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Unformatted text preview: pokharel (yp624) HW11 Radin (56520) 1 This printout should have 15 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Find the height of the largest rectangle that can be inscribed in the region bounded by the xaxis and the graph of y = radicalbig 25 x 2 . Correct answer: 3 . 53553 units. Explanation: Since the graph is upper half of the circle x 2 + y 2 = 25 having radius 5 and center at the origin, the rectangle is the one shown in P ( x, y ) 5 5 5 Thus the area of the rectangle is A = 2 xy = 2 y radicalbig 25 y 2 , so we have to maximize A ( y ) on the interval [0 , 5]. Now A ( y ) = 2 radicalbig 25 y 2 2 y 2 radicalbig 25 y 2 = 2(25 2 y 2 ) radicalbig 25 y 2 . Thus the critical points of A ( y ) occur at y = 5 2 , 5 2 , only one of which lies in [0 , 5]. But A (0) = 0 , A parenleftBig 5 2 parenrightBig = 25 , A (5) = 0 . Consequently, the maximum area = 25 sq. units , and this occurs when the rectangle has height = 5 2 = 3 . 53553 . 002 10.0 points A 6 6 square sheet of metal is made into an open box by cutting out a square at each corner and then folding up the four sides. Determine the maximum volume, V max , of the box. 1. V max = 36 cu. ins. 2. V max = 31 cu. ins. 3. V max = 26 cu. ins. 4. V max = 16 cu. ins. correct 5. V max = 21 cu. ins. Explanation: Let x to be the length of the side of the squares cut from each edge. Then the volume of the resulting box is given by V ( x ) = x (6 2 x ) 2 . Differentiating V with respect to x we see that dV dx = (6 2 x ) 2 4 x (6 2 x ) . The critical points of V are thus the solutions of 3 x 2 12 x + 9 = 0 , i.e. , x 1 = 1 , x 2 = 3 , where the second one can be disregarded for practical reasons. At x = x 1 , therefore, V ( x ) becomes V max = 16 cu. ins. . pokharel (yp624) HW11 Radin (56520) 2 003 10.0 points The canvas wind shelter is to be constructed for use on Padre Island beaches. It is to have a back, two square sides, and a top. If 150 square feet of canvas are available, find the depth of the shelter for which the space inside is maximized assuming all the canvas is used. 1. depth = 5 2 feet 2. none of these 3. depth = 10 feet 4. depth = 11 2 feet 5. depth = 5 feet correct Explanation: Let x be the be the length of the shelter and y its depth. Then, since the ends of the shelter are squares of side length y , the volume V of the shelter is given by ( ) V ( x, y ) = xy 2 , while its area is given by ( ) 2 xy + 2 y 2 = 150 . Using ( ) to eliminate x from ( ) we see that V ( x ) = y 2 parenleftBig 150 2 y 2 parenrightBig = 75 y y 3 ....
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This note was uploaded on 06/05/2010 for the course PHYS 92515 taught by Professor Tsoi during the Spring '10 term at University of Texas at Austin.
 Spring '10
 TSOI

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