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HW 11 Solutions

# HW 11 Solutions - Mathematics 408K Homework 11 Gagan Tara...

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Mathematics 408 K: Homework 11 Gagan Tara Nanda 02 nd November, 2004 HW 11 : # s 6 , 8 , 9 6. Let the side of the squares be a . Then the canvas covers four separate pieces. The sides surfaces are squares, each of side length a . The top is a rectangle, of width a and length x . The back is also a rectangle, again with length x and width a . So the total surface area is S = a 2 + a 2 + ax + ax = 2 ¡ a 2 + ax ¢ . We are told that a total of 294 ft 2 of canvas is available. So S = 294 a 2 + ax = 147 . This gives ax = 147 a 2 . We wish to maximize the space inside the shelter. This is the same as the volume of a box of sides a, a, x . So V = a 2 x = ( ax ) a . Writing V in terms of a only, we get V = ¡ 147 a 2 ¢ a = 147 a a 3 . We wish to maximize this quantity. So V 0 = 147 3 a 2 . This has critical points when 147 3 a 2 = 0 a 2 = 49 a = ± 7 . Clearly, a > 0 , so a = 7 . We check that this gives a maximum: V 00 = 6 a < 0 whenever a > 0 . We wish to fi nd the length x , so we get x = 147 7 2 7 = 14 ft .

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