, determine the concavity of the greph of fatx=—2 ond x— 4, SOLUTION ["2) = Sx" 4241 end f(z) =6x-4 Evalvating /*(=) at x= —2 gives f'(-2) = 6-2) 4-16 ‘Since f”(—2) <0, the graph of fia concave down at x= ~2. To determine the concavity nt ==, we Bind f"(3) = 6(3) ~4= 14 Since /”(3) > 0, the graph of fis concave up at x = 3. Determint the concavity of the gragh of f(x) = x4 at z~ 0. SOLUTION aa? f(x) ae? 710) = 12(0)8 =340 CHAPTER 7 OPTIMIZATION: METHODOLOGY ri =0 Figure 73 No conclusions regarding concavity According to relationahip 111, we can make no statement about the concavity at x= 0 However, substituting @ sufficient number of values for x inta f and plotting these ordered pairs, we see that f has the shape shown in Fig. 16.5. From this sketch itis obvious that the graph is coveave up at x = 0. Qa LOCATING INPLECTION POINTS 1 Fost ull poinss a where ("tay = 0 Th If P" 1x) changes sign when passing through x= a, there és am inflec tion point at x™ a. A necessary condition (something which must be true) for the existence of an inflection point at x = ais that f"(a) = 0. That ia, by finding all values of x for which f(x) = 0, candidate locations for inflection points are identified." The condition #"(a) = 0 does not guarantee that an inflection point exists at x = a (see Example 4). Step Il confirms whether a candidate location is an inflection point. The essence of this teat is to choose points slightly to the left and right of x =a and determine if the concavity is different on each side. 1ff“(x) is positive to the left and negative to the right, or vice versa, there has been a change in concavity when. passing through x= a. Thus, an inflection point exista at x= a In Example 4, (°(0) = 0 implies that x = 0 is a candidate location for an inflection point (atop 1). © Step II To verify that an inflection point does not exist at «= 0 for f(z) = 24, #2) is evaluated vo che left at x= ~0.1 and to the right at x = +0.1. * Other candidates for inflection points occur where f%(z) ia discontinuous. However, we will not encounter auch carididaten i hie book,DERIVATIVES: ADDITIONAL. INTERPRETATIONS 341 £(- 0.) = 12(-0.1)" =0.12>0 and #(+0.1) = 1240.1 =012>0 Since the second derivetive has the same sign to the Jn no inflection point at x = 0, fand right of x = 0, there To determine the location(s) of oll inflection pointa on the graph of fan 2 42410 we find and Step 1/72) is wot equal to 0 in order to find candidate locations! ir +2—0 or (2-1-2) 0 Therefore, /"(x) = 0 when x= 1 and x = 2. G Stop M1 Fors = 1, ("x re Ll is evalusted to the left and right of x= 1 at x= 0.9 and £09) = (0.8) ~ 3009) + 2 081-2742 ~011>0 f(a) = (1.4 = (1.1) +2 | 121-9342 i} 0.09 <0 Since the sign of /“(x) changes, an inflection point existe at x= 1. When valuen of x= 1.9 and x= 21 are chosen for evaluating /“(x) to the left and right of x = 2 £19) =—0.09 <0 fA) 011 >0342 CHAPTER 7 OPTIMIZATION; METHODOLOGY Becaune /"(x) changes sign, we conclade that an inflection point also exists at PRACTICE EXERCISE Determine the locations of a: flection points on the graph of i= | 29/3 + 2? — Bux, Answer: inflection point (Tracking an Epidemic: Revisited) The Motivating Scena rend of « flu epidemic. The function n= /(t) = 0.30 + 100? + B00 + 256 0 entiqate the number of persona afflicted with the days since initial detection by health department officiala, ¢ Determine any inflection pointe and interpret their meaning in thie application , 4, ab & function of the| number o SOLUTION 2 Step t f(t) = 0.907 + 208 + 300 10) = 1.88 + 20 8 "4E) =O emults sn o7ie candidate for an inflection point at ¢—= 11.11 Step HI Because /"(11) = 0.2 and /“(12) = ~1.6, the sign of /"(t) changes when passing through the candidate location, Thus, one inflection point exints for the function 2 Interpretation ‘The inflection point can be interpreted as representing the point fh pernons are being afflicted by the fui decreases. Prior 1 (= 11.11, additional parsons are being ailicted at an inereasing rate. After t = 11.11, additionsd persons are being afflicted, but at a decreasing rate in tiene when Concavity from a Different Perspective We will use the ter 8 coneave up und concave down to describe the curvature attribute which we call concavity, Other terminology may be used to describe this attribute. For example, many writers distinguish between strictly concave func- tions and atrictiy convex functions. DEFINITION: STRICTLY CONCAVE (CONVEX) FUNCTION A function shich js strictly concave (convex) has the following graphical froperty: Givers any {wo points and Bwhich lle on the curve representing th ion, f the wo points are connected by a straight line, the entire line segment AA will lie below (above) the curve except at points A and BDERIVATIVES: ADDITIONAL INTERPRETATIONS \ Concave and conver f ‘This definition can be loosened someswhat to define a concave function and a convex function [ax opposed to str sagment AB ia allowed to lie below (above) the curve. or to fie on the curve, the function is termed & concave (convex) function. Figure 7.6 illustrates these defi+ nitions. concave (convex) functions). If the line Section 7.1 Follow-up Exercises For sack of the following functions, (a) determine whethar fia incroming or decreasing at x 1, Determine the values of x for which f is (d) an increasing fanetion, (c) a decreasing function, and (d) either increasing nor decreasing. 1 f(z) 20—4e 2 fix) = 10s + 8 fiz) x? ~ Be +20 4 fa) = 32" + Wet 9 5 fs) m= xt/3+3"/2 6 fiz) = 31/3 + 24/2 ~ Ox 7 fiz) xt + Det 8 ft) ax* 8 f(s) = (+3) 10 f(x) = 32%/(x* AL fiz) 29 +42 +16 12 fix) = 13 f(x) 62! + 40% +60 14 f(s) =2 15 fix) = (2-4) 16 f(x) = (x 5)* 17 (a) = (2s — 10) 18 f(x) = (Bx + 24)" 19 (x)= 20 f(x) = (2x + 18) For each of the following functions, use /(x) to determine the concavity conditions at x=-Qands=+1 21 f{x)==32"+ 2-3 22 [x)= +122 +1 23 fix) 2-42 +9 4 fla) a9 + Bx 25 f(x) Ve F 10 26 fix)mix+0)!344 CHAPTER 7 OPTIMIZATION, METHODOLOGY 27 fix) = x + ae! 29 (x) = Bx? ~ 4x94 10x ST f(z) = 29/3 — 21/2 + 0x 33 f(x) = (2-1) 88 f(x) = (3x'+2)¢ 87 f(x) = Ve 10 39 f(s) =e 41 fiz) lox Ifa>0, 6>0, and c>0, determine the values of x for which / in (a) increasing, 28 f(z) — x44 + 9) 30 f(x) = x + 2x" — 10st ‘B2 f(x) = 5x"/3 + 329/2— 6x + 25 34 f(x) = (20— 86 f(x) = (2x ay 3B f(x) = 24/1 — 2) 40 f(x) =e" 42 f(x)——Inz (b) decreasing, (c) concave up, and (d) concave down, if: "43 fis) =axtb #45 f(x) ax? + bebe AT f(z) 3 ax? For each of the following functions, identify the locations of any inftection points. 49 f(x) 24-93! 61 fiz) = 34/12 — 29/3 — 7.6" 53 f(x) = (x—5) 55 f(x) = —10x+ +100 87 fix) = x*+ 6e"— 18 5 fix) = 4/12 + 21/8— ae 61 f(x) = 1'/4— 92/2 + 100 83 /lx) = (3e— 199M 65 f(z) = (x~ 51 67 fix) =e" 69 f(x) = Inx 44 f(x) =b— or 746 f(x) =~ ax $48 f(x) = ox* 50 fx) — 29+ 2427 52 f(x) = (@—x)* 54 fis) = 24/20 — 24/8 86 fiz) = (x—1)/s 8B /(s) = ~2*— 30x 80 f(x) = 24/12 + 72°/6 + Bx" 02 f(x) = £*/30 — 4x*/2 Ad fin) = (9 86 f(x) = (x +2)" 88 f(x) me 70 f(s) = —Inx 71 Given the function shown in Fig 7 7, indicate the valves of x for which / is (a) increaning, (6) decreasing, and (c) naither increasing nor decreasing. 72 Given the function shown in Fix. 7 7, indicate the values of x for which fis (a) conceve up, (6) concave down, (c) changing concavity, (d) concave, and (e) convex. sp 'IQENTIPICATION OF MAXIMA AND MINIMA Figure 72 ps) 73 Given the function shown in Fig. 77, indicate the values of x for which / is (a) increasing at an increasing rate, (b) increnaing at a decreaning rate, (c) decreasing at waging rate, and (d) décteaaing at an increasing rate. 74 Given the function shown in Fig. 7.8, indicate the values of x for which / in (a) increasing, (4) decreasing, and (c) neither increasing nor decreasing 75 Given the function shown in Fig. 7,8, indicate the valuee of x for which /is (a) concave up, (6) concave down, (c) changing concavity, (d) concave, and (e) convex. 76 Given the function shown in Fig 7.8, indicate the values of x for which / is (a) increasing at an increusing rate, (b) increasing at a decreaning rate, (c) decreasing ot ‘a decreasing rata, and (d) decreasing at an increasing rate IDENTIFICATION OF MAXIMA AND MINIMA In this section we will examine functions with the purpose of locating maximum and minimum values Relative Extrema DEFINITION: RELATIVE MAXIMUM If fis defined on an Mtervai (8, chwhich contanis x a, fis sand to reach joe local maximom x= aif fiat % fa) Ine all a within the interv! DEFINITION: RELATIVE MINIMUM Hi fis defined on an interval 46, cl which contains xe a, Fis saul to ren. ths relative flocal) méntmum at x= .1\{ 10,3 (4x) for sll x within the interyst (iba Both definitions focus upon the value of f(x) within an interval. A relative maximum refers toa point where the value of f(x) is greater than the values for any pointa which are nearby. A relative minimum refers to a point where the value of346 CHAPTER 7 OPTIMIZATION: METHODOLOGY Figore 729 fat Relative extrema. _ Both raatve mentor {Both rtetive mink Fad ebeoksts rnin 7(z)’is lower than the values for any points which are nearby. If we use these definitions and examine Fig. 7.9, / has relative maxima at x= a and x Similarly, fhos relative minimaatx = bundx = d. Collectively, relative maxima and minima are called relative extrema. DEFINITION: ABSOLUTE MAXIMUM A function fis said to reach an absolute maximum at x= aif ((a) > f(x) for any other xin the domain of f DEFINITION: ABSOLUTE MINIMUM A function fis said to reach an absolute minimum at x= aif (a) < fix) for any other x in the domain of £ If we refer again to Fig, 7-9, /{x) reaches an absolute maximum et x~c. It reaches an absolute minimum at x= 6. [t should be noted that a poing on the graph ofa function can be both arelative maximum (minimum) and an absolute maximum (minimum), Critical Points We will have a particular interest in relative marima and minima, It will be important to know how to identify and distinguish between them. NECESSARY CONDITIONS FOR RELATIVE MAXIMA (MINIMA) Given the function £ necessary conditions for the existence of a relative maxi: mum or minimum at x= a (a contained in the domain of F) are 1. f(a) = 0, or 2. f(a) is undefined. Points which satisfy either of the conditions in this definition are candidates for relative maxima (minima). Suck points are often referred to as eritieal pointe,IDENTIFICATION OF MAXIMA AND MINIMA 347 Critical points |_— Tero ines = 0 fade £0} ot dein 1) Slope = 0 {1 Slope not defines (eonaten 1) (condition 2) Points which satisfy condition 1 are those on ths graph of f where the slope equals. Points satisfying condition 2 are exemplified by discontinuities on or points where f(z) cannot be evaluated. Values of x in the domain of f which satisfy either condition 1 or condition 2 are called critical values. These are denoted with an asterisk (x*) in order to distinguish them from other values of . Given a critical value for f, the corresponding critical point ia [x*, /(x*)]. Figure 7.10 illustrates the graphs of two functions which have critical points at (7, 0). For the function f(x) = x?, shown in Fig. 16.10a, /’(x) = 2x and a critical value occurs when x= 0 (condition 1), where the function achieves a relative minimum. For the function f(x) = 3%, f(x) = 2/x4*. Note that condition 1 can never be satisfied since there are no pointa where the tangent slope equals 0, However. a critical value of x =O exiats according to condition 2. The derivative is undefined (the tangent line ia the vertical line x ~ 0 for which the slope is undefined). How- ever, /(0) in defined and the critical point (0, 0) is @ relative minimum, as shown in Fig. 7.108. In order to determine the location(s) of any critical points on the graph of f= 6x +100 the derivative f’ is found. Iz) = Owhen . zt 6-0 or (2— Ble +2)—0 When the two factors are net equal to 0, twocritical values are x = Sand x= —2. WhenCHAPTER these values are sul patted into f, and x= — PRACTICE EXERCISE 7 OPTIMMATION: METHODOLOGY reoulting critical pointa are (~ ‘The only statements we can make about the bebavior of fat these pointe i that the elope equals 0. Furthermore, nowhere else on the graph of / doss tha alopa equal 0, Additional teeting in neceesary to determine whether there is relative mnaxizoum of rinimum at x = 3 ,107}) and (3, 86) Determine the location(s) of any critical points on the graph of f(x) = 29/3-+ 8 = Bx, Ansiner: Critical points a (= 4, 2641 and (2, ~ 94, THOUGHT. 2 functions (ivr example. fix) = 10)? DISCUSSION POINT FOS §— What comment can be: made regarding thee \istene ofc (points 5h instant Figure 7.11 Critical pbints where (702) ! Figure 7.11 ilhustrates the different possibilities of critica) points where f(z) = 0, Figures 7.11a and b illustrate relative maximum and minimum pointa, whereas Figs. 7.11¢ andd illustrate two different types of inflection points. In Fig.IDENTIFICATION OF MAXIMA AND MINIMA a 7.1lc the graph of the function has a slope of 0 at point a, and it is also changing from being concave down toconcave up. In Fig.7 .L1dthegraph hasa slope of 0 and in changing from being concave up to concave down. Any critical point whare f'(x) = 0 will be a relative maximum, @ relative mini mum, oF an inflection poine POINT FOR = Far polynomial fun ions fof degree n, the lamgest possible number of critscal THOUGHT & points where /(r) = 0) is 7 — 1 Thus, a function fof degree 5 cam have as DISCUSSION — many 2s four points of zero stape Why is this sol The First-Derivative Test In an effort to locate relative maximum or minimum points, the first step is to locate all critical pointa on the graph of the fmetion, Given that acritical point may be either a relative maximum or minimum or am inflection point, some test must be devised to distinguish among these. There are a number of tests available. One test which is easy to understand intuitively is the Aret-derivative test. After the locations of critical points ave identified, the first-derivative test requires au examination of slope conditions to the left and right of the critical point. Figure 7.12 illustrates the four critical point possibilities and their slope fist fl fw | | 1 cin resem, |iomreara} | | oucrmara TAZ First-derivative test350 CHAPTER 7 OPTIMIZATION: METHODOLOGY conditions to either side of x*. For a relative maximum, the alope in positive to the left (x,) and negative to the right (x,). For a relative minimum, the alope is negative to the left and positive to the right. For the inflection points, the alope has the same sign to the left or the right of the critical point. Another way of describing this test is the following: 1 Fora relative maximum, the value of the function is increasing to the left and decreasing to the right 2 Forarelative minimum, the value of the function is decreasing to the left and increasing to the right. 8 For inflection points, the value of the function is either increasing both to the left and right or decreasing both to the left and right. A summary of the first-derivative test follows. FIRST-DERIVATIVE TEST Locate alt critical values x* For any critical value x*, determine the value of (x) to the left (x) and right (x,) of x* (x;) > O and ' (x,) < 0, there is a relative maximum for/ ot [x tx), (6) IF Fx) <0 and £'s,) > 0. there és a relative minimum for ut [xt (x?) (ch If tx) has the same sign at both x, and x,, an inflection point exists at (x*, £0 RARDIN Determine the location(s) of any critical points on the graph of f(x) = 2x*~ 12x ~ 10, and determine their nature, SOLUTION ‘The firnt derivative in Pa) 4-12 When the first derivative in set equal to 0, 4r-12=0 or dred and there is a critical value at Since /(3) = 2(97) — 12(3) — 10 —28, there is critical point located at (3, —28).IDENTH MAXIMA AND MINIMA 351 To teat the critical point, te y= 29 and x, 3.1 (2.9) = 412.9) = 118-124-04 fi) = 491) — 12 1-12=+04 he laft of x = 3 and positive (+0.4) to th 2 on /(note that fina quadratic function which Because the first derivative is negative (0.4 right, the po graphs as o parabols that is concave up). a A CAVEAT When selecting 2,and.x,, one must stay reasonably close to the critical value x7 If you stray (00 far left or right, you may reach an errongous result, asin Fig. | 16.13, where a relative minimum could be judged to be a relative maximum Some latitude does enist in selecting x) and x,. However, when more than one point exists, %, and, should be chosen in such a way that they fall the cr alve being examined and any adjacent critical values fess ss Vigure 7.1 ne 2 = 6+ 100 hae critical points at (3, 864) and (~2, 107}). To detormine the nature of these critic! points, we examine the first denivatSana CHAPTER 7 OPTIMIZATION: METHODOLOGY Pia) = 29-3 In testing the critigal point at x 3, let's select s;— 2 and x, = 4, r= (Qa - 2-6 4 faye a4 6 6 Since ’(x) is negntive (fis decreasing) to the left ofx ~ 3 and positive (/is increasing) to the right, a relative minirouin occurs for f when x ™ 3 In testing the critical point at x ~~2, let's select xj~ —9 and x, ~—1 3) = (~a)*~(-3) -6 “6 Since /’(x) is positive (fin inereasing) to the left of x = ~2 and nogative (fia decreasing) to the right, a relative maximum occurs for f(x) when x= ~2 The first-derivative test also is valid’ for thove critical pointa where /“(x) in undefined Consider the function f(x) = 3x2", the kraph of which 1# shown in Fig. 7 106. For this function, /'(x) = 2/x!. Because /'(x) is undefined when x = 0, acritical value exiats accord ing to condition 2. Since /(0) = 3(0)** 0, there is a critical point at (0, 0), Uning the firat-derivative toat, let's wolect x, = —1 and x, = 1 Since /’(z) is negative to the left of x = Onnd ponitive to the right, a relative minimum occurs at the critical point (0, 0) Q PRACTICE EXERCISE In the Practice Exercise on page critical points of (~4, 264) and (2, —9}) were identified for the function f(x) = 22/3 + x? ~ Bx. Determine the nature of these critical points using the first-derivative test, Ansiner Rel maximum at (~4. 264) and relative micimum at (2, ~ 9) The Second-Derivative Test For critical points, where /’(«) ~ 0, the moat expedient teat is the seecond-deriva- tive test. Intuitively, the second-derivative test attempts to determine the con-IDENTIFICATION OF MAXIMA AND MINIMA cavity of the function at a critical point (x*,/(x*)]. We concluded in Sec. 16.1 that if /"(z) <0 at a point on the graph of f, the curve is concave down at that point. If /"(x) > ata point on the graph of f, the curve is concave up at that point. Thus, the second-derivative test suggests finding the value of /"(x*). Of greater intereat, though, is the sign of f”(x*). [ff”(x*) > 0, not only is the slope equal to 0 at x* but the function fis concave up at x*. If we refer to the four critical point possibilities in Fig 7.11, only one is concave up at x*, that being the relative minimum in Fig 7b. If/"(x*) <0, the function is concave down at x*. Again referring to Fig. 7 11, the only ctitical point accompanied by concave-down conditions is the relative maximum in Fig. °7 1a. As stated in Sec. 7.1, iff"(x*) = 0, no conclusion can be drawn regarding the concavity at [x*,/(x*)]. Another test such as the firet-derivative test is required to determine the nature of these particular critical points. A summary of the second- derivative test follows. SBCOND-DERIVATIVE TPST 1 Pind all critical values x", such that {"(x*)= 0 Hh For ony critical role x", determine the value off" tx"). fa) WP" tx") > 0, the function is concave wp at x? and there is a rela- tive mininuem (ar fat f°, 110") (bi If" (x*)< 0, the function is concuve down jat x* and there is a relative maximum for { at /x*, f(x"); te) Hf" (<4i= 0, no conclusions can be draws about the concavity at 2? wor the nature of thy critical point, Ayother test such as (he frat-derivatlie 1st is necessary. Examine the following function for any eritical points and determine their nature [lz) =~} + bx 20 SOLUTION We abauld recognize this ux u quadratic function which graphs as a parabola that isconcave down. There should he one critical point which ia x relative masimum. To confirm this, we find the firet derivative Pla)m 92 +6 IF //(2) in net equal to 0), ar+6=0 and one critical value occurs whenCHAPTER 7 OPTIMIZATION: METHODOLOGY ‘The value of f(x) when x = 2 is 112) = —4(2") + 6(2) ~ 20 ~-6+12-20—-4 ‘The only critical point occurs at (2, ~14), Using the socond-derivative text yinlda Mie=—3 and ’Q)=-3<0 Since the second derivative is negative at x= 2, we can conclude that the graph of / is concave down ot this point, and the critical point in a relative maximum. Figure 16.14} prasenta a sketch of the function : tae | st fix)* =332 +20 Figure 7.14 Relative maximum at (2, —14) FRENTE EG Examine the following function for any critical pointa and determine their nutur. SOLUTION lef’ w identified,DENTIFICATION OF MAXIMA AND MINIMA 12) equals zero when 2-960 x{x7— 9) = 0 or when x(x+3)ix-3)=0 Ifthe three factorn are met equal to 0, critica! valuen ure found when ye0 xe-8 xm ‘Substituting thease criti 81/4), and (3, ~B1/4) The pecand derivative is, 3! values into f, critical pointa occur on the graph of fat (0, 0),(~3, Pl) = 32~9 To test the critical point (0, 0), #*(0) = 40") — 9 =-9<0 1¢/"(0) ia negative, the function is concave down at x = Onda relative maximum occurs at (0, 0), To test the eritical point (~ 3, 1/4) f-3) = 3-3) 9 27—9=18>0 ‘The graph of fis concave up when x~~3 and a relative mininium occurs at (~3,—81/4). To teat the critical point (3, ~81/4). ‘The graph of fis concave up when x= 3 and w relative minimum occurs at (3, —81/4), ‘To summarize, relative minima occur on f at the points (~3, 81/4) and (3, ~81/4): and a relative maximum occurs at (0, 0). Figure 7.15 ia a sketch of the grapb of f PRACTICE EXERCISE In the Practice Exercise on page 752, you were asked to determine the nature of two critical points using the first-lerivative test. Use the second-derivative test to confirm these results. FP) Examine the following function for any critical pointe and determine their nature fla) = ~ 10,000e-6" — 1908 + 10,000CHAPTER 7 OPTIMIZATION: METHODOLOGY SOLUTION If we find /’ and set it equal to 0, 1/02) = —10,000(-0.03)¢-*™ — 120 = 000 -e#%* — 190 8002 °% ~ 120 = 0 when 3006-8 = 120 or when ott 04 ‘To solve for x, take the natural logarithm of both rides of the equation. ~ 0.08: = In 0.4 Therefore, 0.032 = 0.9168 when x= —09163/~0.03 A critical value occurs when = 3054 The only critical point occurs when £ = 30.54. Continuing with the second-derivative teat, we have 17(2) = 300(—0.08)e*™ = —9¢-0mm 1°(30.54) = —De-eevonse, == 99-280 = -0(04) =-38<0IDENTIFICATION OF MAXIMA AND MINIMA 357 Stel = ="0,0008-**% — 120" + 10,000 70 70 30 40 50 @0 70 \ 90 100 500 Therefore, « relative maximum occurs when x= 30.54. The corresponding value for /(2) ia 1(30.54) = 10,000e-°9%44 — 120(90,54) + 10,000 10,000(0.4) ~ 3,664.8 + 10,000 = 2,335.2 ‘The relative maximum occurs at (30.54, 2,335.2). Figure 16.16 contains a aketch of the Bineston Q When the Second-Derivative Test Fails If /"(x*) = 0, the second derivative does not allow for any conclusion about the behavior of / at x*. Consider the following example. Examine the following function for any critical pointe and determine their nature. EXAW fe) soLUTION I(x) = 63 Setting /' equal to 0, ~br'=0 when x=0 ‘Thus, acritieal valuo existe for fwhen x = Oand theres a critical point at (0, 0). Continuing swith the second-derivative test, we got (x) = — 2028358 17 CHAPTER 7 OPTIMIZATION: METHODOLOGY Atr=0 ro) —20(0" =0 Using the second. derivative test, there is no conclusion about the nature of the critical point. ‘We can use the firet-derivative test to determine the nature of the critical point, If x= —1 and x, 1, then srt) = 51-1" =-5 ra)y=—s0)* Since /’(~1) and /'(1) are both negative, an inflection point occurs at x= 0. Figure 7.17 presents 0 sketch of the graph of the function. fo _Infiaccton point Higher-Order Derivative Test (Optional) ‘There are several ways to reach a conclusion about the nature of a critical point when the second-derivative test fails. One such method is the higher-order de- rivative test, Although not as intuitive as the first- or second-derivative tests, it is generally conclusive. HIGHER-ORDER DERIVATIVE TEST 1 Given a enitical point /x*, f(")) on f, find the lowest-order derivative whose value is nonzero at the critical value x*. Denote this derivative as P(x), where n is the order of the derivative,IDENTIFICATION OF MAXIMA AND MINIMA IL If the onder n of this decivalive is even, Féx*) is 0 relative maxtmin it Pix") < O and a relatioe minimum if x") > 0. ‘ MLW the onder n of this dertoative is udd, the critical point és an inflec- tion point. fn Identify any critical pointe and determine their nature if Hla)= r= 9) SOLUTION First, we find /' fa) = 4x ~ 291) = A(z— 2)" If we vet /! equal to 0 when At thin critical value f(2) = (22) =(0%=0 ‘Thus a critical point occurs at (2, 0) To determine the nature of the critical point, the second derivative is F(x) = 4(3)(x — 2° = 122-2)" Evaluating /* at the critical value, (*(2) = 12(2- 2" =o ‘There is no conclusion based upon the second-derivative test, If we proceed using the higher-order derivative test, the third derivative is (x)= r= 2) ani 17 (2)— 24(2—2) 0 ce /’"(2) = 0, there is no conclusion based on the third derivative ‘The fourth derivative is\PTIMIZATION: METHODOLOGY pita) = 24 and f9(2) = 24 This is the lowest-order derivative not equaling 0 when x ~ 2. Since the order of the deriva ive (ns 4) ineyen, a relative maximum or minimum existe at (2, 0),"To determine which ia the case, we look at the sign of 42). Because /(2) > 0, we ean conclude that there is « relative minima at (2,0). Figure 16.18 contains a sketch of the function. Q NOTE — The second denvative testis actually a special case of the higher-order derivative =the case where the lowest-order derwative not equating 0 is the second depvative (9 = 2) PRACTICE EXERCISE Apply the higher-order derivative test to determine the nature of the critical point in Example 15, Section 7.2 Follow-up Exercises For each of the following functions determine the location of all eritical points and deter: mine their nature 1 f(z) = B24 = 48s + 100 2 f(z) = 29/9 — Bz" + 18x + 100 8 fiz) = 100745 4 (x)= x82 +473 CURVE SKETCHING G f(x) = 29/3 — 2.617 + 4x 7 fix) Bat/4 — 1539/2 f(x) = Sx" ~10 AL fz) — 2/24 6x +9 AG f(x) = 424/15 +4 1S f(x) = 2x4/3 ~ 21/2 — 10s UT f(a) = 44/6 ~ 424x 18 f(x) = 0/6 + 227 Bi f(z)— ~ 228 + 24/4 23 f(x) = 22"/5 ~ 21/4 ~ x" 2B fix) = xt/5—« 27 fix) = 24/6 x1/242 20 fx) = (x + 10)* BR fx) (ae +2? 93. f(x) = - (228) 36 file 37 f(x) = 800° + Sox 38 f(x) ~e* AD f(x) =e" 49. (a) = A0e-8™ + 6x— 10 4B fix) = 104 Inx 47 f(s) = Intx? +1) 30h 6 f(x) = 5x4 — 20 8 fx) = —xt/4 + 99/2 10 fx}= 4-2 12 f(x) ~ — 6x? ~ 96: +10 14 fix) == 29/10 16 f(x) = 2/0 + Bs! + 60x 1B fix) ~224/6 + 924 20 f(x) = 2°/3 + 37/2 ~ 202 22 f(s) =—x'/A + Brt +5 24 fix) = 24/5 + 3x44 ~ 42/3 26 f(z) =~ 2s° + 10.5x* + 12r 28 f(x) = 424/39 — 6 30 f(x) = (22+ 97 32 fix) = (2e~ 8) 34 f(x) = (x? ~ 167 36 f(x) =e" 3B f(x) =~ 450-8 — 18x +10 40 f(z) = —100e-*™* ~ 50x 42 [(z) = —BOe-9™ — 40x 44 fir) = 20 0% 442-3 46 f(s) In x 0.52 4B fix) =n x 29/4 48 f(x) =4z Ins BO fix)e x? Ins BI fix) = In 5x ~ 10: G2 fix) =n Me ~ 2? BS fix) = 27+ BA /(x) = 051? +7x—30In x #5B f(x) = x/ HG f(s) = x(x +2)" 857 [{x) = ax? + bx +6, where a > 0,6<0,6>0 #58 f(x) = ax? + bx +6, where a <0,b< 0,6 <0 *59 Original Equation Test {t was mentioned in the lant section that other techniques ‘exint for determining the nature of critical points. One teat involves comparing the value of f(x*) with the values of f(x) juat to the left and right of 1°, Refer to Pig. 16.11, and determine a vet of rules which would allow one to distinguinh among the four critical point possihiti 60 Compare the relative efficiencies associated with performing the original equation teat and the Brat-derivative tent of critical points, 61 Compare the relative efficiencies associated with performing the first-derivative, wee ‘ond-derivative, and higher-order derivative tests of critteal pointa *62 When the Second-Derivative Test Fails—An Alternative Given # critical value determined when /"(x) = 0 and failure of the sscond-derwvative test to yield» conclusion, determine a net oF rules which will yield w conclusion based upon checking. the concavity conditions to the left and right of the critical value. CURVE SKETCHING Sketching functions is facilitated with the information we have acquired in this chapter. One can get # feeling for the general shape of the graph of a function without determining and plotting # large number of ordered pairs, This section discusses some of the key determinanta of the shape of the graph of a function and illusteates curve-sketching procedures,362 CHAPTER 7 OPTIMIZATION: METHODOLOGY Key Data Points In determining the general shape of the graph of a function, the following atteiimabes are the most significant: G@ Relative maxima and minima Q Inflection points Q sand y intercepts Q Ultimate direction ‘To illustrate, condider the funetion Hay my At + 128 +6 1 Relative Maxinia and Minima To locate rélative extrema on /, aii the firat detivative Ps) 2*— Be +12 Setting /’ equal to 0 yields sta +120 or (x 22-6) <0 Critical values occur at x = 2.and x = 6. If we subetitute these critical valusowagiiy, (Q)= . 4(2) + 12(2) +6 “$= 16 +24 +6 = 15} 7 aid 16) = 5 = 46) + 1216)+ 5 = 72-1444 724+5=6 ‘This, critieal pointe exist at (2, 15}) and (6, 6). Graphically, we know that exe slope conditions exiét «! these points, aa ahown in Fig. 16.194 ‘The second derivative of fis ["(s) = 22=8 ‘To test the nature of the critical point (2, 164), P'(2)=2Q)-8=-4<07.3 CURVE SKETCHING 363 {) Crikeal pina on (8) Flatts exscoma for / (e) Raletve extrema and inflection goin (6) Fina semch off © 7.19 Development of sketch of Ax) x!/3 ~ Ax? + 12245. ‘Therefore, a relative maximums occurs at (2, 164).'To test the nature of the critical point (6, 5), £6) = 6) 8 =4>0 ‘Thorefore a relative minimum occurs at (6, 5). The information we have devel thus far allows us to develop Lhe sketch of f to the degree shown in Fig. 7.19CHAPTER 7 OPTIMIZATION: METHODOLOGY 2 Inflection Points Inflection point candidates are found when /” ix aot equal to 0, or when or s=4 If we substitute x = 4 int we con state that the only candidete for an inflection point occurs at (4, 104). Without checking the sign of /” to the left and right of = 4, we can conclude that the point (4, 10) is the only inflection point on the graph of f. The renson for this is that there must be an inflection point between the relative maximum at (2, 15%) and the relative minimum at (8, 5). For a continuous function, the concavity of the function must change between any adjacent critical points, The only candidate identified lien between the two critical points; thus, it must be an inflection point. The information developed to this point allows us to enhance the sketch of /, as shown in Fig. 19¢, 3. Intercepts The y intercept is usually an easy point to locate. In this case 10) =5 ‘The y intercept occurs at (0, 5) Depending on the function, the x intercepts may or may not be easy to find. For this function they would be rather difficult to identify. Our sketch of f will not be affected significantly by not knowing the precise location of the one x intercept which exists for /. 4 Ultimate Direction For /, the highest-powered term is x*/3. To deter: mine the behavior of / as x becomes more und more positive, we need to observe the behavior of x4/3 as x becomes more and more positive. As Therefore, as 4s? + 12x +6 +0 Similarly, as and Figure 7 ,19¢ incorporates the intercopts and the ultimate directions into our sketch,CURVE SKETCHING 26 (Motivating Sconario) Sketch the graph of the function SOLUTION 1 Relative Maxima and Minima To locate relative extrema of /, we find the Est derivative: f(s) x8 ~ Bat + 16x ‘Setting /’ equa} to 0 yield or x(x- M4) 0 Ifthe factora are vet equal to 0, critical values are found at x = O and x = 4. The correspond ing values of f(x) are j40) =0 Bay end r~4 +814) ro = 64— 170} + 128-~ 214 ‘Therefore, critical points occur at (0, 0) and (4, 21}). ‘eating x= 0, we find and 0 A relative minimum occurs at (0, 0) Testing x= 4 F(a) 364)? 164) +17 48-64 +16=0 No conelusion can be drawn about x = 4 based upon the second derivative, Continuing with the higher-order derivative teat, f(x) = 6-16 4) 6(4) — 16 =3>0 and Since the order of the derivative is odd, un inflection point occurs at (4, 214).366 CHAPTER 7 OPTIMIZATION: METHODOLOGY 2 Inflection Points Candidates for inflection points are found by setting / equal to 0, or when 3s? ~ 16x + 16=0 (3x 4x4) =0 emt and xed We have already verified that an inflection point occurs at (4, 214). Confirm for yourself that ({, 8.69) ia also an inflection point. 3 Interoepta By computing /(0) = (0)'/4 — 8(0)*/8 + 8(0)* = 0, we conclude that the _y intercept occurs at (0, 0). To locate the + intercepts, when One root to this equation ia x = 0, suggesting that one x intercept is located at (0, 0) (earlier, wwe sbould have observed that the y intercept in also an x intercept). Use the quadratic formula to verify that there are no roots for the equation ar | To yteno Por /, the point (0, 0) represents tbe only intercept. 4 Ultimate Direction The ultimate behavior of /(x) is linked to the behavior of the term 24/4. As +e and f(x) Ae re Sate and. f(x)-++@ Uding the information gathered, we can sketch the approximate shape of / as shown in Fig 7.20, Qa Section 7.3 Follow-up Exercises Sketch the graphs of the following functions. 1 f(x) = 29/3 — 5x" + 16x ~ 100 2 fymx'- 6x46 B f(x) 24/4 2529/2 4 f(z) = 29/3 — 25x" + de 5 f(x) = (6x~ 12? @ f(x) = x"/6—x*—10 7 f(x) =—G@— 5) B f(x) (x! 16)720 Ta RESTRICTED-DOMAIN CONSIDERATIONS 367 He: 9 fz) = 4/3 — 7134/2 10 fi) = AA f(z) = ~8r* + 100 12 fz) =x 1B f(z) = 22°/3 + 29/2 ~ 105 14 fz) = 21/3 — Bx? + 60x 16 f(z). — 44/5 + 32Ax ~ 250 16 f(z) = 24/5 — 82x AT fia) 07? — 10 18 f\x) = x4/4—959/2 19 f(z) in (x? + 25) 20 Az)m 2/3 — 3,52? — 90x RESTRICTED-DOMAIN CONSIDERATIONS In this section we will examine procedures for identifying absolute maxims and minima when the domain of # function is restricted. When the Domain Is Restricted Very often in applied problems the domain is restricted. For example, if profit Pie stated as a function of the number of units produced 2, it is likely that x will be reatricted to values auch that 0 0, a relative minieausn occurs at (6, —13), Because f’(s) is defined for all real x, no other critical valves exitt the domain are endpoints 0 BStep U1 The values of f(x) at ti 2 12) i 3 P-bit 5g +62) +5 f00) = = +5 000 70 ae + 5 = 48h D Step UIT Comparing /(2), (6), and /(10), we find the absolute minimum of 15 occurs when x = 6 and the abwolute maximum of 484 occurs when x= 10. Figure presents a sketch of the function, a Section 7.4 Follow-up Exercises Inthe following exercises, determine the locationa and values of the absolute maxiraumn and absolute minimum for /. 1 /(x) = 2x? —4e +5, where 25258 2 f(x) = —x* + Bx — 100, where 25454 3 f(x) = x*— 12a", where 2 = x= 10CHAPTER 7 OPTIMIZATION: METHODOLOGY tie Feros, 200510 6 4 {(x) =~ 2x" — 15s + 10, where 8 5x5 +2 5 lx) = x*/5~ 2-25, whore sx 5B 6 f(x) = 8/5 ~ 8xt/4 + 22/3 — 20, where 0's x <5 7 lx) ™ xt ~ 2° + 25s", where 0 S454 B fle) = £5 +10, where 1255 9 f(z) = —42" + 6x — 10, where O= s<10 10 f(x) = 3/1 ~ 29/2 ~ 6x, where 0<=6 LI [As) = 24/4 — 4x" + 16, whore 5 = x= 10 12 fla) = xt/4 — 154/3 + Be, where O'S x54 13 flx) ~ 29/6 ~ 5xt/4 ~ 1429/3 — 10, where OK a <6 Ld f(s) = 51/4 ~ 8x" +25, where x20 15 f(z) In (x*+20), where ~1 5254 16 f(x) = 2, whore Osas4 17 /(x) =x", where 4s x= 16 1H f(z) = (x ~ 2), where 0 x102 ADDITIONAL EXERCISES Section 7.1 For the following exercises, determine the intervals over which / in (a) incrmawing, (6) creasing, (¢) neither increasing nor decreasing, (d) concave up, end (e) concay 1 fla) 3 fiz 5 Hx) 7 fiz) = 4s? s 7 loca i ection pointe 14 fix) = a4 1G f(x) = 24/12 + 29+ ax" 20 /\ + x¥/6 — Sx* +120 ast 4 f(x) = 2x°—x 26 / 1” Section 7.2 For the following functions, determine tbe location of all critical points and determine the 7 fia) = de — 0x + 20 fix) = BI f(z) 8 — 4e* +40 39 fiz) = x"—x* BB f(x) = Sx*— 16s" + 242" + 10 87 fa) 9 f(x) 41 fix) 43 is 45 1421 47 f(s) 49 fix) 5} fiz) 59 f(x 56 fix) 57 fix) 59 fix)CHAPTER 7 OPTIMIZATION: METHODOLOGY Section 7.3 Sketch the gr ha of the following functions. 1 f(s) = x49 — 739/24 12x 62 f(x) = (10-2) 3 f{x) = (x+ 6)" 84 fx) = 228/56 +24/4— 2441 6 f(x) = — 22? + Tx! 4x-9 6 f(x) = x4/h txt eat +6 87 f(xh= (x43) 8B f(x) = x4 — 2524/2 Section 7.4 For the following functions, determine the location and values of the absolute maximum and absolute minimum. GD /(x) = 3x" — 48x + 90, where O 0, and (b) f(x) > O and /"(x) > 0. 3 For the following functions, determine the locations of all critical points and determine their nature. (a) f(z) =~ 2x4 = 215+ 5 (b) fyaernCHAPTER TEST 4 Given aot fh -F ae identify the locations of ail inflection points 5 Determine the locations and values of the absoluce inm minimum for fis)=-~ ~ © + ox ~ 10, ox a 6 Sketch the function f(x) = (x + 4)*. a73CHAPTER .8 OPTIMIZATION: APPLICATIONS 8.1 REVENUE, COST, AND. PROFIT APPLICATIONS 8.2 ADDITIONAL APPLICATIONS ADDITIONAL EXERCISES CHAPTER TEST MINICASE: THE EOQ MODEL.Illustrate a wide variety of applications of optimization procedures UW Reinforce skills in problem formulation (Reinforce skills of interpretation of mathematical results “ill = i N } z CF ix ——SORIA ‘ajor oi! company is planning to construct a pipeline to deliver crude oll SCENARIO: from a major well site to a point where the crude will be loaded on tankers MIR und shipped to reineries. The pipeline aust be constructed through (wo ditferent types of Verrain, relatively barren land und dense forest, C« etruct, costy vary significantly depending upon the terrain, The com pany wishes tdtetdrmane a construction plan u buch wil minnie the cose of constructing the pipeline (Example 17) as Chapter 16 provided the tools of classical optimization. That is, it gave us amethod for examining functions in order to locate maximum and minimum points. This chapter will be devoted to illustrating the use of these procedures in a variety of applications. As you begin this chapter, remember that these applied problems usually require a translation from the verbal statement of the problem to an appropriate mathematical representation. Care must be taken to define variables (unknowns) precisely. Once a mathematically derived solution has been found, an conential element in the problem-solving process is the interpretation of the math- ematical reault, within the context of the application setting. As you proceed through this chapter, you will utilize some or all of the stages of this problem-solv ing process, as shown in Fig. 8.1 process Het spies © ‘ YS arti a Pn REVENUE, COST, AND PROFIT APPLICATIONS Revenue Applications ‘The following applications focus on revenue maximization. Recall that the money which flows into an organization from either selling products or providing servicesREVENUE, COST, AND PROFIT APPLICATIONS in referred to us revenue. The most fundamental way of computing total revenue from selling a product (or service) is Total revenue = (price per unit)(quantity sold} An assumption in this relationship is thet the selling price is the same for all unite sold. ‘The demand for the product of « firm varies with the price thet the firm charges for the product, ‘The firm estimates that annual total revenue R (stated in $1,000») in a function of the price p (stated in dollars). Specifically, R=f(p) = —50p* + 500p (a) Devermine the price which should be charged in order to maximize total revenue, (b) What is the maximum value of annual total revenue? SOLUTION (0) From Chap. 6 we should recognize that the revenue function is quadratic, It will graph a aparabola which is concave down. Thus the maximum value of R will occur at the vertex. ‘The first derivative of the revenue function is I'(p) = —100p + 500 I we vet ("(p) equal to 0, —100p + 500 = 0 —100p = ~500 or a critical value cocurs when Pp Although we know that « relative maximum occurs when p = 6 (because of our know! edge of quadratic functione), lat's formally verify this using the aecond-derivative tent: f(p)m 100 and /*(6)=-100<0 Therefore, « relative maximum of f occurs at p =. ‘The maximum value of R is found by substituting p= 6 into f, or 50(54) + 50015) 1,250 + 2,500 = 1,250 ‘Thus, annual total revenue is expected to be maximized at $1,260 (1,000) or $1 million when the firm charges $5 per unit. Figure 8.2 presents a sketch of the reven function. Qa8 OPTIMIZATION: APPLICATIONS 260) oe Ravers maxinieation point Quadratic revenue function. ‘This has been mentioned earlier in the text when we have dealt with applications; however, itisworth repeating: ftis yuite common for students to work through a word problem, find the solution, but have no ability to interpret the results within the framework of the application. Ifyou becume caught up in the mechanics of finding a sulution\ and temporarily fose your frame of reference regarding the ‘onginal problem, reread! the problem. making special note of Nav the variables. are defined. Also review the specific questions asked in the aroblein. This showid assist in reminding vau of the objectives and the direction in whicl+ou s!ould he heading. (Public Transportation Management) ‘The transit authority for a major metropolitan area has experimented with the fare atructure for the city’s public bun system. It has aban: doned the zone fare structure in which the fare varies depending on the :uzber of rones through which « passenger jmases. The new system is a fixed-fare ayxter in which a paasen: ger may travel between any two points in the city for the anme fare ‘The transit authority han surveyed citizens to determine the number of persons who would use the bus nyatew ifthe fare was fixed at different amounts. From the survey rewults, ss analysta have determined an approximate demand function which exprevaes the ss a function of the fare charged. Specifically, the demand function is a= 10,000 — 125 where q equals the average number of riders per hour and p equals the fare in cents (a) Determine the fare which should be charged in order to maximize hourly bus fare (6) What ie the expected maximum revenue” (c) How many riders per hour are expected under this fare?REVENUE, COST, AND PROFIT APPLICATIONS SOLUTION (a) ‘The firat step is to determine a function which states hourly revenue ae a function of the fare p. The reason for selecting p as the independent variable is that the question was to determine the fare which would result in maximum total revenue. Also, the fare is.a decision variable—n varinble whose value can be decided by the transit authority management. The general expression for total revenue is, ax mated before, But in this form, 2 is stated as.a function of two variables —p and. AC hia time we.can not deal with the optimization of functions involving more than one independent vari- able. The demand function, however, establishes a relationship between the variables p and q which allows us to transform the revenue function into one where R in stated an a function of one independent variable p.’The right aide of the demand function states q in terme of p. If we mubstitute this expression for q into the revenue function, we get R= fp) = p(10,000 ~ 125p) or R= 10,000p — 125p* The firat derivative is 1p) = 10,000 ~ 250p If the derivative in set equal to 0, 10,000 ~ 250p = 0 10,000 = 250p and n critical value occurs when 40=p ‘The second derivative ie found and evaluated at p = 40 to determine the nature of the critical point: 1p) = =250 £"(40) = -250 <0 ‘Thus, o relative maximum occurs for / when p= 40. Since / in everywhere concave downward, the interpretation of this result is that hourly revenue will be maximized when a fixed fare of $0.40 (40 centa) in charged @) £(40) = 10,000(40) ~ 125(40)* = 400,000 — 200,000 = 200,00¢380) 8 VUPTINIZATION APPLICATIONS Figure 17.3 Quadratic revenve function Since the fare it stated in conte, the maximum expected hourly revenue is 200,000 cents, or $2,000. (c) The average number of riders expected each hour with thia fare in found by substivuting the fare into the demand finction, ot ¢ = 10,000 ~ 125(40) 10,000 — 5,000 ~ 6,000 riders per hour Figure 17.3 presente a sketch of the hourly revenue function, Q Cost Applications ‘As mentioned earlier, costs represent cash outflows for an organization. Most organizations seek ways to minimize these outflows. This section presents applice- tiona which deal with the minimization of some measure of cost (aventory Management) A common problem in organizations is determining how much of s needed item should be kept on hand. For rotailere, the problem may relate to how many units of each product abould be kept in stock. For producers, the problem may involve how much of each raw material should be kept available. This problem is identified with an ares called inventory control, or inventory management, Concerning the question of how much "“inventory"*to keep on hand, there may be costs ansociated with having toa little or too touch inventory on hand. A retailer of motorized bicycles has examined cost data and has determined a cost function which expresses the annual coat of purchasing, owning, and mé tuna function of the aize (number of function is ntaining inventory a) of each order it places for the bicycles. The cout sin cm) = $88 + 151+ raREVENUE AND P PP where C equals annual inventory coat, 9 Joliars, and q equals the number af cycles irdered wach time the ret the supp (a) Devermine the order mize annual inventory coat. (6) What is minim xpected to equal? SOLUTION (a) The first derivativ 1'(q) = —4,860q-* + 15 fin net equal to 0, 4.8609"? + 16 = 0 60 when 5 Multiplying both sides by y/ and dividing both sides by ~15 yielde 4.6 Taking the square root of both sides, critica) values exist ‘The value ¢ =~ 18 it meaningless in this application (negative order quantities are not possible), The nature of the only meaningful critical point (q= 18) is checked by finding / Note that /"(q) > 9 for g > 0 « the graph of fis everywhere concave upward. ‘Thus, the minimum value for / occurs when q= 18, Annual inventory costa will be dered each time the retailer repleniahee the supply. ating /(18), oF minimized when 18 bicyel >) Minow p aniaual inventory co ermined by calcul 381CHAPTER 8 OPTIMIZATION: APPLICATIONS (18, 798,580) Cow miniaation point 4 6 10 18 2025 90 a5 AO umber of bicyelavorder Figure 8.4 Inventory cost function = 270 + 270 + 750,000 = $750,540 Figure 5.4 presents # sketch of the cost function. (The end of chapter minicase dis ‘cusses assumptionn which underlie the inventory cost function in this example.) (Minimizing Average Cost per Unit) The total coat of producing q unite of « certain product in described by the function C= 100,000 + 1,6009 + 0.2¢% where Cia the total cost stated in dollars. Determine the number of unita of that should be produced in order to minimize the average cost per unit. SOLUTION Average coat per unit ia calculated by dividing the total coat by the nuraber of unita produced. For example, if the total cost of producing 10 units of a product equals $275, the average cost par unit is $275/10 = $27.50. Thus, the function representing average cost per unit in thie example is E=1q=E= 200.000 + 1,500 + 0.8 ‘The first derivative of the average cost function ia 1(q) = ~ 100,009"? + 0.2 If" ia sot equal to 0,VENUE, COST, AND PROFIT APPLICATIONS The vahw q = —707.11 is meaningless in this application since ction level, ¢, must. be nonnegative The nature of the one relevant critical point is tested with the secand-darivative teat: #"\q) = 200,0009-4 rivative /*(p) ia positive fe of finconeave upward for q > 0. Thus, the minimum v 11. ‘This minimum 100,00 (707.41) = 141.424 Figure & 5 in m aketch of the average © Average cos function, ral oon hg 100 200 06 400 G00 GOH Yoo B00 G00 1.08CHAPTER 8 OPTIMIZATION, APPLICATIONS PRACTICE EXERCISE For Example 4, what is the total cost of production at this level of output? What are two different ways in which this figure can be computed? Answer $1,260,663.90 Profit Applications This section contains two examples which deal with profit maximization. (Bales Force Allocation) Example lin Chap. 6 discussed the law of diminishing returns ma ‘an illustrotion of a nonlineer function. A major cosmetic and beauty supply firm, which specializes in a door-to-door sales approsch, bas found that the response of ualea to the allocation of additional sales representatives behaves according to the law af diminishing returna. For one regional uales district, the company estimates that annual prafit P, stated in ‘hundreds of dollars, is a function of the number of sales representatives x assigned to the district. Specifically, the function relating these two variables is P= f(x) =~125x" + 1,975x - 1,500 (0) What number of representatives will result in maximum profit for the district? (6) What is the expected maximum profit? SOLUTION (a) The derivative of the profit function is 113) = = 25x 41,375 If /* is wet equal to 0, or a critical value occurs ==55 Checking the nature of the critical point, we find (xy—=-25 and (55) = nee the graph of / is everywhere concave downward, the maximum value of f occurs when x= 55 (b) The expected maximum profit is (68) =~ 12.5(55)* + 1,375(65) — 1,500 = —87,812.6 + 75,625 — 1,500 = 38,9125 We can conelude that annual profit will be mo. 1d at m value of $36,312.5 (1008), oFREVENUE, COST. AND PROFIT APPLICATIONS: 385 85,051,260, if 65 representatives ore ossigned to the district, Figure 8 6 presents a sketeb of the profit funetion a) POINT FOR THOUGHT & DISCUSSION What do the xuntercepts represent in Fig, 17 6? Interpret the meaning of the » intercept, Discuss the law of diminishing returns ast pertains to the shape of this profit function XAMPLE (Solar Energy) A manufacturer has developed a new design for solar collection panels Marketing studies have indicated that annual demand for the panels will depend on the price charged. The demand or the panels has been estimated as q = 100,000 = 200p (8.1) here q equals the number of units demanded each year and p equals the price in dollars, Engineering studies indicate that the total coot of producing ¢ panels is estimated function, 150,000 + 100g + 0.0034) 18.2) Formulate the profit function P = /(q) which stater the annual profit P as a function of the number of unita ¢ which are produced and sold. SOLUTION We have been asked to develop a function which states profit Pas a function of 9. Axopposed to Example 5, we must construct the profit function. The total cont function {Fs stated in terms of 7. However, we need to formulate a total revenue function state: of g. The basic structure for computing total revenue in YO 20-3 40 90 60 70 BO 90100110CHAPTER 8 OPTIMIZATION: APPLICATIONS. (2.3) “Pq Because we want & to be stated in terme of g, we can replace p in Eq, ( 8.3) with an equivalent expression derived from the demand function, Solving for p in Eq | 8.1) give 200p = 100,000 ~ q or p= 600 ~ 0,006q (8.4) We can substitute the right side of this equation into Eq. | 8.3) to yield the revenue function| R= (600 ~ 0.0069)q = 5009 — 0.0069" Now that both the reveriue and cost functions have beon stated in terms of g, the profi function ean be defined as = 600q ~ 0.00593 — (150,000 + 100g + 0.0034) = 500g — 0.0059" — 150,000 — 100q — 0.003" or P= ~0,00894-+ 4009 — 150,000 PRACTICE EXERCISE For Example 6, determine (a) the number of units.q that should be produced to maximize annual profit; (5) the price that should be charged for each panel to generate a demand equal to the answer in part a; and (c) the maximum an- nual profit. Answer la) 9 25,000 units, (61 p= $275, (c $4,850,000 (Restricted Domain) Assume inthe last example that the manufacturer's annual produc- IA ir, capacity is 20,000 unita. Re-eolve Example 6 with this added restriction. SOLUTION With the added restriction, the domain of the function is dafined as 0 ¢ = 20,000, From Soc, 8 4,it should be recalled that we must compare the values of f(q) st the endpoints of the domain with the values of /(q*) for any q* value, where 0 = g* = 20,000. ‘The only critical point on the profit function occurs at q = 25,000, which is outaide the domain. Thus, profit will be maximized at one of the endpoints. Evaluating /(q) at the endpoints, we find (0) = — 150,000 and (20,000) = .008( 20,000)" + 400(20,000) — 160,000 200,000 + 8,000,000 ~ 150,000 = 4,650,000 Profit is maximized at a value of $4,650,000 when g= 20,000, or when the manufscturer| operates at capacity.REVENUE, COST, AND PROFIT APPLICATIONS. sponser pea 1.008? +4009 - 180,000". 5 3,000,000 one <4 520.000 09.000 750m 20000 75,000 W000 %000 «0000 ” Figure 8.7 Profit function /restricted domain. The price which should be charged ix found by mibatituting g = 20,000 into Bq. ( 8.4), or p= 500 ~ 0.005(20,000) 500 — 100 = $400 Figure 8.7 presenta a sketch of the profit function a Marginal Approach to Profit Maximization An alternative approach to finding the profit maximization point involves mar- ginal analysis. Popular among economists, marginal analyais examines incre mental effecta on profitability. Given that « firm is producing a certain number of unita each your, marginal analysis would be concerned with the effect on profit if one additional unit is produced and sold. ‘To utilize the marginal approach to profit maximization, the following condi tions must hold ‘REQUIREMENTS FOR USING THE MARGINAL APPROACH 1 It must be possible to identify the total revenue function and the told cnst function, separately. 11 Phv renenue and cost functions must be stated in terms of the level vf futpul or number of units produced and sold. 387HAPTER 8 OPTIMIZATION: APPLICATIONS Marginal Revenue One of the two important conc arginal analysis is marginal revenue, Marginal revenue is the additional revent ved from selling one mare unit of a product or service. If each unit of a product sells at the same price, the marginal revenue is always equal to the price. For example, the linear revenue function fepresents @ situation where each unit sells for $10. The marginal revenue from selling one additional unit is $10 at any level of output q In Example 6 a demund function for solar panels was stated as 100,000 = 200 From this demand-functio R= fila) = 500g — 0.00592 (8.5) Marginal revenue for this example is not constant. We can illustrate this by com puting total revenue for different levela of output. Table g 1 illustrates these calculations for selected values of g, The third column repi margina revenue associated with moving from one level of output t although the differerices are ulight, the marginal revenue vplues are changing at each different level of ovtpui Total Revenve Marginal Revenue A) AR= fil) ~ Ala 1) $498.00 $498.985, $498,976 Fora total revenue function £(q), the derivative (a) represents the instant neous rate of change in revenue given a change in the number of units sold, 2’ also represents a general expression for the slope of the graph of the total revenue function. For purposes of marginal analysis, the derivative is used to represent the marginal revenue, or MR= FQ) | (8.6) ‘The derivative, as discussed in Chap.6_, provides an approximation to actual changes in the value of a function, As such, &’ can be used to approximate the margins] revenue from gelling the next unit. If we find R’ for the revenue function in Eq ( 85),approximate the marginal reven ling the 101at unit, we 100, or R‘(100) = 500 — 0,010(100) 500 499 This is a very close approximation to the actual value ($498,995) for marginal revenue ahown in Table 8.1 Marginal Cost The other important concept in marginal analysis is mar: ginal cost. Marginal coat is the additional cost incurred as a result of producing and selling ane more unit of a product or service. Linwar cost functions asaume that the variable cost per unit is constant; for such functions the marginal cost is the same at any level of output. An example of this is the cost function C= 150,000 + 4.5¢ where varinble coet per unit is constant at $3.50. The marginal cost for this cost function is alwaye $3.50. A nonlinear cost function is characterized by variable marginal costs. This ean be illustrated by the cost function © = fxlq) = 150,000 + 1009 + 0.0039" (6.7) which was used in Example 6. We can illusteate that the marginal costs do fluctuate at different levels of output by computing marginal cost values for selected values of ¢q. This computation is illustrated in Table 17.2. For a total cost function C(q), the derivative C’(g) represents the instanta neous rate of change in total cost given a change in the number of units produced €'(g) alao representa a general expression f of the graph of the total cost function, For purposes of marginal anal vative is used to represent the marginal cost (8.8) Computatios of Marginal Cost Level of Outpat Total Cost Marginal Cost « Ato AC= Ae) ~ Gle— 1) 100 $160,030.00 101 $160,190.64 102 $160,291.212 10a390) ATION: APPLICATIONS ‘An with R’, C’ can be used to approximate the marginal cost associated with producing the next unit. 'The derivative of the cost function in Eq. ( 8.7) is €(q) = 100 + 0.0064 To appro: q~ 100, of pate the marginal cost from producing the 101at unit, we evaluate C at (100) = 100 + 0.006(100) = $100.60 Ifwe compare this value with the actual value ($100,603) in Table 8.2, we see that the two values are very close, : Marginal Profit Analysis As indicated earlier, marginal profit analysia is concerned with the effect on profit if one additional unit of a product is produced and sold. Aa long as the additional revenue brought in by the next unit exceeds the coat of producing and selling that unit, there is a net profit from producing and selling that unit and total profit increases, If, however, the additional revenue from selling the next unit is exceeded by the cost of producing and selling the additional Unit, there is 8 net loae from that next unit and total profit decreases. A rule of thumb concerning whether or not to produce an additional unit (assuming profit is of greatest importance) is given next RULE OF THUMB: SHOULD AN ADDITIONAL UNIT BE PRODUCED? 1 2EMR> MC, produce the ment nil li MR < MC. do not produce the next unit For many production situations, the marginal revenue exceeds the marginal cost at lower levels of output, As the level of output (quantity produced) increases the amount by which marginal revenue exceeds marginal cost becomes smaller. Eventually, a level of output is reached ac which MR = MC. Beyorid this point MR < MC, and total profit begina to deerease with added output. Thus, if the point can be identified where MR = MC for the last unit produced and sold, total profit will be maximized, This profit maximization level of output can be identified by the following condition PROFIT MAXLMIZATION CRITERION Produce (0 the level of output where MR= MC Stated in terms of derivative where thia criterion suggests producing to the pointREVENUE, COST, AND PROFIT APPLICATIONS 391 RQ) = C9) ( 8.10) This equation is s natural result of finding the point whare the profit function is maximized, i.e., set the derivative of P(y) = Rig) — Cla) equal to 0 and solve for g. P(g) = Rg) —C'(g and Pig)=0 when Rg) — 0%) =0 or R(q) = CQ) Let g* be # value where R’(g) ** C'(g). The second de RY(q) ~C’(q). By the second derivative test, profit will be maximized at = q' provided vative of P in PY(g) = Pgh) <0 or Rg’) — C*(q") <0 or Rg") < C%q") IER"(q) < C"(q) for all values of q > 0, then profit has an absolute maximum value of g=g*. SUPFICIENT CONDITION FOR PROFIT MAXIMIZATION Given » level of output 9* where Ril) tg) ior MR MC), producing 2° will result in profit maximization if Rie cr@) (8.10) R= 600y — 0.0069" and C= 150,000 + 1009 + 0.00897 Because the revenue and cost functions are distinct, and both are stated in terme of the Level of output ¢, the two requirements for conducting marginal analysis are satisfied. We have already detormined thatCHAPTER 8 OPTIMIZATION: APPLICATIONS 1g) = 00 — OIG and C'(q) = 100 + 0.006q Therefore, R@=C@ when 600 ~ 0.019 = 100 + 0.0069 0.0169 = —400 a? = 28,000 R"G")=—001 and CQ") = Reg’) < Cg") 0.01 < 0.006 and there is relative maximum on the profit function when q— 25,000, Figure 8 & presents the graphs of R(q) and C(q) 4,000,000 12,000,000 10,000,000 pre | 2,000,000 | 180,000 + 1064 + 0.00%)7 16,000,000 4,906,000 2,000,000) Figure 8.8 Marginal analysis: profit maximization ‘Take w moment to examine Fig. 8 8. The following observations are worth nating: 1 Points C ond D represent paints where the revenue and cost functions intersect. These represent break-even point 2 Between points C and D) the revenue function is above the cost function, indicating that total revenue is greater than total cost and a profit wil! be earned within this interval. For levels of output to the right of D, the cost function lies above the revenue function, indicating that total cost exceeds total revenue and a negative profit (loss) will reauttREVENUE, COST, AND PROFIT APPLICATIONS 393 8 The vertical distance separating the arapha of the tue functions represents the profit or loas, depending on the level of output. 4 Inthe interval 0 = q = 25,000, the slope of the revenue function is positive and greater than the siope of the cout function. Stated in terms of MR and MC, MR > MC in this interval 5 Also, in the interval 0 = q = 25,000, the vertical distance separating the two curves becomes greater, indicating that profit is increasing on the interval. 6 At q= 25,000 the siopes at points A and Bare the same, indicating thot MR = MC: Also, at q™= 26,000 the vertical distance separating the two curves is greater than at ‘any other point in the profit region; thus, this is the point of profit maximization. 7 Fora > 25,000 the slope of the revenue function is positive but less positive thon that for the coat function. Thus, MR < MC and for each additional unit profit decreases, actually resulting in o loss beyond point D. In Example 5 we were anked to determine the number of sales representatives x which would reoult in maxiroum profit P for a commetic and beauty supply firm. The profit function way tated an P=jix)= 75x ~ 1,500 Uning the marginal approach, determine the number of representatives which will result in maximum profit for the firm. SOLUTION ‘We cannot use the marginal approach in this example because we cannot identify the total revenue and total coat functions which were combined to form the profit function! Require. ment 1 for using marginal analysis is not satisfied. Figure 8 9 illustrates « sketch of a linear revenue function and a nonlinear cost function, To the left of g*, the slope of the revenue function exceeds the slope of the cost function, Maximum prot ——=4 Unies produced and sokd © 8.4 Linear revenue / quadratic cost functions.CHAPTER 8 OPTIMIZATION: APPLICATIONS indicating that MR > MC. Atq* the slopes of the two functions are the samme. The vartical itance separating the two functions ia greatar at q* than for any other value of q between points A and B, Points A and B are break-even points Section § .1 Follow-up Exercizes 1A firma has datermined that total revenun in a function of the price charged for ite product. Specifically, the total mvenus function is Re=f(p)=—10p"+ 1,750p where p equals the price in dollars (0) Determine the price p which results in maximum total revenue. (©) What ia the maximum yalue for total revenue? 2 The demand function for 4 firm's product @ 150,000 ~ 15p where q equals the number of units demanded and p equals the price in dollars (@) Determine the price which uhould be charged to maximize total revenue. (©) What io the maximum value for total revenue? (©) How many unite are expected to be demanded? The annual profit for 8 firm depends upon the number of unite produced. Specifically, the function which describes the relationship between profit P (stated in dollars) and the number of units produced x is — 0.0127 + 6,000 ~ 25,000 (o) Determine the number of unite ¥ which will result in maximum profit. (b) What is the expected maximum profit? Beach Management A community which is located in resort area is trying to decide ‘on the parking fee to charge at the town-owned beech, There are other beaches in the area, and there is competition for bathers among the different beaches. The town hus determined the following function which expresses the average number of cars per day q ‘as 6 function of the parking fee p stated in centa @= 5,000 — 129 {a} Determine tne fee which should be charged to maximize daily beach revenues, {b) What is the maximus daily beach revenue expected to be? {c) How many care are expected on an average aay? Import Tax Monagement The United States government ia studying the import tax strocture for color televinion seta imported from other countries into the United States. ‘The government is trying to determine the amount of the tax to charge on each TV ast, The government roalizen that the demand for imported TV seta will be affected by the tax. Itentimates that the demand for imparted seta D, meamured in hundreds of TV seta, will be related to the import tax ¢, mezauired in centa, according to the functionREVENUE, COST, AND PROFIT APPLICATIONS 05 D = 80,000 — 12,58 {a} Determine the import tax which will ros ing TV so%s (0) What ia the maximum revenue? (c) What will the demand for importad color T'V neta equal with thin tax? 8 A manufacturer bas determined cost function which expresses the annual cort of purchesing, owning, and maintaining ita raw material inventory aa a function of the size of each omar. The cost. function ts in maximum tax revenues from import- 61,200 + 80y + 750,000 where q equals the size of each order (in tona) and C equals the annual inventory cost. (a) Determine the order size g which minimizes annual inventory cost. (b) What are minimum inventory costs expected to equal? 7 In Exercise 6 anaume that the maximum amount of the raw material which can be accepted in any one ¢hipment is 20 tons. (@) Given thia restriction, determine the order nize q which winimizes annual inventory cont (b) What are the minimum annual inventory costs? (c) How do these remults compare with those in Exercise BA major distributor of raccquetballe is thriving, One of the distributor's major problema is keeping up with the demand for racquetballs. Balls are purchased periodically from a sporting goods manufacturer. he annual cost of purchasing, owning, and maintaining Ure inventory of racquetialls is described Ixy ite functor = FE. 5 6159 + 2,000,000 q whore ¢ equals the order size (in dozens of racquetballs) and C equals the annual (a) Determine the order size 9 which minimizes annual inventory cont (6) What are the minimum inventory costa expected to equal? 8 The diniributorin Exercise & has storage fncilitiento accept up to 1,200 dozens of balls in any ove shipment. (a) Determine the order aize q which minimizen annual inventory conta (6) What are the minimum inventory coste? (¢) How do these reulta compare with those in Exercine 87 10 The total cost of producing g units of a certain product is described by the function C= 5,000,000 + 250g + 0.0024" where C is the total cout wtated ir dollars (o) How many unita should be produced in order to minimize the average cost per unit? 4b) What is the minimum average cont per unit” (c) What ia the total cost of production at this level of output? 11 The total cost of producing q unita of a certain product is described by the function (C= 360,000 + 7.5009 + 0.258 OPTIMIZATION: APPLICATIONS CHAPTER Where C is the total cost stated in datlars, {a} Detertsine how many unita q should be produced in order to minimize the average cost per unit. (b) What is the minimum averoge cost per unit? {c) What is the total cost of production at this level of output? 12 Re-solve Exorcie 11 if the maximum production capacity in 1,000 units 13 Public Utilitie A cable TV antenna company has determined that its profitability depends upon the monthly fee it charges ita customers. Specifically, the relationship which describes annual profit P (atated in dollars) as a function of the monthly rental fee 7 (stated in dollars) ia P= ~50,000r' + 2,750,000” — 5,000,000 (a) Determine the monthly rental fee r which will lead to maximum profit. (b) What is the expected maximum profit? 14 In Exercise 13 annume that the local public utility commission has restricted the CATV company to « monthly fee not to exceed $20. (a) What (ee leads to a maximum profit for the company? (b) Whats the effect of the utility comminsion’s ruling on the profitability of the firm? 15 A company estimates that the d’mand for ita product fuctuntes with the price it charges. The demand function in ¢ = 280,000 — 400p where q equals the number of units demanded end p equala the price in dollars. The total cost of producing q units of the product ia estimated by the fumetion C= 350,000 + 3004 + 0.00154" (a) Determine how many unita q should be produced in order to maximize annual profit (b) What price should be charged” (©) What is the annual profit expected to equal? 16 Solve the previous exerciee, using the marginal approach to profit maximization 17 If annual capacity is 40,000 unite in Exercise 15, how many unite g will result in maximum profit? What is the loss in profit attributable to the restricted capacity? 18 An equivalent way of solving Example 2 is to atate total revenue wa a function of 9, the average number of riders per hour. Formulate the function & = 4(q) aud determine the number of riders g which will result in maximum total revenue, Verity that the maxi mum value of @ and the price which should be charged are the same ax obtained in Example 2 19 The total cost and total revenus functions for a product are Clq) = 600 + 100g + 0.59% R(q) = 500 (a) Using the marginal approach, determine the protit-maximizing level of output. (0) What is the maximum profit?REVENUE, COST, AND PROFIT APPLICATIONS 20 A firm sells euch unit of a product for $50. The total cost of producing x (thousand) un: {a described by the function C(x) = 10 = 2529+ 24 where C(x) is measured in thousands of dollars. (a) Use the marginal approach to determine the profit-maximixing level of output (b) What in total revenue at this level of output? Total cost? Total profit” 21 The profit function for a firm ia P(q) =~ 4.59" + 36,0009 — 45,000 (@) ‘Using the margioal appronch, determine the profit-maximizing level of output (6) What is the maximum profit? 22 The total cost and total revenue functions for a product are C(g) = 6,000,000 + 250g + 0.0024* Ria) = 1.2500 — 0.0059 (a) Using the marginal approach, determine the profit-maximizing level of outpu (b) What is the maximum profit? 23 The total cost and total revenue functions for a product are Cla) = 40,000 + 28q + 0.0029" Fiqh = 76 ~ 0.0084 (a) Using the marginal approach, determine the profit-maximizing level af outpul (®) What is the maximum profit? 24 Portrayed in Fig. 8.10 is « total cost function C(g) and a total revenue function Ry Dincuns the economic significance of the four levels of output 4. ds, dy» ANA Figure 8.10 4 ow) Rig) 9398 CHAPTER 8 OPTIMIZATION: APPLICATIONS. 8.2 ADDITIONAL APPLICATIONS ‘The following examples are additional applications of optimization procedures. (Real Eatate) A large multinetional conglomerate is interested in purchasing some pritne boardwalk real estate at @ major ocean resort. The conglomerate is interested in acquiring @ rectangular lot which is located on the boardwalk. The only restriction ie that the lot have an ‘area of 100,000 square feet. Figure § 11 presents a sketch of the layout with x equaling the boardwalk frontage for the lot and y equaling the depth of the ot (both measured in feet) The seller of the property ia pricing the lota at $5,000 per foot of frontage along the boardwalk and $2,000 per foot of depth away from the boardwalk. The conglomerate is interested in determnining the dimensions of the lot which will minimize the total purchase 4= 300000 |y Ovo1n Pferta Pig, 811. Total purchage cost for a lot having dimensions of x feet by y C= 5,000: + 2, where C is cost in dollars ‘The problem is to determine the valuew of x and y which minimize C, However, C ia stated as a function of two variables, and we are unable, as yet, to handle functions which have two independent variables, Because the conglomerate has specified that the area of the lot must equal 100,000 square feet, a relationship which must exiat between x and y it 00,000 8.13) Given this relationship, we can solve for either variable in terms of the other, For instance, 100,000 8.14) We can substitute the right side of t y appears, oF equation into the cost function wherever the variableom fle 6,000: + 2,000 —* = 5,000: + 8.18) Equation ( 9.15) is « restatement of Eq. ( only in terms of one independent variable. We can now soak the value of x which minimizes the purchase cost C. The first derivative ix IEC’ in not equal to 0, or critical valuon occur at ‘The critical point at x= —200 is moaninglevs. To tent x ~ 200, C(x) = 400,000,000 Sapa x Since C(x) > 0 for x > 0, the graph of C is concaye upward for.x> 0, Thus, the minimum value of C occurs nt x = 206 ‘Total conta will ba minimized when the he Jot equals 200 feet, The depth of the Jot can be found by substituting x= 200 into E 4),0r 100,00¢ If the lot is 200 feot by 500 foot, votal coxt will be minimized at « valua ofCHAPTER 8 OPTIMIZATION: APPLICATIONS C= $5,000(200) + $2,000(500) = $2,000,000 (Emergency Response: Location Model) Example 13 in Chap. 6 discussed s problem in which three resort cities had agreed jointly to build and support an emergency response facility which would house rescue tricks and trained paramedics. ‘The key question dealt with the location of the facility. The criterion velected wax to choose the location s0 as to minimize S, the um of the product of the summer populations of each town and the square of the distance between the town and the facility, Figure 17.12 shows the relative locations of the three cities Figure 8.12 Wi Ow? Gy _———— ° ar oer) “The criterion function to be minimized was determined to be Sm [(2) = 45023 — 19,600 + 241,600 there x is the location of the facility relative to the zero point in Fig. 8.12. (You may want to reread Example 1:1 on page 234.) Given the criterion function, the first derivative ix £2) = 900 — 19,600 If/" is set equal to 0, ‘and a critical value occurs at Checking the nature of the critical point, we find [°(a) = 900 for x > 0 In particular, 1°(21.71) = 900 >0 ‘Thus, f is minimized when x= 21.77. The criterion S ia minimized at x™ 21.77, and the facility should be located as shown in Fig, 8.13.ADDITIONAL APPLICATIONS 404 (Equipment Replacement) A decision faced by many organizations is devermining the ‘optimal point in time to replace a major piece of equipment. Major pieces of equipment are often characterized by two coat components — capita! cost and operating cost. Capital coat in purchase cost lous any salvage value. Ifa machine costs $10,000 und is luter sold for 82,000, ‘the capital cost in $8,000. Operating cost includes conts of owning and maintaining a piece of equipment. Gasoline, oil, insurance, and repair conta associated with owning and operat ing a vehicle would be considered operating costs Some organizations focus on the average capital cost and average operating cost when, they determine when to replace a piece of equipment. These costs tend to trade off against one another. That is, as one cost increases, the other decreases, Average capital cost for a piece of equipment tends to decrease over time. For # new wutomobile which decreases in value from $12,000 to $9,000 in the firat year, the average capital cost for that year is $3,000 If the automobile decreases in value to $2,000 after 5 years, the average capital cont is Average operating cost lands to increase over time as equipment becomes less efficient and ‘more maintenance is required. For example, the average annual operating cont of a car tends to increase as the ages. A taxi company in a major city wants to determine how long it should keep ite cabs, ich cab comes fully equipped at n cost of $18,000. The company estimates average capital cost and average operating cost to be a function of z, the number of miles the car ia driven. ‘The nalvage value of the car, in dollars, ie expressed by the function 3(x) = 16,900 — 0.10% ‘This means that the car decreases $2,000 in value aa soon aa the cab is driven, and it decreases in value at the rate of $0.10 per mile. ‘The average operating cost, stated in dollars por mile, is estimated by the function O(x) = 0.000008 + 0.15 Determine the number of miles the car should be driven prior to replacement if the objective is to minimize the sum of average capital coat and average operating cost. SOLUTION Average capital cost per mile equals the purchase cost lees the salvage value, all divided by the number of miles driven, or ._ 18,000 ~ (16,000 ~ 0.103) _ 2000 + 0.10% Ct) 2,000 - +010 ‘The sum of avernge capital cost and average operating cost is402 CHAPTER 8 OPTIMIZATION: APPLICATIONS 16x) = Of) + Cla) = 0.000008 + 05+ 22 + o.10 = 0.0000008 + 0.25 + 222° 1"€2) = 0,0000008 ~ 2,000x-* ff’ in vet equal to 0, = 9,0000008 6,566,566,666.67 or critical values occur when = $81,009.68 Again, & negative value for x is meaningless. Checking the here For x > 0, "(2) > 0 (Le, the graph of fis concave upward for x > 0). ‘Thorefore fis minimized when x = 81,648.6, /(81.649.8) = 0,0000009(81,649.6) + 0.25 + -2.000_ 1.6496 — 0.02450 + 0.25 + 0.02450 0.298 Average capital and operating costa are minimized at a value of $0.299 per mile when a taxi in driven 81,649.6 miles. Total capital and operating coats will equal (cost/mile) - (number of miles), oF ($0.299)(81,649.6) = $24,413.23 Figure 17.14 illustrates the two component cost functiona and the total cost function. Notice that the average operating cost per mile (x) increases with incroaaing values of and that average capital cost per mile C(x) decreasea with increaaing values of x.ADNITIONAL, APPLICATIONS 403 ‘Au operating caile POINT FOR © Given what you understand THOUGHT & critique the assumptions used in this movel. W DISCUSSION _ tiens are not accounted for when using the re Jecisions of when to replace equipment, relevant factors or considera- model? {Bill Collection) Example 11 in Chap. 7 discussed the collection of accounts receivable for credit waned to people who use a major credit card. ‘The financial inatitution determined that the percentage of accounts receivable P (in dollars) collected ¢ months after the credit was iasved in ‘The average credit isaved in any one month is 4: ceatimates that for each $100 million in new credit iaaued in any month, collection efforts cost $1 million per month. That is, iferedit day, it costs $1 million for avery month the institution attempts to collect these accounts receivable, Determine the umbe: the net collections N (dollars collected tninus collection coats) 00 million. The Gnnancial inatitution $100 million is isaued of months that collection efforts should bu continued if the objective in to maximize SOLUTION Given that $100 mullion of credit is ianued, the amount of receivables collected (in millions of dollars) equalsCHAPTER 8 OPTIMIZATION: APPLICATIONS (Amount of credi¢ isaued)(perventage of accounts collected) (200)(0,98)(1 — ‘Therefore, net collections N are described by the function Net collections — amount collected — collection coats Nese = (100(0.95)(1 ~ e-") ~ (1) =95(1—e-8")—t = 95 — %5e-*"—¢ where t equals the number of months during which collection efforts are conducted. ‘The first derivative is £(t) = 66.5e-*" — 1 IE/' is eat equal to 0, B65e-O% = 1 8% = 0.01503 mee 4 = 0.0150 ‘Thus, ¢-®"* 0.01603 when and # critical value occurs at ‘The only critical point on f occurs when t = 6, Since /"(t) = —48.55e-*" < 0 for all t> 0, 1°(8) <0 and fis maximized at ¢ = 6. Maxisvum net collections are 116) = 96 — De*7 — 6 = 95 — 95(0.0160) 96 — 1.425 — 6 = 87.575 or $87.575 million. For each $100 million of credit iasued, net collections will be maximized at a value of $87.575 million if collection efforta continue for 6 months. QADDITIONAL APPLICATIONS 405 PRACTICE EXERCISE (a) Verify that the critical point at f= 6 is a relative maximum. (b) What is the total (gross) amount collected over the 6-month period? An sae: (b) $93.575 million (Welfare Management) A newly created stato welfare agency is attempting to determine the number of analysts to hire to process welfare applications. Efficiency experts estimate that the average cost C of processing an application ina function of the number of analysts z Specifically, the cost function is C= fla) = 0.001x 5 inx + 60 Determine the number of analysts who should be hired in order to minimize the average cost eeu. SOLUTION spt P(x) = 0.002s = ~oome-! If f’ in sot equal to 0, tO =~ F002 =a Eileen cach viae (The root x ~ —50 is meaningless.) ‘The value of /(x) at the critical point is 1450) = 0.001(60) — 5 In 50 + 60 = 0.001(2,500) — 5(9,912) + 60 = 2.5 ~ 19.58 + 80 = 842.94 To check the nature of the critical point,CHAPTER 8 OPTIMIZATION: APPLICATIONS © 0.001e4—Binx + 60 fe{s) = 0.002 + Bx™# =0.002+5>01orx>0 in particular (60) = 0.002 + — I lar, f*160) = 0.002 + = 0.002 = 0,002 + 0,002 = 0,004 > 0 0.002 + 5555 = 0.002 + 0.002 = 0.004 Therefore, fis minimized when x = 60, Average processing cost per application is minima at 8 value of $42.94 when 60 analysta are employed. Figur’ 8 15 presents a nketch of th average co function: (Compensation Planning) A producer of a perishable product offers a wage incentive (4 dirivern ofits tracks. A standard delivery takes an average of 20 hours. Drivers are paid at th rate of $10 per hour up to a masimum of 20 hours. Ifthe trip requires more than 20 hours, the drivera receive compensation for only 20 hours, There is an incentive for drivers to mak the trip in less (but not ton much lees!) than 20 hours, For each hour under ! wage increases by 31 (a) Determine the function w = (x) where w equals the hourly wage in dollars and x equaia the number of hours required to complete the trip (b} What trip time « will comxisize the driver's salary for a trip? ) What is the hourly wage associated with thia trip time? (@) What ia the maxicoum salary? (e) How does ry compare with that received for « 20-hour trip? ), the hourly SOLUTION (a) The hourly wage function munt be stated in two parts $10 + 81 X (no. of hours trip time ia loss than 20) Hourly wage = (when trip time in lens than 20 hours) 10 (when Lip time is 20 hours or mare)ADDITIONAL APPLICATIONS Given the variable definitions for x and w, this function can be restated ux we pig)= [104 20-2) Os 2<20 (3.16a) 3) 119 x22 (8,165) (b) Acdriver's salary 5 for a trip will equal $10/hour X 20 hours = $200 iftrip time is greater than or equal to 20 hours. If the trip time ia levs than 20 hours, Smale) ux = [10 + 1(20 ~ x)}x = (30 -2)x = 30r— 27 (8.17) ‘We need to compare the $200 salary for x = 20 with the highest salary for a trip time of Jean than 20 hours ‘To examine g for a relative maximum, we find the derivative 4/2) = 30-25 Setting 4’ equal to 0, aid a critical value occurs whon as To check the behavior of g(x) when x= 15, x(a) = —2 for 0525 20 and a"(16) = —2<0 ‘Therefore, a maximum value of g occurs when x= 16, or when a trip takes 16 hours (¢) The hourly wage associated with « 16-hour trip ia w= 10+1(20~16) 10+ 5— 915 {d) The driver salary associated with # 16-hour trip is found by evaluating 4(16). If we wubstitute x= 15 into Eq ( 8.17), S = 30(15) — 15 = 450 — 226 = $225 We also could have arrived at this answer by multiplying the hourly wage of $15 times the trip time of 15 hours.8 CHAPTE DPTIMIZATION: APPLICATIONS (e) The $225 salary for « 16-hour trip ia $26 more than the salary for # trip time of 20 hours (Pipeline Construction; Motivating Scesario) A major oil company ia planning to construct a pipeline to deliver crude oil from a major well nite to a point where the crade will be loaded on tankers and shipped to refineries, Figure 6 .16 illuotrates the relative locations of the well site A and the destination point C. Points A and C are on opposite rides of a denne tely 25 miles wide. Point Cis also 100 miles south of A. The oi company is proposing a pipeline which will run south along the east side of the forest, and at some point 2 will cross through the forest to point C. Construction costa are $100,000 per le along the edge of the forest and $200,000 per mile for the section crossing through the forest, Determine the crossing point x which will reeult in minimum construction coata for thé pipeline. EXAM foreat which is approxin Figure 8.16 rs smten Propoime pie ine SOLUTION Construcyjan coats will be computed according to the formula Cont = $100,004/mile (miles of pipeline along the edge of the forest) +$200,000/mile (miles of pipeline crossing through the forent) ' 8-18) ‘The distance from point A to the crossing point x is (100 ~ +) miles PYTHAGORRAN THEOREM Given a right tnangle with base «. height 4. and hypotenuse ¢, tie ahh) or are See Fig. 8.17. Using the Pythagorean theorem, the length of the section of the pipeline from C to xis Vers eH409 Fins 8.17 P Using Eg | 18), the total cost of construction of the pipeline, C (stated in thousands of dotlars), i 100(100 — x) + 20027 > ( 10,000 ~ 100 + 20039 + 6: = 10,000 — 1005 + 200s" + 6 (8.19) ‘To examine / for any relative minima, we find the derivative F(x) = —100 + 200())(x* + 625)""7(2e) 100 + 200s(e# + 62614 100 + — 200s Setting /* equal to 2ro, 200 1004+ — 0 200 Bi ion, oe ax VF OR ( 8.20) It-we square both 6 8 20), tx 208.33 and a (relevant) critical value occurs when x = 14,43. (A negative root is meaningless.)410 CHAPTER 8 OPTIMIZATION: APPLICATIONS PRACTICE EXERC! Use the second-derivative test to verify that ¢ has a minimum when x= 14.43, The practice exercine above should verif relative minimum occurs when x == 14.43, or the pipeline should cross through the foreat aftar the pipeline has come south 85.57 miles. Total construction costs (in thousands of dollars) can be calculated by aubatituting x= 14.43 into Eq. (8.19), or C= 10,000 — 100(14.43) + 200 = 10,000 = 1,443 + 200 835. = 10,000 — 1,443 + 200(28.86) 10,000.— 1,443 + 5,772 = 14,329 ($1,0008) = $14,329,000 TAA) F625 Q POINT FOR What alternative procedure could be used to compute the minimum total cast THOUGHT & — figure of $14,329,000? DISCUSSION ‘The following example, although not an optimization application, is one of particular importance in economics. {Elasticity of Demand) An important concept in economics and price theory is the price - elasticity of demand, or more simply, the elaaticity of demand. Given the demand function for a product q = /(p) and a particular ploint (p, q) on the function, the elasticity of demanf is the ratio Percentage change in quantity damanded Percentage change in price (8.21) This ratio is a mendure of the relative response of demand to changes in price. Equation (17.21) can be expressed symbolically mx Aa a 8.22) e (8.22) P ‘The point elasticity of demand is the timitot Bq ( 8 22) as 4p —* 0. Using the Greek letter 7 (eta) to denote the point clasticity of demand ot « point (p, @)ADDITIONAL APPLICATIONS. Fo (8.28) Given the demand fumetion g~ /(p) = 500~25p, let’ calculate the point elasticity of demand at prices of (a) $15, (b) $10, and (e) $5. For p= $15 16 ~ Boo = 25115) 5 125 ~ ‘The interpretation of y= —3 is that at aprice of $15, an increase in price by 1 percent would result in a decrease in the quantity demanded of approximately 3 percent. The percent change in demand is estimated to be three Limes the percent change in price. For p= $10 10 io ° es atsT 500 — 25010) =250 ~ 350 Thei erpretation of y= ~ Lin that ‘price of $10, an increase in price by 1 peroant would result in x decrease in quantity demanded of approximately | percent. The percent change in demand is estimated to be the same aa the percent change in price.412 CHAPTER 8 OPTIMIZATION: APPLICATIONS For p= $5 ‘The interpretation of 7 = — 1/3isa that at aprice of $5, an increase in price by | percent would result in « decreute in quantity demanded of approximately 0.33 percent. The patcent change in demand is estimated to be lene than the percent change in price, Q Economists classify point elasticity valuea into three categories. Q Case Z (\nj> 1): The percentage change in demand is greater than the percentage change in price (e.g, a 1 percent change in price results in greater than 1 perces.t change in demand). In these regions of ademand ) function, demand is said to be elastic. D Cave 2 {Ini < 1): The percentage change in demand is less than the per: centage change in price, In these regions of the demand function, demand in anid to be inetastic. Q Cave 8 (inj = 1): The percentage change in demand equals the percent age change in price. In these regions of the demand function, demand ia anid to be unit elastic. Section 8.2 Follow-up Exercises 1 A person wishew to fence in « rectangular garden which is to have an area of 1,600 square fect. Determine the dimensions which will create the desired area but will require the minimum length of fencing. 2 An owner of s ranch wishes to build a rectangular riding corral baving an area of 6,900 fequare meters Ifthe corral appears as in Fig: 8 18, determine the dimensions x and y which Will require the minimum length of fencing. (Hint: Set up @ function for the total length of fencing required, stated in terma of x and y. Then, remembering that xy 5,000, restate the length function in terme of either x or y.) 3 A small bench club has been given 900 meters of flotation barrier to enclose a swimming ares. The desire is to crente the Inrgest rectangular awim area given the 300 meters of flotation barrier. Figure 8 19 illustrates the proposed layout. Note that the tlotationADDITIONAL APPLICATIONS 43 barrier is required on only three aides of the ewimming area. Determine the dimenaians = end y whieh result in the largest swim area. What in the maxirour area? (Hint: Re member that x + 2y = 300.) 4 An automobile distributor wishes to create a parking area neer a major U.S. port for new cars from Japan, The parking area is to have a total area of 1,000,000 square meters and will have dimensions ns indicated in Fig. §.20. Because of security concerns, the section of fence acrosa the front of the lot will be more heavy-duty and taller than the fence used along the vides and rear of the lot. The coat of fence for the front ia $20 per runing meter and that sed for the other three sides costa $12 per running mete) Datermine thedimensions rand y which result in xminimnus total cont of fence, What in the minimum cost? (Hint. cy = 1,000,000.) Pigure 8.20 ia | 5 Correc fenved within a the total area in half, [£3,600 feet of fence are available, determine the dimensions x end jons Management Figure 8 21 jlluateaten « recreation yard which ia to be won. In addition to encloving tbe area, section of fence sbould divide y which result in the mazimum enclosed a: 2x + By = 3,600.) area? (Hint Figure 8 2 SS 2 8 Warehouse Location A manufacturer. wishes todocate a warehouse between three cities, The relative locations of the cities are shown in Fig. 6.22. The objective is to locate the warehouse ho as Lo tninimize Lhe sum of the squares of the distances aeparat ing each city and the warehouse, How far from the reference point should the warehouse be located? Figure 8 22 A ®414 CHAPTER 8 OPTIMIZATION: APPLICATIONS. 7 Hoalth Maintenance Organization Figure 21 illustraten the relative loeationa of three cition. A largo health maintenance organitation (HMO) wishes to build a natallite clinic to service the three cities. The location of the clinic x ahould be such that the wu of the squares of the distances between the clinic and each city is minimized. This criterion can be stated as Minimice S53 (3-27 fi ‘where x, in the location of city j and x is the location of the clinic. Determine the location which minimizes S. *8 HMO, continued In Exercise 7, suppose that cities 1,2, and 3 have 10,000, 6,000, and 3,000 pervons, respectively, who are members of the HMO, Assume that the HMO has ‘established ita location criterion aa the minimization of 3 separating city j EN incity and the clinie i (ime srshenir) (Se of “') or SB nlz 2? where n, equala the number of members residing in city j. Determine the location x which minimizes S, ® A police department purchases new patrol cars for $26,000. The department estimates average capital cost and average operating cost to be a function of x, the number of miles the car is driven, The salvage value of « patrol car (in dollars) is expressed by the function ‘S(x) = 22,500 — 0,16: Average oparating cost, atated in dollars par mile, is estimated by the function Ot) = 0.000006: + 0.20 (a) Determine how many miles the car should be driven prior to replacement if the objective ia to minimize the sum of average capital cost and average ope per mile (0) What ia the minimum coat per mile? {c) What ia the salvage value expected to equal? 10 Commeretal Aircraft Replacement A major airline purchases a particular type of plane at a coat of $40,000,000. The company estimates thet average capital cost and average operating cost are a function af x, the number of hours of fight time. The salvage value of a plane (in dollars) is expressed by the functionADDITIONAL APPLICATION (x) = 28,000,000 — 10,0002 Average operating cost, stated In dollars per hour of fight time, is estimated by the function O{x) = 500 + 0.402 (@) Determine how snany houte a plane should be Bown before replacement if the objective is to minimi: uso of average capital and average operating cost per hour. (0) What is the minimum cost per hour’? (c) Whnt is the walvage value expected to equal? 11 A university aki club is organizing & weekend trip tom aki lodge. The price for the trip is {$100 if 50 or fewer persons sign up for the trip, Por every traveler in excess of 50, the price for all will decrease by $). For instance, if 5} persons sign up, each will pay $99, Let x equal the nomber of travelers in excesa of 60. (a) Determine the function which staten (®) In part o, is th (6) Formulate the fu A(x), which states total revenue R aa a function of x (4) What value of « results in the maximum value of R? (@), How many persona should sign up for the trip (f) What is the maxiroum value of R? (a) What price per ticket n (h) Could the club geners 12 A national charity is planning w fund-raisiny having = population of 2 million, The porventaye of the population whe will mal donation is extimated hy the function ice per person p ea a function of x ‘ction on the tw in the maximum reven 0 or min a major United States city whore R equale the per campaign is conducted. Pa 2 per donor C many days abould the campaign be tage of the population and x equals the number of days the experience indicates that the average contribution in this n are estimated at $10,000 per day wucted if the objective is to ranximize the campal net proceeds (total contributions minus total costa) from the campaign? (@) Whot ore maximum net proceeds expected to equal? What percentage of the popu ation in expected to donate 18. A national distribution company pelle CD's by mail only. Advertising ia done on local TV stations. A promotion pr a being p 4 fora major metropolitan area for e jurn recording, The Larget audiences — thoas who might be interested in ¥),000. Past experience indicates that for this ng the percentage of the target market R actually purchasing vertiving campaign f, atated in days. Speci this type of rec: city and thie type of rec #CD ik function of the length of th cally, thiw sales reponse ding — is entimated at The profit m and a variabl in on each CD js 1.60, Advertising costa include a fixed cost of $16,000416 CHAPTER 8 OPTIMIZATION. APPLICATIONS {@) Determine how long the campaign should be conducted ifthe goal profit (gross profit minus advertising coma) (6) What is the expected muxitsuzn net profit? (c) What percentage of the target market is expected to purchase the CD? 14 Assume in Example 14 that the average umount of credit issued each month in $50 million and monthly collection costa equal $0.6 million. Re-solve the problem, 16 A police department has determined that the average daily crime rate in the city de pends upon the number of officers assigned to each shift, Specifically, the function deseribing this relationship is N= f(s) = 500 ~ 10cm | where Nequals the average daily crime ate und x equals the number of oficers unsigned to ench shift. Determine the number of officers which will result in minimum average daily crime rate, What ia the minimum avernge daily crime rate? 16 A firm's annual profit is stated os a function of the number of salespersons employed. ‘The profit function i P= 20(s1¢ where P equals profit stated in thousands of dollars and x equals the number of nalew persons, (a) Determine the number of salespersqns which will maximize annual profit {b) What io the maxiraur profit expected to equal? 17 A company is hiring persons to work in its plant. For the job the persona will perform femictency experts estimmve that the average cost C of performing the task is w function of the number of persins hired s, Specifically, C= {i = 0.003¢? = 0.218 Ine +5 (a) Determine the number of persona who should be hired to minimize the average cont (©) What is the minimum average cost? 18 A company is hiring people to work in its plant. For the job the people will perform, efficiency experts estimate that the average cowt C of performing the task is « function of the number of people hired x. Specifically, C= f(s) = 0005s" — 0.49 In 245 (0) Determine the number of people who should be hired to minimise the average coat (®) What is the minimum avernge cost” 19 Wage Incentive Plan A manufacturer offers a wage incentive to persons who work on one particular product. The standard time to complete one unit of the product is 15 hours. Laborers are paid at the rate of $6 per hour up to a maximum of 15 hours for each unit they work on (if laborer takes 20 hours to complete aunit, he or shen only paid for the 15 hourn the unit should have taken). There is an incenitive for laboreru vo complete n unit in less than 45 hours, For each hour under 16 the hourly wage increases by $1.50. Let x equal the number of hours required to complete a unit (o) Determine che function w= f(x) where w equals the hourly wage in dollars (8) What length of times will maximize « laborer's total wages for completing one unit? (c) What is the hourly wage associates with this time per unit 2?ADDITIONAL EXERCISES a (4) What is maximum wage per unit? (e) How does this aalary compare with the wages oarned hy taking 15 or more hours per unit? 20 Pipeline Construction A major oil com deliver crude oil fram # mejor well site to « point where the crude will he loaded on tankers and shipped to refineries. Figure 8 24 illustrates the relative locations of the well nite A and the destination point C. Pointa A and € are on opposite a which ia 20 miles wide, Point C in also 200 miles south of A along the Inke The oil company is proposing a pipeline which will run south along the eaat side of the lake and at some point, x, will cross the lake to point C. Construction costa are {$50,000 per mile along the bank of the lake and $100,000 per mile for the eection crossing the lake, Determine the crossing point x which leada to minimum construction coats What in the minimum construction cont” ws planning to construct # pipéline to of a lake —— - — les ia | 1 4 — “e- —____. — + In the following exercisas, (o) determine the general expression for the point elasticity of Gemand, (b) determine the elnsticity of demand at the indicated prices, also classifying demand ax elantic, inelastic, or unit eluatic, and (c) interpret Ube meaning of the elasticity values found in part (6). 21 q™{(p) = 1200 ~ 60p, p= $5, p= $10, and p= 22 q~f(p) ~ 180 — 2.5p, p = $15, p = $30, and p = $45 23 q = f\p) = 12000 — 10p%, p = $10, p = $20, and p = $30 24 q=f(p) = 900 ~p*, p= 35, p= $15, and p = $25 25 Given the demand fonction q = f(p) = 900 ~ op, determine the price at which the demand in unit elantic 26 Given the demand function q=/(p) = 80 — L.6p, determine the price at which the demand in unit elaatic ADDITIONAL EXERCISES 1 A firm sells each unit of e product for $250. The coat function which describes the total cost Cay a function of the number of unita produced and sold x is, C(x) = 50x + 0.12! + 150 (@) Formulate the profit function P = f(x)418 CHAPTER 8. OPTIMIZATION: APPLICATIONS (b) How many wnita should be produced and sold in order to maximize total profit? (c) What i total revenue at this level of output? (2) What is total cont at this level af output? 2 Re-solve Exercise 1 using the marginal approach. A local travel agent is organizing a chartar flight to a well-known resort. The agent bas «quoted 1 price of $300 per person if 100 or fewer sign up for the fight. For every person aver the 100, the price for all will decrease by $2.50. For instance, if 101 people sign up, each will pay $297.50, Let x equal the number of parsons above 100. (a) Determine the function which stntos price per pervon p as a function of x, oF p= fis). (b) In part a, ia there any reatrietion on the domain? {e) Formulate the function R= A{3), which states total ticket revenue R as a function of x (d) What value of + results in the maximum value of R? {e) Whot is the maximum value of R? (/) What price per ticket resulta in the meximum R? ‘The total cost of producing q unite of a certain product is described by the function 12,500,000 + 1009 + 0.0247 (a) Determine how many units g should be produced in order to minimize the average cost per unit (8) What is the minimum average cost per unit? (c) What is the total cost of production at this level of output? 5 A principle of economics atatea that the average cost per unit is minimized whan the marginal cost equals the average cost, Show that this is true for the cost function in Exorcie 4 41 The quadratic total cont fimetion for a product is Cm ar’ +brte where x equals the number of units produced and sold and C is stated in dollars. The product nella at m price of p dollars per unit (a) Construct the profit function stated in terms of x (o) What value of z resulta in maximum profit? (c) What restriction aasucea that « relative maximum occure at this value of x? (d) What restrictions on a, b,c, and p assure that x > 0? 7 Ancoil field currently has 10 wells, each producing 300 barreln of oil per day, For each new well drilled, it is estimated that the yield per well will decrease by 10 barrels por day, Determine the number of new wells to drill in order to maximize total daily output for the oil field. What is the maximum output? 8 A. small warehouas in to be constructed which isto have a total aree of 10,000 square foot. The building into be partitioned as shown in Fig. 8 25. Coste have been entimnted based on exterior and interior wall dimensions, The costs are $200 per running foot of exterior wall plus $100 per running foot of interior wall (0) Determine the dimensions which wil minimize the construction coste (0) What are the minizourn conta?ADDITIONAL EXERCISES 49 ee a ee © An open rectangular box is to be constructed by exiting square comers from a 40 X 40: inch piece of canthoard and folding up the fiape as shown in Fig 8.2 (a) Determine the vrluve of x which will yield the box of mas (8) What in the maximum volume? qu volume, Figur, 8.26 10 The demand function for a product ia @ = f(p) = 25,0000- 9 bere ¢ is quantity demanded (in unita) and p is the prica (in dollars) (a) Determine the value of p which will result in maximum total revenue. (®) What in the maximum total reverse? 11 A marketing research organization believes that if a corpazy spends « million dollars on TV advertising, total profit can be estimated by the function Pom f(x) = 40x70 where P in measured in millions of dollars (a) How much should be spent on TV advertising in order to maximize total profit? (b) What ia the maximum profit? 12 Memory Retention An experiment was coryduxcted to determine the effecta of elapsed time on a person's memory. Subjecta were anked to look at a picture which contained many different objecta. After atudying the picture, they were asked to recall as many of the objects us they could, At different time intervals following this, they would be maked$20. 4 +15 Pe WPTIMIZATION: APPLICATIONS 0 recall as many objects as they could. Based on the experiment, the following function wan developed: Re j{t) = 84—25 In i For this function 2 represents the average percent recall ax a function of time since studying the picture (menaured jn hours). A value of R = 50 would indicate that at the rreaponding tire f the average rocell for the atucly group was 60 percent (0) What is the average percent recall after 1h (b) After 10 hours (€) Find the expression for the rate of change in R with respect to time (d) What is the maximum percent recall? Minimum? A new atste weltare agency wants to determine haw many analysts to hire for proces. ications, Tt ia estimated that the avernge cost C of processing ar application ia 8 function cally, the ¢ of wellare app Specificall at Function i (a) If the of nuinber f analysts who show! (6) Whatis the airauws averoge cost of processing an application expected to equal A firm han estimated that the average production coat per unit © fluctuates with the umber of unita produced x. The average cowt function is &= 0.0022" — 1,000 In x + 7,500 where © js wtated er unit and x in wtated in hundreds of uy number of duction cast p nita whic mit? unm average Coal expected to equal? What are total production costa expected to equal? i be produced in order to minimize the average (o) What is the ( Pontal Service The U.S, Postal Service requires that y specified dimensio inches. fa) Find the dimensions (r ond) af the package. (Hint: See Fie (0) What ie the girth of the package (e) What in th pont packages.coi ngth plus girth must be in. Specifically, th no greater than & acyl 8 drial package which ov ? and remember that VWhat ia th22 (APTEN 8 OPTIMIZATION: APPLICATIONS hy hot axe che optimal dimenuione uf ube lot? Woot in the iainimum lot aren? 18 Deters ne two ouznbers x and y whose sum ia 50 and whow product is aa large as What ié the masimum product of the two numbern? 28 Deter.>:no two pouitive numbers whose product equals 40 and whose suum ih as stall as sible. Woet fs the minimum suns? #20 Binthion A woman is going to participate in» rua-ewim biathlon, The courve in of variable langth and is shown in Fig. § 30, Each participant must zun (or walk) from the starting poiut along the river."They must cross the river by swimming; however, the crossing point x may be chosen by each participant. This contestant eatimaton that she will average 6 miles per hour for the running portios of the biathlon an? 1 mile per hour for the swimming portion. She wishes to minimise her time in the event (a) Lesernine the eroaing point x which will ronalt in the minimum tune (6) What ia the minimum time expected to equal {c) How does thia tine compare with that if she :bose torun the full point in at x= 0)? Hint: Time = distance + mpeed or hours = tailed + milea/nour; amt ~ bo + bar “21 Sold Waste Management A local city is planning to construct u nolidl waate treat- ment favility, One of the major somponeuta of the plant isa nolid waste agitation pool This pool ix to be circulur on ahape aud is supposed to have a voluine capacity of 2 million cubic feet. Municipal exyineess have estimaicd construction couts as a func tion of rhe nurface area of the base ad wall of the pool, Coustruction costa are est alee at $80 per squury foot for te base of the poo! and $20 yer mare fact of wall surface, Figur 8 31 presents a sketch of the pool. Note that r equals tbe radius of the pool in feet and h equals the depth of the pool in fost. Deter uiivie the dimenions r and / which provide a capacity of ? million cubic feet at. a wininusa cost (Hint The aren A of a circle baving radiuar in A = xr¥, the wurtace arew A ofa circular der buving radiva r tod height h in A= 2xeh, and the volume V is V= nr’h.) lowing exercises, (a) deternnine the general expression for the point elasticity of demand, (6) determine the elasticity of demand et the indicated prices, also classifying domand a4 elastic, inelastic, or unit elastic, and (c) interpret the meaning of the elnsticity values found in part 6.CHAPTER TEST 423 Figure 8,31 22 q=/(p) = 2,500 — 80p, p = $6, p+ $16, p= $24 23 q=/(p) = 816 — p~ 0.05p%, p ~ $50, p = $70, and p= $100 24 Given the demand function in Exercine 22, devermine the price(s) at which demand in (a) elastic, (b) inelastic, and (c) unit elas 25 Given the demand function q = f(p) = 2400 ~ 40p. determine the price(s) at which demand is (a) elastic, (b) inelastic, and (c) unit elastic. O CHAPTER Test 1. The demand function for » product is @= Mp) = 60,000 - ip where g equals the quantity demanded and p equi late the total revenue function R = 9(p) he price in dollars. Formu- 2 The total revenue function for a product is R= fx) = — 4x94 30x where Ris meneured in hundreds uf dollars and x equals the number of unite sold {in 1008). The total cost of producing x (hundred) units is described by the function C= g(x) = 2? — 150s + 5,000 where C is measured in hundreds of dollars, (a) Formulate the profit function P= h (2) {b) How many units should be produced and sold in order to maximize total profit? (e) What ie the ras um profit? 3 An importer wants to fence in a storage area near the local shipping docks. The ‘area will be used for temporary storage of ehipping wntainers, The area ia to be rectangular with an ares of 100,000 square feet. The fence will cost $20 per running foot.HAPTER 8 OPTIMIZATION APPLICATY (a) € termine the dimensions of the area which will result in fencing costs being zed 16) What in the miniroum coat? 4A retailer ha determined that the annual cost ( of purchasing, owning, and maintaining one of ita products behaves according to the function 20,000 + 0.59 + 80,000 he size (in units) of each order purchased from suppliers 0) Whot order quantity g results in minimurn annual east” 'b) What is the minimum annual cost? 5 The demand function for a product is q=/(ph = 35,000e~9e%" where q s quantity demanded (in units) and p is the price (in dollars), (a) Determine the value of p which will result in maximum total revenue (4) What js the maximum total revenue? 6 A company is hiring persons 9 work in its plant, For the job the persons will perform, efficiency eaperts estimate that the average cost ( of performing the 1 of the number of persons hired x. Specifically tank ix a functis C= fx) = 0,005x? — 0.49 In x +5 (a) Determine the number of persons who should be hired to minimize the average ost, (6) What is the minimum average cost? THE E0Q MODEL | The economic order quantity (EOQ) model is a classic inventory model, The pur pose af the EOQ model is to find the quantity of an item to order which minimizes total inventory coats, The model agaurnes three different cost components: ordering vst. carrving cost, and purchase cost. Ordering costs are those xasoc ated with plac ng and teceiving an order, These costs are largely for salaries of persons in requisitioning goods, processing paperwork, receiving goods, and placing inventory, Ordering costs are assiimed to be incurred each time an order 1s placed. Carrying costs, sometimes referred to an holding custs, are the costs of owning and maintaining inventory. Carrying costs inclide such ‘age space, inqurance, salaries of inventory control personnel, obsolescence, and ‘opportunity costa assccinted with having the investment in inventory. Carrying costs are often exprensed as « percentage of the average value of inve (e.g. 25 percent per year), Purchase cont Although there are variations on the ponentaas cost ol stor ory on hand simply the cont of the inventory OQ model, the basic model makwing assumptions: (1) demand for items ie known and iat a constan) (or near constant) rate, (2) time between placing and receiving an order (/ead rime) is known with certainty, (3) order quantities are always the same size, and (4) inventory replenishment is instantaneous (ie., the entire order is received in une batch) If we ansume 4 time frame of | year, total inventory costs are TC = annual ordering cost + annual carrying cost + annual purchase cost = (seer Se) (costae) average + (maton) wate \(_caroiae_) > ) per unit) (cost in percent ia unite + (ant) (mocha pice) demand, per unit D = onnwal demand, units = ordering cost per order C, = carrying cost (stated as a percentage of average valle of inventory on hand) p= purchase price per unit q™ order quantity annual inventory conts can be expressed an a finction of the order quantity g follows: Tema) = 2.4 20, +d Requirements 1 For « given inventory item, D = 5,000, C, = $125. p= $100, and C, = 0.20, De _ termine the value of g which minimizes total annual mventory costs. What are minimum annual inventory costs? How many orders must be placed each year? What are annual ordering costa? Annual carrying costa? 2 The order quantity which minimizes annua) inventory costs is termed the “eco nomic order quantity” or EOQ. Using the generalized cost function, determine the general expression for the order quantity g which minimizes annual inventory cost. (Hint: Find the derivative with reepect to 9, assuming thst D, C,..p,and Cy are constant.) $ Prove that the critical value for q does reeult in a relative minimum on the cost funetion. 4. Using the expression for ¢ found in part 2, show that annual ordering cont equals annual carrying cost when operating at the OQ level 5 Annual inventory costa can be expressed in terms of the number of orders placed per year, N, recognizing that N= D/q. Rewrite the generalized cont function in terms of N rather than q. Determine the general expreasion for the value of N which minimizes annual inventory costs. Confirm that the critic] value for does result in the minimum value of the cost function,