Review_Exam_3-solutions

Review_Exam_3-solutions - bornstein(cjb2383 – Review Exam...

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Unformatted text preview: bornstein (cjb2383) – Review Exam 3 – Fouli – (58320) 1 This print-out should have 33 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A rectangle is inscribed between the y-axis and the parabola y 2 = 27 − x as shown in Determine the maximum possible area, A max , of the rectangle. 1. A max = 111 sq. units 2. A max = 112 sq. units 3. A max = 108 sq. units correct 4. A max = 109 sq. units 5. A max = 110 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 = 54 y − 2 y 3 . Differentiating A ( y ) with respect to y we see that A ′ ( y ) = 54 − 6 y 2 . The critical points of A are thus the solutions of 54 − 6 y 2 = 0 , i . e ., y = − 3 , 3 ; the solution y = − 3 can obviously be disre- garded for practical reasons. Substituting for y = 3 in A ( y ) we get A max = 108 sq. units . 002 10.0 points Circuit City has been selling 80 television sets a week at $480 each. A market survey indicates that for each $20 rebate offered to a buyer, the number of sets sold will increase by 5 per week. How large a rebate should Circuit City offer a buyer in order to maximize its revenue? 1. rebate = $65 2. rebate = $75 3. rebate = $80 correct 4. none of these 5. rebate = $70 6. rebate = $85 Explanation: Let $20 x be the rebate offered to a buyer. Then the price of a TV will be $(480 − 20 x ) and the number of sets sold at this price will be 80 + 5 x . The revenue with this rebate is thus R ( x ) = (480 − 20 x )(80 + 5 x ) = 100(24 − x )(16 + x ) = 100(384 + 8 x − x 2 ) . But then R ′ ( x ) = 100(8 − 2 x ) , while R ′′ ( x ) = − 100 × 2 < . bornstein (cjb2383) – Review Exam 3 – Fouli – (58320) 2 Consequently, the Revenue is maximized at x = 4, in which case the rebate = $80 . 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 and applying one iteration. 1. estimate = − 1 10 2. estimate = − 1 5 3. estimate = 0 4. estimate = 1 10 5. estimate = 1 5 correct 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, x 1 = 0 − 1 − 8 5 2 + 1 = 1 5 . 004 10.0 points Find the value of f (0) when f ′′ ( t ) = 2(6 t + 1) and f ′ (1) = 4 , f (1) = 2 ....
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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.

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Review_Exam_3-solutions - bornstein(cjb2383 – Review Exam...

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