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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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 Spring '10
 TSOI

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