# Y 1 and y 1 let f y y 4 y 1 then f y 4y 3

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Unformatted text preview: x)x (b) P (x) = R(x) − C (x) = (1000 − x)x − (3000 + 20x) = −3000 + 980x − x2 (c) 44. (a) P (x) = 980 − 2x, P (x) = 0 for 0 < x < 500 when x = 490; test the points 0, 490, 500 to ﬁnd that the proﬁt is a maximum when x = 490. (d) P (490) = 237,100 (e) 45. p = 1000 − x = 1000 − 490 = 510. The proﬁt is P = (proﬁt on nondefective) − (loss on defective) = 100(x − y ) − 20y = 100x − 120y but y = 0.01x + 0.00003x2 so P = 100x − 120(0.01x + 0.00003x2 ) = 98.8x − 0.0036x2 for x > 0, dP/dx = 98.8 − 0.0072x, dP/dx = 0 when x = 98.8/0.0072 ≈ 13, 722, d2 P/dx2 < 0 so the proﬁt is maximum at a production level of about 13,722 pounds. Exercise Set 6.2 46. 190 The total cost C is C = c · (hours to travel 3000 mi at a speed of v mi/h) 3000 3000 = (a + bv n ) = 3000(av −1 + bv n−1 ) for v > 0, =c· v v dC/dv = 3000[−av −2 + b(n − 1)v n−2 ] = 3000[−a + b(n − 1)v n ]/v 2 , dC/dv = 0 when v = a b(n − 1) 1/n . This is the only critical point and dC/dv changes sign from − to + at this point so the total cost is least when v = a b(n − 1) 1/n mi/h. √ 47. The distance between the particles is D = (1 − t − t)2 + (t − 2t)2 = 5t2 − 4t + 1 for t ≥ 0. For convenience, we minimize D2 instead, so D2 = 5t2 − 4t + 1, dD2 /dt = 10t − 4, which is 0 when t = 2/5. √ d2 D2 /dt2 > 0 so D2 and hence D is minimum when t = 2/5. The minimum distance is D = 1/ 5. √ 48. The distance between the particles is D = (2t − t)2 + (2 − t2 )2 = t4 − 3t2 + 4 for t ≥ 0. For convenience we minimize D2 instead so D2 = t4 − 3t2 + 4, dD2 /dt = 4t3 − 6t = 4t(t2 − 3/2), which is = 0 for t > 0 when t = 3/2. d2 D2 /dt2 √ 12t2 − 6 > 0 when t = 3/2 so D2 and hence D is minimum there. The minimum distance is D = 7/2. 49. 50. 51. Let P (x, y ) be a point on the curve x2 + y 2 = 1. The distance between P (x, y ) and P0 (2, 0) is √ (x − 2)2 + y 2 , but y 2 = 1 − x2 so D = (x − 2)2 + 1 − x2 = 5 − 4x for −1 ≤ x ≤ 1, D= 2 dD which has no critical points for −1 < x < 1. If x = −1, 1 then D = 3, 1 so the = −√ dx 5 − 4x closest point occurs when x = 1 and y = 0. √ x, then the distance D between P and (2, 0) is √ D = (x − 2)2 + y 2 = (x − 2)2 + x = x2 − 3x + 4, for 0 ≤ x ≤ 3. For convenience we ﬁnd the 2 extrema for D2 instead, so D√= x2 − 3x + 4, dD2 /dx = 2x − 3 = 0 when x = 3/2. If x = 0, 3/2, 3 √ 2 then D = 4, 7/4, 4 so D = 2, 7/2, 2. The points (0, 0) and (3, 3) are at the greatest distance, and (3/2, 3/2) the shortest distance from (2, 0). Let P (x, y ) be a point on y = Let (x, y ) be a point on the curve, then the square of the distance between (x, y ) and (0, 2) is S = x2 + (y − 2)2 where x2 − y 2 = 1, x2 = y 2 + 1 so S = (y 2 + 1) + (y − 2)2 = 2y 2 − 4y + 5 for any y , dS/dy = 4y − 4, dS/dy = 0 when y = 1, √ d2 S/dy 2 > 0 so S is least when y = 1 and x = ± 2. 52. The square of the distance between a point (x, y ) on the curve and the point (0, 9) is S = x2 + (y − 9)2 where x = 2y 2 so S = 4y 4 + (y − 9)2 for any y , dS/dy = 16y 3 + 2(y − 9) = 2(8y 3 + y − 9), dS/dy = 0 when y = 1 (which is the only real solution), d2 S/dy 2 > 0 so S is least when y = 1, x = 2. 53. If P (x0 , y0 ) is on the curve y = 1/x2 , then y0 = 1/x2 . At P the slope of the tangent line is −2/x3 so its 0 0 2 2 3 3 1 equation is y − 2 = − 3 (x − x0 ), or y = − 3 x + 2 . The tangent line crosses the y -axis at 2 , and x0 x0 x0 x0 x0 9 3 9 x0 . The length of the segment then is L = + x2 for x0 > 0. For convenience, 2 x4 40 0 9 9 2 dL2 36 9 9(x6 − 8) 0 2 2 = − 5 + x0 = , which is 0 when x6 = 8, we minimize L instead, so L = 4 + x0 , 0 x0 4 dx0 x0 2 2x5 0 22 √ dL √ x0 = 2. > 0 so L2 and hence L is minimum when x0 = 2, y0 = 1/2. dx2 0 the x-axis at 191 Chapter 6 54. If P (x0 , y0 ) is on the curve y = 1 − x2 , then y0 = 1 − x2 . At P the slope of the tangent line is −2x0 so 0 its equation is y − (1 − x2 ) = −2x0 (x − x0 ), or y = −2x0 x + x2 + 1. The y -intercept is x2 + 1 and the 0 0 0 12 13 1 x-intercept is (x0 +1/x0 ) so the area A of the triangle is A = (x0 +1)(x0 +1/x0 ) = (x0 +2x0 +1/x0 ) 2 4 4 for 0 ≤ x0 ≤ 1. 1 1 dA/dx0 = (3x2 + 2 − 1/x2 ) = (3x4 + 2x2 − 1)/x2 which is 0 when x2 = −1 (reject), or 0 0 0 0 0 0 4 4 √ √ 1 when x2 = 1/3 so x0 = 1/ 3. d2 A/dx2 = (6x0 + 2/x3 ) > 0 at x0 = 1/ 3 so a relative minimum and 0 0 0 4 hence the absolute minimum occurs there. 55. At each point (x, y ) on the curve the slope of the tangent line is m = dy 2x =− for any x, dx (1 + x2 )2 √ 2(3x2 − 1) dm dm = = 0 when x = ±1/ 3, by the ﬁrst derivative test the only relative maximum , dx (1 + x2 )3 √ dx occurs at x = −1/ 3, which is the absolute maximum because lim m = 0. The tangent line has x→±∞ √ greatest slope at the point (−1/ 3, 3/4). 56. Let x be how far P is upstream from where the man starts (see ﬁgure), then the total time to reach T is t= √ (time from M to P ) +...
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## This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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