Lecture for Week 12 Secs 5 5 and 5 7 Optimization Problems and Antiderivatives 1 We are concerned this week with finding the maximum and minimum values of a function in practical problems Extremum is the generic term for either type of extreme value Optimum is the generic term for whichever type is considered desirable All these words form plurals by changing um to a 2 The main principle Maximum and minimum values of a function f occur at places say x c where f x 0 i e the tangent line is horizontal f x c 3 x But there are Complications 1 A local extremum of f may occur at c even if the tangent line does not become horizontal there This happens if f c doesn t exist c 4 2 If c is an endpoint of the interval I then f c may be the absolute extremum of f on I even if f c exists and f c 6 0 c I 5 Remark According to what I consider the standard terminology an absolute maximum is automatically a local maximum However in Exercise 12 a which I assigned last week Stewart seems to be reserving the term local maximum for an interior maximum occurring where f c is either 0 or undefined By the way relative and global can be used instead of local and absolute 6 3 f c may be 0 even if f c is not an extremum even a local one This happens for f x x3 c 0 for example 7 Procedure for finding the absolute extrema of a function on an interval 1 Find all critical numbers in the interval i e places where f c 0 or f c is undefined 2 Calculate f c at the critical numbers and at the endpoints of the interval 3 Compare the results pick out the largest and smallest 8 Example Your dream house will be built on a rectangular lot Along the side facing the street you will have a stone wall that will cost 10 per foot measured horizontally along the street The other three sides will be enclosed by a steel fence costing 5 per foot You have 2500 to spend on wall and fence together Find the dimensions of the lot with the maximum area consistent with your plans 9 I will work out this example then go back and point out the general principles and strategies that it illustrates Let L be the length of the lot the dimension parallel to the street and W be the width from front to back The area is A LW The cost of the fence is 2500 C 10L 5 2W L 15L 10W Let s solve this equation for W 10 2500 15L 3 W 250 L 10 2 Then 3 2 A LW 250L L 2 Now find critical numbers 250 0 A L 250 3L L 3 We should also consider the endpoints of the interval but what is the interval Well it 11 would make no sense for L to be negative so the interval should start at L 0 Also if L is too big then the constraint cost equation will force W to be negative So the interval ends at the value of L that makes W 0 L 2500 15 500 3 At these two endpoints A 0 whereas in the interior A is obviously positive Therefore the endpoints are absolute minima of A and the absolute maximum must occur at the critical point L 250 3 Calculate the other quantities 12 W 250 3 250 125 2 3 A 250 31250 125 3 3 Alternative argument that the critical point is the minimum A L 3 0 everywhere hence in particular at the critical point So the function is concave downward there and that extremum must be a maximum So there was no real need to study the endpoints in detail 13 Strategy for optimization problems 1 Read the problem carefully Understand What quantity is to be extremized Let s call it Q What other quantities can vary What quantities are fixed 2 Introduce notation Draw a diagram if appropriate 14 3 Write down the relations among the variables They may be given in the problem or deducible from general knowledge You need 1 an objective function expressing Q in terms of other variables 2 constraint equations relating those other variables so that you can write Q as a function of just one independent variable let s call it x 15 4 Solve the constraints and substitute the results into the objective function Don t differentiate the constraints Don t differentiate Q until you have eliminated all variables but x 5 Differentiate Q x to find its critical points 16 6 Verify that your favorite critical point is the correct extremum Is it a max or a min or neither Is it in the physically allowed interval Remember to check endpoints as possible candidates 17 Exercise Redo the dream house example but this time suppose that the budget is flexible and you want a lot of exactly 10 000 ft2 Find the dimensions of the lot with the minimal wall fence cost and find that cost 18 We still have the area and cost formulas A LW C 15L 10W But now the roles of objective and constraint formulas are interchanged The constraint solved is W 10000 L so the objective is C L 15L 100000 L In either problem I could have eliminated L in favor of W instead of the reverse That would 19 change the intermediate algebra but not the answers 100000 0 C L 15 2 L q q 2 100 L 100000 15 3 q W 100 32 q q C 1500 23 1000 32 20 And now something completely different Antiderivatives There is not a whole lot to say here because 1 we ve been talking all along about the problem of finding a function whose derivative is a given function and 2 such problems are either very easy if you know all the differentiation formulas or hard and the hard part is postponed to Sec 6 5 and next semester 21 Given f find a function or all functions F such that F x f x Let s go ahead and write this problem in the notation you will be using for the rest of your life even though it won t be explained till next week Z Find F x f x dx 22 There is one crucial theoretical point Any two antiderivatives of a function on an interval differ only by a constant Example We know that one antiderivative of 1 x2 1 2 is sin 1 x Therefore a formula for all possible antiderivatives of 1 x2 1 2 is Z 1 x2 1 2 dx sin 1 x C where C stands for an arbitrary constant 23 There is one exception to the rule that only a constant is needed to get all antiderivatives It is handled by the disclaimer on an interval in the theorem on the previous slide If f is defined on two or more disjoint intervals 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