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# Quiz 10 - Version OWL/AAABA w DGQlOc —— Gilbert...

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Unformatted text preview: Version OWL/AAABA w DGQlOc —— Gilbert — (78759} i This print—out should have 8' questions. Multipie—choice questions may continue on the next coiurnn or page M ﬁnd all choices before answering. III-Inn Eln- BACK 001 10.0 points Use Lagrange Multipliers to determine the maximum vaiue of fan y) 3 m2 —- 2:12 + 3 subject to the constraint 9135,29): m2+y2w1= 0- 1. maximum m 3 2. maximum :2 2 3. maximum = 1 4. maximum m 8 5. maximum = 4c0r‘rect Explanation: The extreme values of f subject to the con— straint g = 0 occur at solutions of (Vfllm, y) m MWXIE: :4)- Now (vf)(\$7 y) w (217: '4y> 3 while (Vexxayl = (233,21!)- Thus 2m :2 2M, may = 2A3}, and so we have that 2m()\~ 1) m 0, 2y()\+2) = 0. From the first equation, we know that either mmOorAml. Now,if}\=1we havefrom the second equation that y = 0, and also that 9(\$=O) = mgml = 0, ore: :2: ii. However, if a: m 0, then we have from the second equation that A m ~2, and also that We) w y2_1= 0, Dry = a1. Consequently, the critical points are (i1, 0) and (0, at). Since ‘ - f(j:1, 0) 3 4 and f(0,:ti)= 1, we thus see that max value = e . keywords: 002 10.0 points A company manufacturing cars at plant A in Detroit and plant B in St. Louis has found that its total cost function is 0 a x2+2y2+2\$y+90m+30y+10 when 3: cars are produced at piant A and y at plant B. What production levei at plant A should the company set to minimize the costs of production when ﬁlling an order for 110 cars? 1. prod plant A 2: 50 cars 2. prod plant A m 80 cars correct 3. prod plant A = 60 cars Version 004/AAABA m DGQlOc _ Gilbert — (787‘59) ' 2 4. prod plant A m? 40 cars 5. prod plant A m 70 cars Explanation: To determine the production level rnirnin— imzing costs we have to minimize C m x2+2y2+2scy+90m+30y+1m a function of two variables, subject to the restriction n+7; 2110. Using this last condition to eliminate y from 0 reduces the problem to optimization of a function C(93) m 3:2 + 2(110 —— 51:)2 + 23:(110 w m) + 90:12 + some m as) + 10 = 3:2 ~ 160\$ + 27510. of one variable. After differentiation, O’(m) = 21m 160, C”(;n) = 2 By the second derivative test for functions of one variable 0 thus has a minimum value at m = 80 t. a, to minimize production costs the company should set the production level at plant/i m 80 cars . keywords: 003 10.0 points Locate and classify the iocal extremum of ' f when 1 = 3x+y+-—+4-, 3 my (333 y > G)‘ flat, 1;) 1. saddie at (3, 3) 2. saddie at (g, 3) 3. local min'at (3, 3) 1 4. locai max at (§’ 3) 5. local max at (3, 3) 1 6. local min at (3" 3) correct Explanation: Differentiating once we see that At a local extremum these ﬁrst partiai deriva» tives are zero. Thus f has a locai extremum at (31 3) I To classify the local extremnm we use the ' second derivative test Now 2 1 2 film 3 "ﬁg: f\$y_ — 3:23“;i fyy“ "' 333\$ But then A 2 fm (£3) = 18 > 0 2 C w fyy (#3) —- 5 >. 0, and = x ‘m 1. fy (an 1 Thus 140nt = 3 > 0. Hence by the second. derivative test3 f has a local min at (—331, 3) ...
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Quiz 10 - Version OWL/AAABA w DGQlOc —— Gilbert...

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