Exam 3-solutions

# Exam 3-solutions - Version 028 Exam 3 Helleloid (58250) 1...

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Unformatted text preview: Version 028 Exam 3 Helleloid (58250) 1 This print-out should have 19 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 = 3- x as shown in Determine the maximum possible area, A max , of the rectangle. 1. A max = 4 sq. units correct 2. A max = 6 sq. units 3. A max = 5 sq. units 4. A max = 3 sq. units 5. A max = 7 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 = 6 y- 2 y 3 . Differentiating A ( y ) with respect to y we see that A ( y ) = 6- 6 y 2 . The critical points of A are thus the solutions of 6- 6 y 2 = 0 , i . e ., y =- 1 , 1 ; the solution y =- 1 can obviously be disre- garded for practical reasons. Substituting for y = 1 in A ( y ) we get A max = 4 sq. units . 002 10.0 points Circuit City has been selling 70 television sets a week at \$720 each. A market survey indicates that for each \$30 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 = \$140 2. rebate = \$130 3. none of these 4. rebate = \$145 5. rebate = \$150 correct 6. rebate = \$135 Explanation: Let \$30 x be the rebate offered to a buyer. Then the price of a TV will be \$(720- 30 x ) and the number of sets sold at this price will be 70 + 5 x . The revenue with this rebate is thus R ( x ) = (720- 30 x )(70 + 5 x ) = 150(24- x )(14 + x ) = 150(336 + 10 x- x 2 ) . But then R ( x ) = 150(10- 2 x ) , while R ( x ) =- 150 2 &amp;lt; . Version 028 Exam 3 Helleloid (58250) 2 Consequently, the Revenue is maximized at x = 5, in which case the rebate = \$150 . 003 10.0 points Use Newtons method to estimate the solu- tion to e 3 x + x = 9 5 starting with the initial guess x = 0 and applying one iteration. 1. estimate = 1 10 2. estimate = 1 5 correct 3. estimate =- 1 10 4. estimate = 3 10 5. estimate = 0 Explanation: If x n is one estimate of a solution to the equation f ( x ) = 0, then Newtons 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 3 x + x- 9 5 , then f ( x ) = 3 e 3 x + 1 , so Newtons method gives the iteration for- mula x n +1 = x n- e 3 x n + x n- 9 5 3 e 3 x n + 1 ....
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## Exam 3-solutions - Version 028 Exam 3 Helleloid (58250) 1...

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