# M 408K - bonci(cab3628 Exam3Practice thian(58380 This print-out should have 33 questions Multiple-choice questions may continue on the next column

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bonci (cab3628) – Exam3Practice – fthian – (58380) 1 This print-out should have 33 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points A rectangle is inscribed between the y -axis and the parabola y 2 = 12 x as shown in Determine the maximum possible area, A max , oF the rectangle. 1. A max = 29 sq. units 2. A max = 28 sq. units 3. A max = 31 sq. units 4. A max = 32 sq. units correct 5. A max = 30 sq. units Explanation: Let ( x, y ) be the coordinates oF the upper right corner oF the rectangle. The area oF the rectangle is then given by A ( y ) = 2 xy = 24 y 2 y 3 . Di±erentiating A ( y ) with respect to y we see that A ( y ) = 24 6 y 2 . The critical points oF A are thus the solutions oF 24 6 y 2 = 0 , i . e ., y = 2 , 2 ; the solution y = 2 can obviously be disre- garded For practical reasons. Substituting For y = 2 in A ( y ) we get A max = 32 sq. units . 002 10.0 points Circuit City has been selling 120 television sets a week at \$320 each. A market survey indicates that For each \$10 rebate o±ered to a buyer, the number oF sets sold will increase by 5 per week. How large a rebate should Circuit City o±er a buyer in order to maximize its revenue? 1. rebate = \$35 2. rebate = \$30 3. rebate = \$50 4. rebate = \$45 5. rebate = \$40 correct 6. none oF these Explanation: Let \$10 x be the rebate o±ered to a buyer. Then the price oF a TV will be \$(320 10 x ) and the number oF sets sold at this price will be 120 + 5 x . The revenue with this rebate is thus R ( x ) = (320 10 x )(120 + 5 x ) = 50(32 x )(24 + x ) = 50(768 + 8 x x 2 ) . But then R ( x ) = 50(8 2 x ) , while R ′′ ( x ) = 50 × 2 < 0 .

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bonci (cab3628) – Exam3Practice – fthian – (58380) 2 Consequently, the Revenue is maximized at x 0 = 4, in which case the rebate = \$40 . 003 10.0 points Use Newton’s method to estimate the solu- tion to e 2 x + x = 8 5 starting with the initial guess x 0 = 0 and applying one iteration. 1. estimate = 1 5 correct 2. estimate = 3 5 3. estimate = 2 5 4. estimate = 3 10 5. estimate = 1 2 Explanation: IF x n is one estimate oF a solution to the equation f ( x ) = 0, then Newton’s method says that x n +1 = x n f ( x n ) f ( x n ) will usually be a better estimate. But when f ( x ) = e 2 x + x 8 5 , then f ( x ) = 2 e 2 x + 1 , so Newton’s method gives the iteration For- mula x n +1 = x n e 2 x n + x n 8 5 2 e 2 x n + 1 . Consequently, with an initial guess oF x 0 = 0, x 1 = 0 1 8 5 2 + 1 = 1 5 . 004 10.0 points ±ind the value oF f (0) when f ′′ ( t ) = 2(9 t 1) and f (1) = 1 , f (1) = 5 .
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## This note was uploaded on 06/15/2011 for the course M 408k taught by Professor Schultz during the Spring '08 term at University of Texas at Austin.

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M 408K - bonci(cab3628 Exam3Practice thian(58380 This print-out should have 33 questions Multiple-choice questions may continue on the next column

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