Lecture for Week 12 (Secs. 5.5 and 5.7)Optimization Probl emsand Antiderivatives1We are concern ed this week with finding themaximum and minimum values of a function inpractical problems.Extremum is the generic term for either typ eof ext reme value. Optimum is the generic term forwhichever type is considered desirable. All thesewords form pl urals by changing “- um ” t o “-a”.2The main principle: Maximum and mini -mum values of a funct ion f occur at places (sayx = c) where f′(x) = 0 (i.e., t he t angent line ishorizontal).......................................................................................................................................................................................................................................................................................................cxf(x)3But t here are Complications1. A local e xtremu m of f may occu r at c e venif t he t angent line does not become hori-zontal there. This happens if f′(c) doesn’texist...........................................................................................................................................................................................................................................................................................................................................................c42. If c is an endpoint of the interval I, th enf(c) may be the absolut e extremum of f onI even if f′(c) exists an d f′(c) 6= 0..............................................................................................................................................................................................................................cI←− −→5Remark: According to what I consider thestandard terminology, an absolu t e maximum isautomatically a l ocal maximum. However, in Ex-ercise 12(a) (which I assigned last week) Stewartseems to be reserving the term “local maximum”for an interior maximum, occurring where f′(c)is either 0 or und efined.By the way, “relative” and “global” can beused instead of “local” and “absolute”.63. f′(c) may be 0 even if f(c) is not an ex-tremu m (even a local one) . This happensfor f(x) = x3, c = 0, for example..................................................................................................................................................................................................................................................................................................................................................7Procedure for finding the ab s ol ute extremaof a function on an interval1. Find all critical numbers in the interval ( i.e.,places where f′(c) = 0 or f′(c) is u nde-fined).2. Calculate f(c ) at the critical numbers andat the e ndpoints of the interval.3. Compare the results; pick out the largestand s mall est.8ExampleYour dream house wil l be built on a rectangul arlot. Along the side facing the s t re et you will havea stone wall that will cost $10 per foot (mea-sured horizontally along the st reet). The otherthree sides will be enclosed by a s t eel fence cost-ing $5 per foot. You have $2500 to s pend on walland fence toget her. Find the dimension s of thelot with the maximum area consi s t ent with yourplans.9I will work ou t thi s example, the n go backand point out the general principles and strate-gies that i t ill ustrates.Let L be the length of the lot (the dimen-sion parallel to the stre et) and W be the widt h(from front to back). The area i s A = LW . Thecost of th e fence is2500 = C = 10L + 5(2W + L) = 15L + 10W.Let’s sol ve th is eq uation for W :10W =2500 − 15L10= 250 −32L.ThenA = LW = 250L −32L2.Now fin d critical numbers:0 = A′(L) = 250 − 3L ⇒ L =2503.We shoul d also consider “the endpoints of theinterval” — but what is the interval? Well, it11woul d make no sense for L to be negati ve, so t heinterval should start at L = 0. Also, if L is toobig, then the constraint (cost ) equation will forceW to be negati ve. So the interval ends at th evalue of L that makes W = 0: L = 2500/15 =500/3. At t hese two end points, A = 0, where asin t he interior A is obviously positive . There-fore, the endpoints are absolu t e minima of A,and t he absolute maximum must occ ur at thecritical point, L = 250/3. Calculate the otherquantitie s :12W = 250−322503= 125, A =2503125 =312503.Alternative argument that the critical pointis the min i mum: A′′(L) = −3 < 0 everywhere,hence i n particular at the critical poi nt. So thefunct i on is concave downward there , and thatextremum mu s t be a maximum. (So there wasno real need to stud y the endpoints in detail.)13Strategy for optimization problems1. Read the problem carefully. Understand –• What quantity is t o be extre mized?(Let’s call it Q.)• What other q uantities can vary? W hatquantitie s are fixed?2. Introduce notation. Draw a diagram ifappropriate.143. Write down the relations among the vari-ables. (They may be given in the problem, ordeducible from general knowledge.) You need1) an objective function exp re s s ing Q interms of other variables;2) constraint equations re l at ing thoseother variable s so that you can write Qas a fu nction of just one independentvariable (let’s call it x).154. Solve the const raint s and substitu t e the re -sults into t he objective fun ction.• Don’t diffe rentiate the c onstraints.• Don’t diffe rentiate Q unt il you haveeliminate d all variables but x.5. Different iate Q(x) to find its critical points.166. Verify that your favorite critical point is thecorrect extremum.• Is it a max or a min? or neither?• Is it in the physi cally allowed interval?• Remember to ch eck endpoints as possi-ble c andidates.17ExerciseRedo the dream house example, but this timesuppose that the bu dget is flexible and you wanta lot of exac t ly 10,000 ft2. Find the dimens ionsof the lot with the minimal wall-fence cost (andfind th at cost) .18We still have the are a and c ost formulasA = LW, C = 15L + 10W.But now the roles of objective and constraint for-mul as are interchanged. The constraint (solved)is W = 10000/L, so the objective isC(L) = 15L +100000L.(In either problem I could have eliminated Lin favor of W instead of the reverse. That would19change the intermediate algebra but not the an-swers.)0 = C′(L) = 15 −100000L2.L =q10000 015= 100q23.W = 100q32.C = 1500q23+ 1000q32.20And now someth ing c ompl etely
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