Week 12 Notes. Aim: How do we use differentiation to maximize/minimize certain values (e.g. profit, cost,

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1 Week 2 Notes ) Optimiztion Problems: Aim: How o we use ifferentition to mximize/minimize certin vlues (e.g. profit, cost, volume, ) Exmple: Suppose you own tour bus n you book groups of 20 to 70 people for y tour. The cost per person is $0 minus $0.25 for every ticket sol. If gs n other miscellneous costs re $200, how mny tickets shoul you sell to mximize your profit? Solution: No picture neee. Vribles re number of trvelers x, n the cost C. The reltion between the vribles is given by: With omin between x = 20 n x = 70. C = (0 0.25x)x C = 0 0.5x x So the criticl point is x = 60, n by using either first or secon erivtive test it is locl minimum. As we hve seen before, since there is only one locl extremum between [20,70], it is lso going to be n bsolute minimum point. Hence I shoul sell 60 tickets to mximize my profit. Exmple: Suppose n irline policy sttes tht ll check-in bggge must be shpe in the shpe of box, whose sum of the with, length n height cnnot excee 08 inches. Wht is the gretest possible volume of squrebse box uner this restriction? Solution: Procee s in the relte rtes question. ) Drw the pictures (if possible). 2) Ientify which vlues re vribles, n which vlues re constnts. ) Write own wht is given in the question, n wht we re ske to fin. 4) Write own the reltions n ifferentite. The picture:

2 The vlues we nee to cre: With, Length, Height, Volume. These vlues cn ll be vrie. Wht is given in the question: w + l + h = 08, Wht we re ske to fin: If w = l, Mximize V. The reltions between the vribles: 08 = w + l + h = 2w + h V = w l h = w 2 h = w 2 (08 2w) So V(w) is function of w, with omin [0,54]. Its first erivtive is V = 26w 6w2 w An the criticl vlue is w = 6. (w = 0 rejecte since criticl points must be interior points) So we compute V(0), V(54), V(6) to etermine the bsolute mximum/minimum points. We get Therefore, the mximum volume is V = 6 (in ) V(0) = V(54) = 0, V(6) = 6 Exmple: A bot on the ocen is 4mi from the nerest point on stright shoreline; tht point is 6mi from resturnt on the shore. A womn plns to row the bot stright to point on the shore n then wlk long the shore to the resturnt. ) If she wlks mi/hr n rows 2mi/hr, t which point on the shore shoul she ln to minimize the totl trvel time? b) If she wlks mi/hr, wht is the minimum spee t which she must row so tht the quickest wy is to row irectly? Solution: The picture: Vribles: the position where the womn gets on shore, time. Wht is given in the question: () it tkes 2 x hr to row, n (6 x)hr to wlk. Wht re we ske to o: For ifferent position she lns, we get ifferent times to rech the resturnt. We wnt to minimize T.

3 So T is function of x, with omin between [0,6] Criticl points: So x = 8 5 or 8 5 (rejecte). One cn use first erivtive test to check x = 8 ccoring to theorem I wrote on the notes, x = 8 Wht is given in the question: T = 2 x (6 x) T x = x (x 2 ) x (x 2 ) = 0 5 x 2 4x = 9 (b) It tkes her (6 x)hr to wlk, n (6 x)hr to row. v Wht we re ske to o: 5x 2 = 64 is locl minimum, n is the only locl extremum. Then 5 is n bsolute minimum. If T(x) hs n bsolute minimum t x = 6, wht re the possible vlues of v? T = v x (6 x) T x = x (x v ) So T = 0 when x = 4v x 9 v2, n check tht it must be locl minimum point. So we wnt to locl minimum point to be beyon our omin [0,6]. We get 4v 9 v 2 6 v 2 8 v 9

4 2) Anti-erivtives Aim: Reverse the process of ifferentition Definition: A function F is n ntierivtive of f on n intervl I if F (x) = f(x) for ll x in I. Exmples: ) Since (sin x) = cos x. F(x) = sin x is n nti-erivtive of f(x) = cos x. 2) An nti-erivtive of x 2 is x. ) An nti-erivtive of e x is e x. 4) An nti-erivtive of is ln x. x 5) An nti-erivtive of is +x 2 tn x. One my sk, is there only one nti-erivtive? We hve nswere this question using MVT: Theorem: Let F be ny ntierivtive of f. Then ll the ntierivtives of f hve the form F + C, where C is n rbitrry constnt. We write f(x) x to be the collection of ll nti-erivtives. For exmple, cos x x = sin x + C. As in the rules for erivtives, we hve list of rules for nti-erivtives: Theorem: Let c be constnt. Then cf(x) x = c f(x) x (f(x) + g(x)) x = f(x) x + g(x) x An we hve the following tbles to fin the nti-erivtives: Polynomils: x p x = xp+ p + + C where p is rel number n C is n rbitrry constnt.

5 Trigonometric functions: (sin x) = cos x x (cos x) = sin x x x (tn x) = sec2 x x (cot x) = csc2 x (sec x) = sec x tn x x (csc x) = csc x cot x x cos x x = sin x + C sin x x = cos x + C sec 2 x x = tn x + C csc 2 x x = cot x + C sec x tn x x = sec x + C csc x cot x x = csc x + C Exponentil n Inverse Trigonometric functions: x (ex ) = e x e x x = ex + C x (sin kx) = x (ln x ) = x x = ln x + C (x 0) x k k 2 x 2 k 2 x x = 2 k sin kx + C k x (tn kx) = + k 2 x 2 + k 2 x 2 x = k tn kx + C k x (sec kx) = x k 2 x 2 x k 2 x 2 x = k sec kx + C Exmple: Fin the nti-erivtives of x 2 2x. x 2 2x x = x 2 x 2 x x (Sum rule + Constnt Mult. Rule) = ( x + C) 2 (x2 2 + C ) (Antierivtives for polynomils) = x x2 + (C 2C ) = x x2 + C (the constnts C n C re rbitrry)

6 Exercises: () Compute the following nti-erivtives: sec 2 x x, tn 2 x x (hint: tn 2 θ = sec 2 θ ), 4 x 9x 2 +, e 2t+ + 2 t t, y2 + 5 y x y, x x 2 00 Answers: tn x + C, tn x x + C, 4 x tn + C, 2 e2t+ + 4 t 2 + C, 2 y2 + 5 ln y + C, 0 sec x 0 + C (2) Prove tht (x ln x) = + ln x x Using the bove result, compute ln x x. (Answer: x ln x x + C)