Exam3_2 4 - available. 1. A max = 721 sq. ft. 2. A max =...

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Taylor, Douglas – Exam 3 – Due: Dec 5 2007, 1:00 am – Inst: JEGilbert 4 2. f 0 ( x ) = 2 1 - e 6 x 3. f 0 ( x ) = 2 e 3 x 1 + e 6 x 4. f 0 ( x ) = 2 1 + e 6 x 5. f 0 ( x ) = 6 1 + e 6 x 6. f 0 ( x ) = 6 e 3 x 1 - e 6 x correct 7. f 0 ( x ) = 2 e 3 x 1 - e 6 x 8. f 0 ( x ) = 6 1 - e 6 x Explanation: Since d dx sin - 1 x = 1 1 - x 2 , d dx e ax = ae ax , the Chain Rule ensures that f 0 ( x ) = 6 e 3 x 1 - e 6 x . keywords: 009 (part 1 of 1) 10 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 88 feet of fencing material is
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Unformatted text preview: available. 1. A max = 721 sq. ft. 2. A max = 729 sq. ft. correct 3. A max = 727 sq. ft. 4. A max = 723 sq. ft. 5. A max = 725 sq. ft. Explanation: Let x, y be the respective parts of the fence shown in the gure 20 ft x y Barn Then 2 x + 20 + 2 y = 88 , so adjacent sides of the fenced area will have length x + 20 , 34-x. Hence, as a function of x , the area enclosed by the fence will be given by A ( x ) = (34-x )( x + 20) . Dierentiating A ( x ) with respect to x we see that A ( x ) =-2 x + 14 ....
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This note was uploaded on 02/14/2012 for the course MATH 408 K taught by Professor Clark,c.w./hoy,r.r during the Spring '08 term at University of Texas at Austin.

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