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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 xaxis 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
xintercept 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.
 Spring '14
 The Land

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