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chapter6ProblemsAndSolutions

chapter6ProblemsAndSolutions - Chapter 6 Optimization 6.1...

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Chapter 6 Optimization 6.1 The sum of two positive number is 20. Find the numbers (a) if their product is a maximum. (b) if the sum of their squares is a minimum. (c) if the product of the square of one and the cube of the other is a maximum. Detailed Solution: Let x be one of the two numbers, then the other number is (20 - x ). (a) The product of the two numbers P is P = x (20 - x ). The maximum of P can be found by making P prime = 0. P prime = (20 - x ) + x ( - 1) = 20 - 2 x = 0 x = 10 Another number is 20 - x = 20 - 10 = 10. (b) The sum S of the squares of these two numbers is S = x 2 + (20 - x ) 2 . S prime = 2 x + 2(20 - x )( - 1) = 4 x - 40 = 0 x = 10 Another number is 20 - x = 20 - 10 = 10. (c) The function to be maximized is f ( x ) = (20 - x ) 2 · x 3 . f prime ( x ) = 2(20 - x )( - 1)( x 3 ) + (20 - x ) 2 (3 x 2 ) = 5 x 2 ( x 2 - 32 x + 240) = 0 x = 0, x = 20 or x = 12. 0 is not a positive number, so the solution x = 0 need not be considered. When x = 20, the other number is 20 - x = 20 - 20 = 0, which is not a positive number. When x = 12 the other number is 20 - x = 20 - 12 = 8. Therefore, the two numbers are 12 and 8. v.2005.1 - September 4, 2009 1

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Math 102 Problems Chapter 6 6.2 A tram ride at Disney World departs from its starting place at t = 0 and travels to the end of its route and back. Its distance from the terminal at time t can be approximately described by the expression S ( t ) = 4 t 3 (10 - t ) (where t is in minutes, 0 < t < 10, and S is distance in meters.) (a) Find the velocity as a function of time. (b) When is the tram moving at the fastest rate? (c) At what time does it get to the furthest point away from its starting position? (d) Sketch the acceleration, the velocity, and the position of the tram on the same set of axes. Detailed Solution: (a) The velocity is v ( t ) = dS dt = 12 t 2 (10 - t ) - 4 t 3 = 120 t 2 - 16 t 3 . (b) We are to maximize v ( t ) over the interval 0 t 10 . Thus dv dt = 240 t - 48 t 2 = 16 t (15 - 3 t ) = 0 ⇐⇒ t = 0 , 5 . Since the tram is at its starting place at t = 0, we know the maximum must occur at t = 5 . We check using the second derivative: v primeprime ( t ) = 240 - 96 t = v primeprime (5) = 240 - 96(5) = - 240 < 0 . Therefore the curve is concave down at t = 5, and t = 5 is a maximum. (c) To determine the maximum distance from the tram’s starting position we solve v ( t ) = 120 t 2 - 16 t 3 = 8 t 2 (15 - 2 t ) = 0 for t. We get t = 0 , 7 . 5 . Since v ( t ) > 0 for 0 < t < 7 . 5 and v ( t ) < 0 for 7 . 5 < t < 10 we see that the tram is farthest from its starting position when t = 7 . 5 . (d) See Figure 6.1 6.3 At 9 A.M. , car B is 25 km west of another car A . Car A then travels to the south at 30 km / h and car B travels east at 40 km / h. When will they be the closest to each other and what is this distance? v.2005.1 - September 4, 2009 2
Math 102 Problems Chapter 6 -4000 -2000 0 2000 4000 0 2 4 6 8 10 t S, v, a Figure 6.1: Figure for solution to Problem 6.2 Detailed Solution: Let D be the distance in km between car A and car B and t be the time in hours that the two cars have travelled since 9 A.M. The distance travelled by car A and car B are 30 t and 40 t , respectively.

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