Business Calc Homework w answers_Part_32

Business Calc Homework w answers_Part_32 - 156 Section 4.4...

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7. Let x be the side length of the cut-out square (0 , x , 4). Then the base measures 8 2 2 x in. by 15 2 2 x in., and the volume is V ( x ) 5 x (8 2 2 x )(15 2 2 x ) 5 4 x 3 2 46 x 2 1 120 x . Then V 9 ( x ) 5 12 x 2 2 92 x 1 120 5 4(3 x 2 5)( x 2 6). Then the critical point (in 0 , x , 4) occurs at x 5 } 5 3 } . Since V 9 ( x ) . 0 for 0 , x , } 5 3 } and V 9 ( x ) , 0 for } 5 3 } , x , 4, the critical point corresponds to the maximum volume. The maximum volume is V 1 } 5 3 } 2 5 } 24 2 5 7 0 } < 90.74 in 3 , and the dimensions are } 5 3 } in. by } 1 3 4 } in. by } 3 3 5 } in. Graphical support: [0, 4] by [ 2 25, 100] 8. Note that the values a and b must satisfy a 2 1 b 2 5 20 2 and so b 5 ˇ 4 w 0 w 0 w 2 w a w 2 w . Then the area is given by A 5 } 1 2 } ab 5 } 1 2 } a ˇ 4 w 0 w 0 w 2 w a w 2 w for 0 , a , 20, and } d d A a } 5 } 1 2 } a 1 } 2 ˇ 4 w 0 w 1 0 w 2 w a w 2 w } 2 ( 2 2 a ) 1 } 1 2 } ˇ 4 w 0 w 0 w 2 w a w 2 w 55 } ˇ 2 4 w 0 0 w 0 0 w 2 2 w a a w 2 2 w } . The critical point occurs when a 2 5 200. Since } d d A a } . 0 for 0 , a , ˇ 2 w 0 w 0 w and } d d A a } , 0 for ˇ 2 w 0 w 0 w , a , 20, this critical point corresponds to the maximum area. Furthermore, if a 5 ˇ 2 w 0 w 0 w then b 5 ˇ 4 w 0 w 0 w 2 w a w 2 w 5 ˇ 2 w 0 w 0 w , so the maximum area occurs when a 5 b . Graphical support: [0, 20] by [ 2 30, 110] 9. Let x be the length in meters of each side that adjoins the river. Then the side parallel to the river measures 800 2 2 x meters and the area is A ( x ) 5 x (800 2 2 x ) 5 800 x 2 2 x 2 for 0 , x , 400. Therefore, A 9 ( x ) 5 800 2 4 x and the critical point occurs at x 5 200. Since A 9 ( x ) . 0 for 0 , x , 200 and A 9 ( x ) , 0 for 200 , x , 400, the critical point corresponds to the maximum area. The largest possible area is A (200) 5 80,000 m 2 and the dimensions are 200 m (perpendicular to the river) by 400 m (parallel to the river). Graphical support: [0, 400] by [ 2 25,000, 90,000] 10. If the subdividing fence measures x meters, then the pea patch measures x m by } 21 x 6 } m and the amount of fence needed is f ( x ) 5 3 x 1 2 } 21 x 6 } 5 3 x 1 432 x 2 1 . Then f 9 ( x ) 5 3 2 432 x 2 2 and the critical point (for x . 0) occurs at x 5 12. Since f 9 ( x ) , 0 for 0 , x , 12 and f 9 ( x ) . 0 for x . 12, the critical point corresponds to the minimum total length of fence. The pea patch will measure 12 m by 18 m (with a 12-m divider), and the total amount of fence needed is f (12) 5 72 m. Graphical support: [0, 40] by [0, 250] 11. (a) Let x be the length in feet of each side of the square base. Then the height is } 5 x 0 2 0 } ft and the surface area (not including the open top) is S ( x ) 5 x 2 1 4 x 1 } 5 x 0 2 0 } 2 5 x 2 1 2000 x 2 1 . Therefore, S 9 ( x ) 5 2 x 2 2000 x 2 2 5 } 2( x 3 2 x 2 1000) } and the critical point occurs at x 5 10. Since S 9 ( x ) , 0 for 0 , x , 10 and S 9 ( x ) . 0 for x . 10, the critical point corresponds to the minimum amount of steel used. The dimensions should be 10 ft by 10 ft by 5 ft, where the height is 5 ft. (b) Assume that the weight is minimized when the total area of the bottom and the four sides is minimized. 2 a 2 1 (400 2 a 2 ) }} 2 ˇ 4 w 0 w 0 w 2 w a w 2 w 156 Section 4.4
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12. (a) Note that x 2 y 5 1125, so y 5 } 11 x 2 2 5 } . Then c 5 5( x 2 1 4 xy ) 1 10 xy 5 5 x 2 1 30 xy 5 5 x 2 1 30 x 1 } 11 x 2 2 5 } 2 5 5 x 2 1 33,750 x 2 1 } d d c x } 5 10 x 2 33,750 x 2 2 5 } 10( x 3 2 x 2 3375) } The critical point occurs at x 5 15. Since } d d c x } , 0 for 0 , x , 15 and } d d c x } .
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Business Calc Homework w answers_Part_32 - 156 Section 4.4...

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