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Spring 2010 Exam 2 Solutions

Spring 2010 Exam 2 Solutions - Exam 2 STOR 112 Spring 2010...

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Exam 2, STOR 112 Spring 2010 Name (please print): Section (circle one): 1 G. Pataki 2 J. Anderson a. The exam consists of 9 questions for a total of 125 points. b. Everything you want us to grade should be written on the exam paper (so you need to make sure that you get the right answers on scratch paper). The instructors give you some scratch paper, which you should also return at the end of the exam. c. Unless otherwise indicated, please carry out all computations resulting in fractions to FOUR decimal places. Please show all work. Unsupported answers receive no credit. d. You may use a calculator. e. For LP and IP formulations, you must give a clear definition of the decision variables, the objective function, and the constraints. Give a descriptive label for the objective AND each constraint(such as ”profit”). Do not make assumptions about the solution based on the problem data (such as, product A makes much less profit than product B, so we will not make any of it, and exclude it from the formulation), even if it is possible to do so. Write all constraints in the standard form: variables on the left side and a number on the right. f. When turning in your exam, please show your UNC OneCard to one of the instructors. Pledge: I have neither given, nor received unauthorized assistance on this exam. —————————————————————— Signature 1
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Question 1. (10 points) Fit an exponential curve to the following two points: (2,1125) and (4, 2531.25). You need to do the calculation rigorously, i.e. guessing gets no credit. Solution We are looking for a function of the form f ( x ) = Ab x . We know Ab 2 = 1125 , Ab 4 = 2531 . 25 . Dividing the second equation by the first we get b 2 = 2 . 25 , i.e. b = 1 . 5. Plugging into the first equation we get 2 . 25 A = 1125 , so A = 500. So the function is f ( x ) = 500(1 . 5) x . Question 2. (15 points) Let f ( x ) = 2 x x + 5 . What is f ( x + 1)? What is f ( x ) + 1? Are they the same for all x , for no x , or for some x ? Solution We have f ( x + 1) = 2( x + 1) x + 6 , f ( x ) + 1 = 2 x x + 5 + 1 = 3 x + 5 x + 5 . Equating the two, we get that f ( x + 1) = f ( x ) + 1 , exactly when ( * ) 2( x + 1) x + 6 = 3 x + 5 x + 5 2( x + 1)( x + 5) = (3 x + 5)( x + 6) . Multiplying out we get 2 x 2 + 12 x + 10 = 3 x 2 + 23 x + 30 x 2 + 11 x + 20 = 0 . The two roots of the last equation are x 1 , 2 = - 11 ± 11 2 - 4 · 20 2 = - 2 . 7161 , and - 8 . 2838 . We still need to check that the 2 roots do not make the denominators of (*) zero, but this is easy.
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