43 let s be the stiffness of the beam then s kwh 3

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43. Let S be the stiffness of the beam. Then, S = kwh 3 , where k is a constant of proportionality. Since w 2 + h 2 = 225 , or h = 225 w 2 , S can be expressed as a function of w,
3.5 Additional Applied Optimization 139 S(w) = kw( 225 w 2 ) 3 / 2 which is the function to be maximized. S (w) = k w · 3 2 ( 225 w 2 ) 1 / 2 ( 2 w) + ( 225 w 2 ) 3 / 2 ( 1 ) = k( 225 w 2 ) 1 / 2 3 w 2 + 225 w 2 = k( 225 w 2 ) 1 / 2 ( 225 4 w 2 ) S (w) = 0 when w = 15 2 (rejecting the solution w = 15 , which is not possible given the diameter) When 0 < w < 15 2 , S (w) > 0 so C is increasing 15 2 < w < 15 , S (x)) < 0 so S is decreasing. So, the dimensions for maximum stiffness are w = 15 2 inches and h = 225 15 2 2 13 . 0 inches. 45. Let x be the number of miles from the house to plant A . Then, 18 x is its distance from plant B , and 1 x 16 . Let P (x) be the concentration of particulate matter at the house. Then, P (x) = 80 x + 720 18 x which is the function to minimize. P (x) = − 80 x 2 + 0 ( 720 )( 1 ) ( 18 x) 2 P (x) = 0 when 80 x 2 = 720 ( 18 x) 2 2 x 2 + 9 x 81 = 0 or, x = 9 2 (rejecting negative solution) P ( 4 . 5 ) = 0 , P ( 1 ) 122 . 4 , P ( 16 ) = 365; So, the total pollution is minimized when the house is 4.5 miles from plant A . 47. Let C(N) be the total cost of using N machines. Now, the setup cost of N machines is aN and the operating cost of N machines is b N . So, C(N) = aN + b N which is the function to minimize. C (N) = a = b N 2 C (N) = 0 when a = b N 2 , or when aN = b N (setup cost = operating cost) C (N) = 2 b N 3 , which is positive for all N in the domain N 1, so there is an absolute minimum when setup cost equals operating cost. 49. Frank is right. In the cost function, C(x) = 5 ( 900 ) 2 + x 2 + 4 ( 3 , 000 x) note where the distance downstream appears. Since it is only part of the constant term in C(x), it drops out when finding C (x) . So, the critical value is always x = 1 , 200 (as long as the distance downstream is at least 1,200 meters). When 0 x < 1 , 200 , C (x) < 0 so C is decreasing x > 1 , 200 , C (x)) > 0 so C is increasing So, the absolute minimum cost is always when the cable reaches the bank 1,200 meters downstream. 51. (a) Let x be the number of machines and let t be the number of hours required to produce q units. The set up cost is xs and the operating cost is pt . Since each machine produces n units per hour, then q = xnt, or t = q nx . The total cost is C(x) = xs + p q nx which is the function to be minimized.
140 Chapter 3. Additional Applications of the Derivative C (x) = s pq nx 2 C (x) = 0 when s = pq nx 2 or, x = pq ns 1 / 2 C (x) = 2 pq nx 3 , so C pq ns 1 / 2 > 0 and there is a relative minimum when x = pq ns 1 / 2 . Further, since C (x) > 0 for all values of x in the domain x 1, it is the absolute minimum. (b) The setup cost xs, at this minimum, becomes xs = s pq ns = pqs n and the operating cost pt , at this minimum, becomes P q n pq ns = pq pqn s = pq s pqn = pqs n So, the setup cost equals the operating cost when the total cost is minimized. 53. (a) Let x be the number of units produced, p(x) the price per unit, t the tax per unit, and C(x) the total cost.

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