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Unformatted text preview: rabbani (tar547) – Review 3 – Fouli – (58395) 1 This printout should have 19 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The rectangle in the figure b ( x, y ) is formed with adjacent sides on the coordi nate axes and one corner on the graph of y = 14 x 2 + 1 Find the maximum possible area of this rect angle. 1. max area = 5 sq. units. 2. max area = 4 sq. units. 3. max area = 7 sq. units. correct 4. max area = 6 sq. units. 5. max area = 3 sq. units. Explanation: The rectangle has side lengths x and y , so its area is given by A = xy . On the other hand, one corner lies on the graph of y = 14 x 2 + 1 so this area can expressed solely as a function by A ( x ) = x parenleftBig 14 x 2 + 1 parenrightBig = 14 x x 2 + 1 . Differentiating with respect to x we see that A ′ ( x ) = 14( x 2 + 1) 2 x (14 x ) ( x 2 + 1) 2 = 14 14 x 2 ( x 2 + 1) 2 . At a maximum A ′ ( x ) = 0, i.e. , 14 14 x 2 ( x 2 + 1) 2 = 0 , i.e. , A ′ ( x ) = 0 when x = ± 1. On practical grounds the positive solution corresponds to maximum area, while the negative solution has no meaning for this problem. Consequently, the maximum area = A (1) = 7 sq. units. . 002 10.0 points A farmer wishes to erect a fence enclosing a rectangular area adjacent to a barn which is 20 feet long. His plan Fenced area 20 ft Barn for the fenced area indicates the fencing in bold black lines; it shows also that the barn will be used as part of the fencing on one side. Find the largest area, A max , that can be enclosed if 104 feet of fencing material is available. 1. A max = 961 sq. ft. correct rabbani (tar547) – Review 3 – Fouli – (58395) 2 2. A max = 953 sq. ft. 3. A max = 959 sq. ft. 4. A max = 957 sq. ft. 5. A max = 955 sq. ft. Explanation: Let x, y be the respective parts of the fence shown in the figure 20 ft x y Barn Then 2 x + 20 + 2 y = 104 , so adjacent sides of the fenced area will have length x + 20 , 42 x. Hence, as a function of x , the area enclosed by the fence will be given by A ( x ) = (42 x )( x + 20) . Differentiating A ( x ) with respect to x we see that A ′ ( x ) = 2 x + 22 . The critical points of A are thus the solutions of A ′ ( x ) = 22 2 x = 0 , i.e. x = 11. On practical grounds (or use the fact that A ′′ = 2), this will yield a maximum value of A ( x ). At x = 11, therefore, the enclosed area will have a maximum value A max = 961 sq. ft....
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This note was uploaded on 04/13/2010 for the course M 408 K taught by Professor Jouve during the Fall '08 term at University of Texas.
 Fall '08
 JOUVE

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