Homework 11 - Vincent(jmvrrr HWll w Gilbert w(57495 1 ‘...

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Unformatted text preview: Vincent (jmvrrr) - HWll w Gilbert w (57495) 1 ‘ This printaout should have 18 questions. Multiplewcholce questions may continue on the next column or page -~ find all choices before answering. '001 10.0 points In the contour map below identify the points P, Q, and R as local minima, local maxima, or neither. A. local maximum-at Q, B. local minimum at P, C. local maximum at R. 1. B and C only correct 2. A and “C only 3. B only 4. A only - 5. none of them 6. A and B only 7. all of them 8. C only Explanation: A. FALSE: the point Q lies on the 0~ contour and this contour divides the region near Q into two regions. In one region the contours have values increasing to 0, while in the other the contours have values decreasing to 0. So the surface does not have a local minimum at Q. B. TRUE: the contours near P are closed carves enclosing P and the contours decrease in value as we approch P. So the surface has a local minimum at P. C. TRUE: the contours near R are closed curves enclosing R and the contours increase in value as we approch R. So the surface has a local maximum at R. keywords: local extrema, True/False, contour map, 002 10.0 points Locate and classify all the local extreme. of flay) == w3wy3w3my+1- 1. local max at; (0, 0), saddle point at (W1, 1) 2. local min at (0, O), saddle point at (”1, 1) 3. local min at (—1, 1), saddle point at (0, 0) 4. local min at (~l, 1), local max at (0, O) 5. local max: at (—1, l), saddle point at (0, 0) correct Explanation: Since f has derivatives everywhere, the crit- ical points occur at the solutions of Vftgay) = fxi+fyj m O~ Butme—Owhen afm 2 _ a$w3m 3y—0, is, y = 2:2, Vincent (jmv777) “- HWil — Gilbert — (57495) 3 the solutions. of which are a: m —1 and 5». ‘ ' ‘ Consequently, the critical points of f occur at 005 (part 3 of 3) 10.0 points 1 3 (iii) Which of the following most accurately de— (—1, 0), (5, WE) scribes the behaviour of f. 1. local maximum at (E, WE) 094 (part 2 of 3) 10.0 points 2 2 (ii) Which of the following most accurately de— ' 1 3 scribes the behaviour of f. 2' saddle—pomt at ("2": “2") 1. 1 w ‘ ' W1 1; sedd e pomt at ( , O) cor-rec 3 local minimum at (W; 23:) 2. saddlepoint at (I, G) 7 , . l 3 4 Eocai maxxmum at "2""? "2"" 3. local maximum at (1,0) 4. local minimum at (—1, U) 5. local minimum at (I, (l) 1 6. localmeximum at (~1, 0) 7. saddle—point at (§’ ———§) Explanation: ‘ _ l 3 After differentiation once again we see that 8. local minimum at (fi’ —§) correct facx=2(4$+2)a fyy=2i fmymga l 3 _ . _ 9. saddle‘point at (we, m) and so at the critical pomt (m1, 0), 2 2 A 2 fm = 4.4“} Explanation: (“119) After difierentiation once again we see that and foam m 2(4$+2): fyy = 2: facy = 2: C = f l m 2 > 0, W (”LG) anti so at the critical point (é, mg , while A = fw${1 3)m 8: B m r...- 2 _ E’W-é fflf’y (—120) end But then, AC—52e—12<e. Cm yy(t,~%)$2>e’ _ Hence by the second derivative test, f has a while saddlepoint at (—1, 0) . B : fey (1 3) = 2. 5!”? vincent (jrnv’???) w HWll w Gilbert w (57495) 5 edge 3 (1,1) occur on the boundary or at a critical point of f inside the square. ' ‘ Follow the next two steps to determine the minimum value of f on the square. (i) Determine the minimum value of f on the boundary of the square. 1. min value m 2 2. min value = 0 3. min value x 1correct 4. min value m —1 5. min value = W2 Explanation: We look at the behavior of f on each of the four edges separateiy. Edge 1: herey $010 S cc S 1, and min f(:1:,0)= min 33.: = 0. {)émgl 05x51 Edge 2: herea:=0,0_<_y§ 1, and _ Z . 2 z _ may“) Dale 9 0 Edge 3: hereyrz 1,0 5 mg1,anci ‘ fl ' — = 2. {12%; f(a3,1) 022E” :1:+2) Edge 4: herem = 1, 0 g y g 1, and Oéngréi f(1, y) m 61315212136 -— 2y) e 3- Consequently, on the boundary f has minimum value = 1 . 009 (part 2 of 3) 10.0 points (ii) Locate the criticai points of f in the square. 3 5 1‘ (2;: ’8') 3 1 2. (2, E) correct 3. none in unit square 5 1 4- (”8* e) ‘ 1 3 5' (a a) Explanation: The critical points of f occur at the solution of 13;: 2 3—4;; = 0, 8f m = 2M4 = 0. 8y :6 Consequently, the only critical point of f is at (3%) and this lies inside the unit square. 010 (part 3 of 3) 10.0 points (iii) Use parts (i) and (ii) to cietermine the minimum value of f on the square. H H :42.th aim nice 1 . min value 2. min value 3. min value Vincent (jmv777) ~ HWii ~ Gilbert w (57495) 7 (ii) on L2 abs min value of f = w23, (iii)on L3 abs min value of f m ~23. Consequently, taking the smaliest of ~7, m8, m23, w23, we see that on D abs min vaiue of f m ~23 keywords: partial differentiation, point, absolute extremum, 012 10.0 points is to be constructed having a volume of 16 (iii. feet. The material for the base costs $3 per square it, while the material for the sides costs $6 per square ft. Find the iowest cost 0min for which the box can be buiit, assuming all material is used. 1. lowest cost $154 2. lowest cost m $144 correct 3. lowest cost $164 $159 H 4. lowest cost 5. lowest cost W $11.49 Explanation: Let :2: be the length of the box, y the width, and z the height as shown in the figure. Then 16 Voiume = zyz = 16, i.€., z = . 33?} criticai Now Cost base = $3563}, while Costésides m $12(2:+y)z w W . Thus, as a function of as, y, the cost of con— structing the box is given by 192 192 C(QE, y) —— 3my+ "—33" + --y--. The cost will be minimized at a critical point of 0, £13., when ' fig _ '192 3 __,______ = 812: y :62 0 and when 80 192 ‘53? W 353* "y? m 0 After simplification these become $211 :2: 64, 33y2 m 64-. rThus x m y m 4. On practical grounds these values of :23, y clearly give a local minimum for C(zc, y). Consequently, lowest cost = $144 . keywords: 013 (part 1 of 3) 10.0 points rThe data points (m8, 3), (0, 5), (8, 6) have been obtained from a certain experi— ment. Theory suggests that the data should be linear. (i) Which expression needs to be minimized if y x mm + b is to be the Regression line for this data set? Vincent (jmv???) “ HWll m Gilbert m (57495) 9 by evaluating y at a: = 4. In this case -.§§ 9'" 12' 016 10.0 points Use the method of Lagrange multipliers to minimize flat, :9) = v3$2+2y2 subject to the constraint m+y m l. . 1 1. min value = ~2— 2. min value = 1 t 3. no min value exists 1 m 3 correct 2 ll 4. min value 5 min value m x/g Explanation: Set 906, y) = Newl- Then by the method of Lagrange multipliers the extreme values of f under the constraint 9 = 0 occur at the solutions of Vf = W9: Way) = 0- But when We 3;) = V3w2 + 2312 we see that 33: 2y _______.+_____.____.‘ V3162 ~§~ 23;? 1 #33:2 "192ng Vf 2 Since V9 2 1+3, the equation V f = AVg thus becomes 1 x/ 33:2 + 293 (333i+2y)j m A(i+j). After comparing coefiicients this reduces to the pair of equations A __ 3$ A M 2y M3582 + 23;? ’ M3232 + 2y? ’ 3 is, y 2 gm. But we still have the constraint equation , my) = HIM/“*1 = 0‘ Substituting y 2 333: gives 3 5 m+§$-l— Emmi—0. Consequently, the only Solution of W = Wei stay) as 0, occurs at 2 3 (337 y) “‘ (g: “5): and at this point 2 3 1 fie ”5“) r 5‘5 But is this a maximum or a minimum value? We can decide algebraically or graphicaliy, the best choice depending on f and 9. Let’s do it graphically because the graphs in Buspace of f and g = 0 are easy to describe. Indeed, the graph of z = flay) = v3w2+292 is a cone Vincent (jun/’77?) -~ HWll m Gilbert w (57495) 11 By investing 23 units of'labor and 1; units of capital, John’s’l‘ees can produce 13(33’ y) = 30$2/5y3/5 T—shirts. Use Lagrange Multipliers to determine the maximum number of Twshirts that can be produced on a budget of $10, 000 if labor costs $100 per unit and capital costs $200 per unit. 1. max# 3 1009 correct 2. max# m 979 3. max# = 989 4. max# 3 969 5. max# m 999 Explanation: When $10, 000 is available and labor costs $100 per unit While capital costs $200 per unit, then 10,000 m 1003: + 200g. To maximize production, therefore, we have to maximize 1’03, y) e 309320.93” subject to the constraint g(.:c, y) = 1005c + 200g —— 10,000 m 0. By the method of Lagrange Multipliers, the maximum value of P occurs at solutions of 07wa, y) = MVQM-t, y), 90313!) = 0- NOW 8P 2 W 2 m —3/5 3/5 85?) 30(53: 9 ), ‘35 _.. E’: 2/5 w2/5 8y m 30(5x y D’ in which case (View, y) = awe—w 2y, as: >— On the other hand, (VgXac, y) = (100, 200). Thus <6x"3/5y"3/5) 2y m 100A, (6xt3/5ng/5)3m = 200A, and so A m 3.12 _ fie tens/5yW2/5 "” 100 “ 200’ 216., y = (3/4)23. But we stiii have the constraint condition g(x, y) 2 10051: ~i— 2001;r — 10,000 = 0. Substituting in for y = (3/ 4):i: and soiving for 3:, we find that "a; = 40, y m 30. Consequently, since 19(40, 30) = 30(40)2/5(30)3/5 1009.76, 22 the maximum number of T—shirts that can be produced is 1009. keywords: ...
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