Embed Size (px)
DESCRIPTION
Â
Citation preview
277
16.
cj Basic 300 400 0 0 0
Variables Quantity x1 x2 s1 s2 s3
0 s1 18 3 2 1 0 0
0 s2 20 2 4 0 1 0
0 s3 4 0 1 0 0 1
zj 0 0 0 0 0 0
cj – zj 300 400 0 0 0
cj Basic 300 400 0 0 0
Variables Quantity x1 x2 s1 s2 s3
0 s1 10 3 0 1 0 –2
0 s2 4 2 0 0 1 –4
400 x2 4 0 1 0 0 1
zj 1,600 0 400 0 0 400
cj – zj 300 0 0 0 –400
cj Basic 300 400 0 0 0
Variables Quantity x1 x2 s1 s2 s3
0 s1 4 0 0 1 –3/2 4
300 x1 2 1 0 0 1/2 –2
400 x2 4 0 1 0 0 1
zj 2,200 300 400 0 150 –200
cj – zj 0 0 0 –150 200
cj Basic 300 400 0 0 0
Variables Quantity x1 x2 s1 s2 s3
0 s3 1 0 0 1/4 –3/8 1
300 x1 4 1 0 1/2 –1/4 0
400 x2 3 0 1 –1/4 3/8 0
zj 2,400 300 400 50 75 0
cj – zj 0 0 –50 –75 0
Optimal
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 277
278
17.
cj Basic 5 4 0 0
Variables Quantity x1 x2 s1 s2
0 s1 150 3/10 1/2 1 0
0 s2 2,000 10 4 0 1
zj 0 0 0 0 0
cj – zj 5 4 0 0
cj Basic 5 4 0 0
Variables Quantity x1 x2 s1 s2
0 s1 90 0 19/50 1 –3/100
5 x1 200 1 2/5 0 1/10
zj 1,000 5 2 0 1/2
cj – zj 0 2 0 –1/2
cj Basic 5 4 0 0
Variables Quantity x1 x2 s1 s2
4 x2 4,500/19 0 1 50/19 –3/38
5 x1 2,000/19 1 0 –20/19 5/38
zj 28,000/19 5 4 100/19 13/28
cj – zj 0 0 –100/19 –13/38
Optimal
18.
cj Basic 100 150 0 0 0
Variables Quantity x1 x2 s1 s2 s3
0 s1 160 10 4 1 0 0
0 s2 20 1 1 0 1 0
0 s3 300 10 20 0 0 1
zj 0 0 0 0 0 0
cj – zj 100 150 0 0 0
(continued)
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 278
279
cj Basic 100 150 0 0 0
Variables Quantity x1 x2 s1 s2 s3
0 s1 20 0 0 1 –16 3/5
100 x1 10 1 0 0 2 –1/10
150 x2 10 0 1 0 –1 1/10
zj 2,500 100 150 0 50 5
cj – zj 0 0 0 –50 –5
Optimal
19.
cj Basic 100 20 60 0 0 0
Variables Quantity x1 x2 x3 s1 s2 s3
0 s1 60 3 5 0 1 0 0
0 s2 100 2 2 2 0 1 0
0 s3 40 0 0 1 0 0 1
zj 0 0 0 0 0 0 0
cj – zj 100 20 60 0 0 0
cj Basic 100 20 60 0 0 0
Variables Quantity x1 x2 x3 s1 s2 s3
100 x1 20 1 5/3 0 1/3 0 0
0 s2 60 0 –4/3 2 –2/3 1 0
0 s3 40 0 0 1 0 0 1
zj 2,000 100 500/3 0 100/3 0 0
cj – zj 0 –440/3 60 –100/3 0 0
(continued)
cj Basic 100 150 0 0 0
Variables Quantity x1 x2 s1 s2 s3
0 s1 100 8 0 1 0 –1/5
0 s2 5 1/2 0 0 1 –1/20
150 x2 15 1/2 1 0 0 1/20
zj 2,250 75 150 0 0 15/2
cj – zj 25 0 0 0 –15/2
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 279
21.
cj Basic 120 40 240 0 0 M M
Variables Quantity x1 x2 x3 s1 s2 A1 A2
M A1 27 4 1 3 –1 0 1 0
M A2 30 2 6 3 0 –1 0 1
zj 54M 6M 7M 6M –M –M M M
zj – cj 6M – 120 7M – 40 6M – 240 –M –M 0 0
cj Basic 120 40 240 0 0 M
Variables Quantity x1 x2 x3 s1 s2 A1
M A1 22 11/3 0 5/2 –1 1/6 1
40 x2 5 1/3 1 1/2 0 –1/6 0
zj 19M + 200 11M/3 + 40/3 40 5M/2 + 20 –M M/6 – 20/3 M
zj – cj 11M/3 – 320/3 0 5M/2 – 220 –M M/6 – 20/3 0
cj Basic 120 40 240 0 0
Variables Quantity x1 x2 x3 s1 s2
120 x1 6 1 0 15/22 –3/11 1/22
40 x2 3 0 1 6/22 1/11 –2/11
zj 840 120 40 1,020/11 –320/11 –20/11
cj – zj 0 0 –1,620/11 –320/11 –20/11
Optimal
20. a. maximization, because cj – zjb. x2 = 10, s2 = 20, x1 = 10, Z = 30c. maximize Z = x1 + 2x2 – x3d. 3e. No, because there are three constraints and
three “slack” variablesf. s1 = 0g. yes, because cj – zj = 0 for s1h. x2 = 3 1/3, s2 = 26 2/3, x1 = 23 1/3, Z = 30
280
cj Basic 100 20 60 0 0 0
Variables Quantity x1 x2 x3 s1 s2 s3
100 x1 20 1 5/3 0 1/3 0 0
60 x3 30 0 –2/3 1 –1/3 1/2 0
0 s3 10 0 2/3 0 1/3 –1/2 1
zj 3,800 100 380/3 60 40/3 30 0
cj – zj 0 –320/3 0 –40/3 –30 0
Optimal
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 280
281
22.
cj Basic .05 .10 0 0 M M
Variables Quantity x1 x2 s1 s2 A1 A2
M A1 36 6 2 –1 0 1 0
M A2 50 5 5 0 –1 0 1
zj 86M 11M 7M –M –M M M
zj – cj 11M – .05 7M – .10 –M –M 0 0
cj Basic .05 .10 0 0 M
Variables Quantity x1 x2 s1 s2 A2
.05 x1 6 1 1/3 –1/6 0 0
M A2 20 0 10/3 5/6 –1 1
zj 20M + .3 .05 10M/3 + .02 5M/6 – .01 –M M
zj – cj 0 10M/3 – .08 5M/6 – .01 –M 0
cj Basic .05 .10 0 0
Variables Quantity x1 x2 s1 s2
.05 x1 4 1 0 –1/4 1/10
.10 x2 6 0 1 1/4 –3/10
zj .80 .05 .10 .0125 –.025
zj – cj 0 0 .0125 –.025
cj Basic .05 .10 0 0
Variables Quantity x1 x2 s1 s2
.05 x1 10 1 1 0 –1/5
0 s1 24 0 4 1 –6/5
zj .50 .05 .05 0 –.01
zj – cj 0 –.05 0 –.01
Optimal
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 281
282
23.
0 4 8 12 16 20 24 28 32x
x
1
4
8
12
16
20
24
2
Firsttableau
Thirdtableau
Second tableauFourth tableau
24.
cj Basic 10 12 7 0 0 0
Variables Quantity x1 x2 x3 s1 s2 s3
0 s1 300 20 15 10 1 0 0
0 s2 120 10 5 0 0 1 0
0 s3 40 1 0 2 0 0 1
zj 0 0 0 0 0 0 0
cj – zj 10 12 7 0 0 0
cj Basic 10 12 7 0 0 0
Variables Quantity x1 x2 x3 s1 s2 s3
12 x2 20 4/3 1 2/3 1/15 0 0
0 s2 20 10/3 0 –10/3 –1/3 1 0
0 s3 40 1 0 2 0 0 1
zj 240 16 12 8 4/5 0 0
cj – zj –6 0 –1 –4/5 0 0
Optimal
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 282
283
25.
cj Basic 6 2 12 0 0
Variables Quantity x1 x2 x3 s1 s2
0 s1 24 4 1 3 1 0
0 s2 30 2 6 3 0 1
zj 0 0 0 0 0 0
cj – zj 6 2 12 0 0
cj Basic 6 2 12 0 0
Variables Quantity x1 x2 x3 s1 s2
12 x3 8 4/3 1/3 1 1/3 0
0 s2 6 –2 5 0 –1 1
zj 96 16 4 12 4 0
cj – zj –10 –2 0 –4 0
Optimal
26.
cj Basic 100 75 90 95 0 0 0 0
Variables Quantity x1 x2 x3 x4 s1 s2 s3 s4
0 s1 40 3 2 0 0 1 0 0 0
0 s2 25 0 0 4 1 0 1 0 0
0 s3 2,000 200 0 250 0 0 0 1 0
0 s4 2,200 100 0 0 200 0 0 0 1
zj 0 0 0 0 0 0 0 0 0
cj – zj 100 75 90 95 0 0 0 0
cj Basic 100 75 90 95 0 0 0 0
Variables Quantity x1 x2 x3 x4 s1 s2 s3 s4
0 s1 10 0 2 –3.75 0 1 0 –.015 0
0 s2 25 0 0 4 1 0 1 0 0
100 x1 10 1 0 1.25 0 0 0 .005 0
0 s4 1,200 0 0 –1.25 200 0 0 –.50 1
zj 1,000 100 0 1.25 0 0 0 .50 0
cj – zj 0 75 –35 95 0 0 –.50 0
(continued)
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 283
284
cj Basic 100 75 90 95 0 0 0 0
Variables Quantity x1 x2 x3 x4 s1 s2 s3 s4
0 s1 10 0 2 –3.75 0 1 0 –.015 0
0 s2 19 0 0 4.6 0 0 1 .002 –.005
100 x1 10 1 0 1.25 0 0 0 .005 0
95 x4 6 0 0 –.625 1 0 0 –.002 .005
zj 1,570 100 0 65.6 95 0 0 .26 .475
cj – zj 0 75 24.4 0 0 0 –.26 –.475
cj Basic 100 75 90 95 0 0 0 0
Variables Quantity x1 x2 x3 x4 s1 s2 s3 s4
75 x2 5 0 1 –1.875 0 .50 0 –.007 0
0 s2 19 0 0 4.6 0 0 1 .002 –.005
100 x1 10 1 0 1.25 0 0 0 .005 0
95 x4 6 0 0 –.625 1 0 0 –.002 .005
zj 1,945 100 75 –75 95 37.5 0 –.30 .475
cj – zj 0 0 165 0 –37.5 0 .30 –.475
cj Basic 100 75 90 95 0 0 0 0
Variables Quantity x1 x2 x3 x4 s1 s2 s3 s4
75 x2 12.7 0 1 0 0 .5 .405 –.006 –.002
90 x3 4.1 0 0 1 0 0 .216 .001 –.001
100 x1 4.9 1 0 0 0 0 –.270 .004 .001
95 x4 8.6 0 0 0 1 0 .135 –.002 .004
zj 2,623 100 75 90 95 37.5 35.7 –.211 .297
cj – zj 0 0 0 0 –37.5 –35.7 .211 –.297
cj Basic 100 75 90 95 0 0 0 0
Variables Quantity x1 x2 x3 x4 s1 s2 s3 s4
75 x2 20 1.5 1 0 0 .50 0 0 0
90 x3 3.5 –.125 0 1 0 0 .25 0 –.001
0 s3 1,125 231.25 0 0 0 0 –62.5 1 .313
95 x4 11 .50 0 0 1 0 0 0 .005
zj 2,860 148.75 75 90 95 37.5 22.5 0 .36
cj – zj –48.75 0 0 0 –37.5 –22.5 0 –.36
Optimal
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 284
285
27.
cj Basic 20 16 0 0 0 M M M
Variables Quantity x1 x2 s1 s2 s3 A1 A2 A3
M A1 6 3 1 –1 0 0 1 0 0
M A2 4 1 1 0 –1 0 0 1 0
M A3 12 2 6 0 0 –1 0 0 1
zj 22M 6M 8M –M –M –M M M M
zj – cj 6M – 20 8M – 16 –M –M –M 0 0 0
cj Basic 20 16 0 0 0 M M
Variables Quantity x1 x2 s1 s2 s3 A1 A2
M A1 4 8/3 0 –1 0 1/6 1 0
M A2 2 2/3 0 0 –1 1/6 0 1
16 x2 2 1/3 1 0 0 –1/6 0 0
zj 32 + 6M 10M/3 + 16/3 16 –M –M M/3 – 8/3 M M
zj – cj 10M/3 – 44/3 0 –M –M M/3 – 8/3 0 0
cj Basic 20 16 0 0 0 M
Variables Quantity x1 x2 s1 s2 s3 A2
20 x1 3/2 1 0 –3/8 0 1/16 0
M A2 1 0 0 1/4 –1 1/8 1
16 x2 3/2 0 1 1/8 0 –3/16 0
zj M + 27 20 16 M/4 – 53/6 –M + 16/3 M/8 – 25/12 M
zj – cj 0 0 M/4 – 53/6 –M + 16/3 M/8 – 25/12 0
(continued)
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 285
28.
0 2 4 6 8 10 12 14 16x
x
1
2
4
6
8
10
12
2
First tableau
Second tableau
Fourth tableau
Fifth tableau
Third tableau
286
cj Basic 20 16 0 0 0
Variables Quantity x1 x2 s1 s2 s3
20 x1 3 1 0 0 –3/2 1/4
0 s1 4 0 0 1 –4 1/2
16 x2 1 0 1 0 1/2 –1/4
zj 76 20 16 0 –22 1
zj – cj 0 0 0 –22 1
cj Basic 20 16 0 0 0
Variables Quantity x1 x2 s1 s2 s3
20 x1 1 1 0 –1/2 1/2 0
0 s3 8 0 0 2 –8 1
16 x2 3 0 1 1/2 –3/2 0
zj 68 20 16 –2 –14 0
zj – cj 0 0 –2 –14 0
Optimal
29. Minimize Z = 8x1 + 2x2 + 7x3 + 0s1 + 0s2 + 0s3– MA1 – MA2 – MA3
subject to
2x1 + 6x2 + x3 + A1 = 303x2 + 4x3 – s1 + A2 = 604x1 + x2 + 2x3 + s2 = 50x1 + 2x2 – s3 + A3 = 20
x1, x2, x3 ≥ 0
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 286
287
30. Minimize Z = 40x1 + 55x2 + 30x3 + 0s1 + 0s2 + 0s3 + MA1 + MA2 + MA3
subject to
x1 + 2x2 + 3x3 + s1 = 602x1 + x2 + x3 + A1 = 40
x1 + 3x2 + x3 – s2 + A2 = 505x2 –3x3 – s3 + A3 = 100
x1, x2, x3 ≥ 0
31.
cj Basic 40 60 0 0 0 0 –M –M
Variables Quantity x1 x2 s1 s2 s3 s4 A1 A2
0 s1 30 1 2 1 0 0 0 0 0
0 s2 72 4 4 0 1 0 0 0 0
–M A1 5 1 0 0 0 –1 0 1 0
–M A2 12 0 1 0 0 0 –1 0 1
zj –17M –M –M 0 0 M M –M –M
cj – zj M + 40 M + 60 0 0 –M –M 0 0
cj Basic 40 60 0 0 0 0 –M
Variables Quantity x1 x2 s1 s2 s3 s4 A1
0 s1 6 1 0 1 0 0 2 0
0 s2 24 4 0 0 1 0 4 0
–M A1 5 1 0 0 0 –1 0 1
60 x2 12 0 1 0 0 0 –1 0
zj –5M + 720 –M 60 0 0 M –60 –M
cj – zj M + 40 0 0 0 –M 60 0
cj Basic 40 60 0 0 0 0
Variables Quantity x1 x2 s1 s2 s3 s4
0 s1 1 0 0 1 0 1 2
0 s2 4 0 0 0 1 4 4
40 x1 5 1 0 0 0 –1 0
60 x2 12 0 1 0 0 0 –1
zj 920 40 60 0 0 –40 –60
cj – zj 0 0 0 0 40 60
(continued)
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 287
cj Basic 40 60 0 0 0 0
Variables Quantity x1 x2 s1 s2 s3 s4
0 s4 0 0 0 1 –1/4 0 1
0 s3 1 0 0 –4 1/2 1 0
40 x1 6 1 0 –1 1/2 0 0
60 x2 12 0 1 1 –1/4 0 0
zj 960 40 60 20 5 0 0
cj – zj 0 0 –20 –5 0 0
Optimal
32.
cj Basic 1 5 0 0 0 –M
Variables Quantity x1 x2 s1 s2 s3 A1
–M A1 25 5 5 –1 0 0 1
0 s2 16 2 4 0 1 0 0
0 s3 5 1 0 0 0 1 0
zj –25M –5M –5M M 0 0 –M
cj – zj 5M + 1 5M + 5 –M 0 0 0
cj Basic 1 5 0 0 0 –M
Variables Quantity x1 x2 s1 s2 s3 A1
–M A1 5 5/2 0 –1 –5/4 0 1
5 x2 4 1/2 1 0 1/4 0 0
0 s3 5 1 0 0 0 1 0
zj –5M + 20 –5M/2 + 5/2 5 M –5M/4 + 5/4 0 –M
cj – zj 5M/2 – 3/2 0 –M 5M/4 – 5/4 0 0
(continued)
288
cj Basic 40 60 0 0 0 0
Variables Quantity x1 x2 s1 s2 s3 s4
0 s4 1/2 0 0 1/2 0 1/2 1Tie
0 s2 2 0 0 –2 1 2 0
40 x1 5 1 0 0 0 –1 0
60 x2 25/2 0 1 1/2 0 1/2 0
zj 950 40 60 30 0 –10 0
cj – zj 0 0 –30 0 10 0
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 288
289
cj Basic 1 5 0 0 0
Variables Quantity x1 x2 s1 s2 s3
1 x1 2 1 0 –2/5 –1/2 0
5 x2 3 0 1 1/5 1/2 0
0 s3 3 0 0 2/5 1/2 1
zj 17 1 5 3/5 2 0
cj – zj 0 0 –3/5 –2 0
Optimal
33.
cj Basic 3 6 0 0 0 0 M
Variables Quantity x1 x2 s1 s2 s3 s4 A1
0 s1 18 3 2 1 0 0 0 0
M A1 5 1 1 0 –1 0 0 1
0 s3 4 1 0 0 0 1 0 0
0 s4 7 0 1 0 0 0 1 0
zj 5M M M 0 –M 0 0 M
zj – cj M – 3 M – 6 0 –M 0 0 0
cj Basic 3 6 0 0 0 0 M
Variables Quantity x1 x2 s1 s2 s3 s4 A1
0 s1 6 0 2 1 0 –3 0 0
M A1 1 0 1 0 –1 –1 0 1
3 x1 4 1 0 0 0 1 0 0
0 s4 7 0 1 0 0 0 1 0
zj M + 12 3 M 0 –M –M + 3 0 M
zj – cj 0 M – 6 0 –M –M + 3 0 0
cj Basic 3 6 0 0 0 0
Variables Quantity x1 x2 s1 s2 s3 s4
0 s1 4 0 0 1 2 –1 0
6 x2 1 0 1 0 –1 –1 0
3 x1 4 1 0 0 0 1 0
0 s4 6 0 0 0 1 1 1
zj 18 3 6 0 –6 –3 0
zj – cj 0 0 0 –6 –3 0
Optimal
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 289
290
34.
cj Basic 10 5 0 0 –M –M
Variables Quantity x1 x2 s1 s2 A1 A2
–M A1 10 2 1 –1 0 1 0
–M A2 4 0 1 0 0 0 1
0 s2 20 1 4 0 1 0 0
zj –14M –2M –2M M 0 –M –M
cj – zj 2M + 10 2M + 5 –M 0 0 0
cj Basic 10 5 0 0 –M
Variables Quantity x1 x2 s1 s2 A2
10 x1 5 1 1/2 –1/2 0 0
–M A2 4 0 1 0 0 1
0 s2 15 0 7/2 1/2 1 0
zj –4M + 50 10 –M + 5 –5 0 –M
cj – zj 0 M 5 0 0
cj Basic 10 5 0 0
Variables Quantity x1 x2 s1 s2
10 x1 3 1 0 –1/2 0
5 x2 4 0 1 0 0
0 s2 1 0 0 1/2 1
zj 50 10 5 –5 0
cj – zj 0 0 5 0
cj Basic 10 5 0 0
Variables Quantity x1 x2 s1 s2
10 x1 4 1 0 0 1
5 x2 4 0 1 0 0
0 s1 2 0 0 1 2
zj 60 10 5 0 10
cj – zj 0 0 0 –10
Optimal
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 290
291
35.
cj Basic 1 2 –1 0 0 0
Variables Quantity x1 x2 x3 s1 s2 s3
0 s1 40 0 4 1 1 0 0
0 s2 20 1 –1 0 0 1 0
0 s3 60 2 4 3 0 0 1
zj 0 0 0 0 0 0 0
cj – zj 1 2 –1 0 0 0
cj Basic 1 2 –1 0 0 0
Variables Quantity x1 x2 x3 s1 s2 s3
2 x2 10 0 1 1/4 1/4 0 0
0 s2 30 1 0 1/4 1/4 1 0
0 s3 20 2 0 2 –1 0 1
zj 20 0 2 1/2 1/2 0 0
cj – zj 1 0 –3/2 –1/2 0 0
cj Basic 1 2 –1 0 0 0
Variables Quantity x1 x2 x3 s1 s2 s3
2 x2 10 0 1 1/4 1/4 0 0
0 s2 30 0 0 –3/4 3/4 1 –1/2
1 x1 10 1 0 1 –1/2 0 1/2
zj 30 1 2 3/2 0 0 1/2
cj – zj 0 0 –5/2 0 0 –1/2
Multiple optimum solution
Alternate solution:
cj Basic 1 2 –1 0 0 0
Variables Quantity x1 x2 x3 s1 s2 s3
2 x2 10/3 0 1 1/2 0 –1/3 1/6
0 s1 80/3 0 0 –1 1 4/3 –2/3
1 x1 70/3 1 0 1/2 0 2/3 1/6
zj 30 1 2 3/2 0 0 1/2
cj – zj 0 0 –5/2 0 0 –1/2
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 291
292
36.
cj Basic 1 2 2 0 0 –M –M
Variables Quantity x1 x2 x3 s1 s2 A1 A2
0 s1 12 1 1 2 1 0 0 0
–M A1 20 2 1 5 0 0 1 0
–M A2 8 1 1 –1 0 –1 0 1
zj –28M –3M –2M –4M 0 M –M –M
cj – zj 1 + 3M 2 + 2M 2 + 4M 0 –M 0 0
cj Basic 1 2 2 0 0 –M
Variables Quantity x1 x2 x3 s1 s2 A2
0 s1 4 1/5 3/5 0 1 0 0
2 x3 4 2/5 1/5 1 0 0 0
–M A2 12 7/5 6/5 0 0 –1 1
zj –12M + 8 –4/5 – 7M/5 2/5 – 6M/5 2 0 M –M
cj – zj 1/5 + 7M/5 8/5 + 6M/5 0 0 –M 0
cj Basic 1 2 2 0 0
Variables Quantity x1 x2 x3 s1 s2
0 s1 16/7 0 3/7 0 1 1/7
2 x3 4/7 0 –1/7 1 0 2/7
1 x1 60/7 1 6/7 0 0 –5/7
zj 68/7 1 4/7 2 0 –1/7
cj – zj 0 10/7 0 0 1/7
cj Basic 1 2 2 0 0
Variables Quantity x1 x2 x3 s1 s2
2 x2 16/3 0 1 0 7/3 1/3
2 x3 4/3 0 0 1 1/3 1/3
1 x1 4 1 0 0 –2 –1
zj 52/3 1 2 2 10/3 1/3
cj – zj 0 0 0 –10/3 –1/3
Optimal
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 292
293
cj Basic 400 350 450 0 0 0 0
Variables Quantity x1 x2 x3 s1 s2 s3 s4
0 s1 120 2 3 2 1 0 0 0
0 s2 160 4 3 1 0 1 0 0
0 s3 100 3 2 4 0 0 1 0
0 s4 40 1 1 1 0 0 0 1
zj 0 0 0 0 0 0 0 0
cj – zj 400 350 450 0 0 0 0
37.
cj Basic 400 350 450 0 0 0 0
Variables Quantity x1 x2 x3 s1 s2 s3 s4
0 s1 70 1/2 2 0 1 0 –1/2 0
0 s2 135 13/4 5/2 0 0 1 –1/4 0
450 x3 25 3/4 1/2 1 0 0 1/4 0
0 s4 15 –1/4 1/2 0 0 0 –1/4 1
zj 11,250 1,350/4 450/2 450 0 0 450/4 0
cj – zj 250/4 250/2 0 0 0 –450/4 0
cj Basic 400 350 450 0 0 0 0
Variables Quantity x1 x2 x3 s1 s2 s3 s4
0 s1 10 –1/2 0 0 1 0 1/2 –4
0 s2 60 2 0 0 0 1 1 –5
450 x3 10 1/2 0 1 0 0 1/2 –1
350 x2 30 1/2 1 0 0 0 –1/2 2
zj 15,000 400 350 450 0 0 50 250
cj – zj 0 0 0 0 0 –50 –250
Multiple optimum solution at x1
Alternate solution:
cj Basic 400 350 450 0 0 0 0
Variables Quantity x1 x2 x3 s1 s2 s3 s4
0 s1 20 0 0 1 1 0 1 –5
0 s2 20 0 0 –4 0 1 –1 –1
400 x1 20 1 0 2 0 0 1 –2
350 x2 20 0 1 –1 0 0 –1 3
zj 15,000 400 350 450 0 0 50 250
cj – zj 0 0 0 0 0 –50 –250
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 293
39. (a)
0 1 2 3 4 5 6x
x
1
Unbounded solution
Z1
2
3
4
5
2
-1-2-3-4
294
38. (a)
0 1 2 3 4 5 6 7x
x
1
Infeasible-no common solution space
1
2
3
4
5
2
(b)
cj Basic 3 2 0 0 –M
Variables Quantity x1 x2 s1 s2 A1
0 s1 1 1 1 1 0 0
–M A1 2 1 1 0 –1 1
zj –2M –M –M 0 M –M
cj – zj M + 3 M + 2 0 –M 0
cj Basic 3 2 0 0 –M
Variables Quantity x1 x2 s1 s2 A1
3 x1 1 1 1 1 0 0
–M A1 1 0 0 –1 –1 1
zj 3 – M 3 3 M M –M
cj – zj 0 –1 –M –M 0
Infeasible solution
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 294
295
(b)
cj Basic 1 1 0 0
Variables Quantity x1 x2 s1 s2
0 s1 1 –1 1 1 0
0 s2 4 –1 2 0 1
zj 0 0 0 0 0
cj – zj 1 1 0 0
Tie for entering variable; if x1 is chosen, the solution is unbounded. Select x2 arbitrarily.
cj Basic 1 1 0 0
Variables Quantity x1 x2 s1 s2
1 x2 1 –1 1 1 0
0 s2 3 1 0 –1 1
zj 1 –1 1 1 0
cj – zj 2 0 –1 0
cj Basic 1 1 0 0
Variables Quantity x1 x2 s1 s2
1 x2 4 0 1 0 1
1 x1 3 1 0 –1 1
zj 7 1 1 –1 2
cj – zj 0 0 1 –2
Unbounded; no pivot row available
40.
cj Basic 7 5 5 0 0 0 0
Variables Quantity x1 x2 x3 s1 s2 s3 s4
0 s1 25 1 1 1 1 0 0 0
0 s2 40 2 1 1 0 1 0 0
0 s3 25 1 1 0 0 0 1 0
0 s4 6 0 0 1 0 0 0 1
zj 0 0 0 0 0 0 0 0
cj – zj 7 5 5 0 0 0 0
(continued)
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 295
Alternate Solution:
cj Basic 7 5 5 0 0 0 0
Variables Quantity x1 x2 x3 s1 s2 s3 s4
5 x2 4 0 1 0 2 –1 0 –1
7 x1 15 1 0 0 –1 1 0 0
0 s3 6 0 0 0 –1 0 1 1
5 x3 6 0 0 1 0 0 0 1
zj 155 7 5 5 3 2 0 0
cj – zj 0 0 0 –3 –2 0 0
296
cj Basic 7 5 5 0 0 0 0
Variables Quantity x1 x2 x3 s1 s2 s3 s4
0 s1 5 0 1/2 1/2 1 –1/2 0 0
7 x1 20 1 1/2 1/2 0 1/2 0 0
0 s3 5 0 1/2 –1/2 0 –1/2 1 0
0 s4 6 0 0 0 0 0 0 1
zj 140 7 7/2 7/2 0 7/2 0 0
cj – zj 0 3/2 3/2 0 –7/2 0 0
Tie
cj Basic 7 5 5 0 0 0 0
Variables Quantity x1 x2 x3 s1 s2 s3 s4
5 x2 10 0 1 1 2 –1 0 0
7 x1 15 1 0 0 –1 1 0 0
0 s3 0 0 0 –1 –1 0 1 0
0 s4 6 0 0 1 0 0 0 1
zj 155 7 5 5 3 2 0 0
cj – zj 0 0 0 –3 –2 0 0
Multiple optimum (continued)
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 296
297
41.
cj Basic 15 25 0 0 0 M M
Variables Quantity x1 x2 s1 s2 s3 A1 A2
M A1 12 3 4 –1 0 0 1 0
M A2 6 2 1 0 –1 0 0 1
0 s3 9 3 2 0 0 1 0 0
zj 18M 5M 5M –M –M 0 M M
zj – cj 5M – 15 5M – 25 –M –M 0 0 0
cj Basic 15 25 0 0 0 M
Variables Quantity x1 x2 s1 s2 s3 A1
M A1 3 0 5/2 –1 1 0 1
15 x1 3 1 1/2 0 0 0 0
0 s3 0 0 1/2 0 0 1 0
zj 3M + 45 15 5M/2 + 15/2 –M 3M/2 – 15/2 0 M
zj – cj 0 5M/2 – 35/2 –M 3M/2 – 15/2 0 0
cj Basic 15 25 0 0 0 M
Variables Quantity x1 x2 s1 s2 s3 A1
M A1 3 0 0 –1 –6 –5 1
15 x1 3 1 0 0 –2 –1 0
25 x2 0 0 1 0 3 2 0
zj 3M + 45 15 15 –M –6M +45 –5M +45 M
zj – cj 0 0 –M –6M +45 –5M +45 0
Infeasible solution
42. a).minimize Zd = 90y1 + 60y2subject to
y1 + 2y2 ≥ 64y1 + 2y2 ≥ 10
y1, y2 ≥ 0
b) y1 = the marginal value of one additional lb of brass = $1.33
y2 = the marginal value of one additional hrof labor = $2.33
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 297
298
cj Basic 6 + ∆ 10 0 0
Variables Quantity x1 x2 s1 s2
10 x2 20 0 1 1/3 –1/6
6 + ∆ x1 10 1 0 –1/3 2/3
zj 260 + 10∆ 6 + ∆ 10 4/3 – ∆/3 7/3 + 2∆/3
cj – zj 0 0 –4/3 + ∆/3 –7/3 – 2∆/3
Solving for the cj – zj inequalities:
–4/3 + ∆/3 ≤ 0∆/3 ≤ 4/3
∆ ≤ 4
Since c1 = 6 + ∆; ∆ = c1 – 6. Thus
c1 – 6 ≤ 4c1 ≤ 10
–7/3 – 2∆/3 ≤ 0–2∆/3 ≤ 7/3
–2∆ ≤ 7∆ ≥ –7/2
Since c1 = 6 + ∆; ∆ = c1 – 6. Thus
c1 – 6 ≥ –7/2c1 ≥ 5/2
Summarizing, 5/2 ≤ c1 ≤ 10.
c2, basic:
cj Basic 6 10 + ∆ 0 0
Variables Quantity x1 x2 s1 s2
10 + ∆ x2 20 0 1 1/3 –1/6
6 x1 10 1 0 –1/3 2/3
zj 280 + 20∆ 6 10 + ∆ 4/3 + ∆/3 7/3 – ∆/6
cj – zj 0 0 –4/3 – ∆/3 –7/3 + ∆/6
c) c1, basic:
Solving for the cj – zj inequalities:
–4/3 – ∆/3 ≤ 0–∆/3 ≤ 4/3
∆ ≥ –4
Since c2 = 10 + ∆; ∆ = c2 – 10. Thus
c2 – 10 ≥ –4c2 ≥ 6
–7/3 + ∆/6 ≤ 0∆/6 ≤ 7/3
∆ ≤ 14
Since c2 = 10 + ∆; ∆ = c2 – 10. Thus
c2 – 10 ≤ 14c2 ≤ 24
Summarizing, 6 ≤ c2 ≤ 24.
d) q1:
x2: 20 + ∆/3 ≥ 0 x1: 10 – ∆/3 ≥ 0∆/3 ≥ –20 –∆/3 ≥ –10
∆ ≥ –60 ∆ ≤ 30
Therefore, –60 ≤ ∆ ≤ 30. Since
q1 = 90 + ∆∆ = q1 – 90
–60 ≤ q1 – 90 ≤ 3030 ≤ q1 ≤ 120
q2:
x2: 20 – ∆/6 ≥ 0 x1: 10 + 2∆/3 ≥ 0– ∆/6 ≥ –20 2∆/3 ≥ –10
∆ ≤ 120 ∆ ≥ –15
Therefore, –15 ≤ ∆ ≤ 120. Since
q2 = 60 + ∆∆ = q2 – 60
–15 ≤ q2 – 60 ≤ 12045 ≤ q2 ≤ 180
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 298
299
e) The marginal value of 1 hr of labor is $2.33.From part d, the sensitivity range for q2, labor,is 45 ≤ q2 ≤ 180. Thus, the company wouldpurchase up to 180 hr at the marginal valueprice.
43. a) Minimize Zd = 500y1 + 800y2subject to
10y1 + 34y2 ≥ 20050y1 + 20y2 ≥ 300
y1, y2 ≥ 0
y1 = $4.13 = the marginal value of oneadditional lb of chili beans; y2 = $4.67 = the marginal value of one additional lb of ground beef.
b)
Point C must become the optimal solution for x2 = 0; therefore the slope of the objectivefunction must be greater than the slope of theconstraint for ground beef, –34/20. Solving thefollowing for the profit, p, of Razorback chiliyields
–p/300 = –34/20p = $510
Thus, if the profit of Razorback chili is greater than $510, no Longhorn chili will be produced.The new optimal solution will be x1 = 23.5 andx2 = 0.
c)
The constraint line for chili beans rotates,creating a new, smaller solution space, and theoptimal solution shifts from point B to point B´where x1 = 21.43 and x2 = 3.57.
0 10 20 30 40 50x
x
1C
BA
10
20
30
40
50
60
2
60
0 10 20 30 40 50x
x
1C
BB´
A10
20
30
40
50
60
2
60
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 299
300
cj Basic 200 + ∆ 300 0 0
Variables Quantity x1 x2 s1 s2
300 x2 6 0 1 17/750 –1/150
200 + ∆ x1 20 1 0 –1/75 1/30
zj 5,800 + 20∆ 200 + ∆ 300 310/75 – ∆/75 70/15 + ∆/30
cj – zj 0 0 –310/75 + ∆/75 –70/15 – ∆/30
d) c1, basic:
Solving for the cj – zj inequalities:
–310/75 + ∆/75 ≤ 0∆/75 ≤ 3310/75
∆ ≤ 310
Since c1 = 200 + ∆; ∆ = c1 – 200. Thus
c1 – 200 ≤ 310c1 ≤ 510
–70/15 – ∆/30 ≤ 0–∆/30 ≤ 70/15
∆ ≥ –140
Since c1 = 200 + ∆; ∆ = c1 – 200. Thus
c1 – 200 ≥ –140c1 ≥ 60
Summarizing, 60 ≤ c1 ≤ 510.
c2, basic:
Since c2 = 300 + ∆; ∆ = c2 – 300. Thus
c2 – 300 ≤ 700c2 ≤ 1,000
Summarizing, 117.65 ≤ c2 ≤ 1,000.
e) q1:
x2: 6 + 17∆/750 ≥ 0 x1: 20 – ∆/75 ≥ 017∆/750 ≥ –6 –∆/75 ≥ –20
∆ ≥ –264.7 ∆ ≤ 1,500
Therefore, –264.7 ≤ ∆ ≤1,500. Since
q1 = 500 + ∆∆ = q1 – 500
–264.7 ≤ q1 – 500 ≤ 1,500235.3 ≤ q1 ≤ 2,000
cj Basic 200 300 + ∆ 0 0
Variables Quantity x1 x2 s1 s2
300 + ∆ x2 6 0 1 17/750 –1/150
200 x1 20 1 0 –1/75 1/30
zj 5,800 + 6∆ 200 300 + ∆ 310/75 + 17∆/750 70/15 – ∆/150
cj – zj 0 0 –310/75 – 17∆/750 –70/15 + ∆/150
Solving for the cj – zj inequalities:
–310/75 – 17∆/750 ≤ 0–17∆/750 ≤ 310/75
∆ ≥ –182.35
Since c2 = 300 + ∆; ∆ = c2 – 300. Thus
c2 – 300 ≥ –182.35c2 ≥ 117.65
–70/15 + ∆/150 ≤ 0∆/150 ≤ 70/15
∆ ≤ 700
q2:
x2: 6 – ∆/150 ≥ 0 x1: 20 + ∆/30 ≥ 0– ∆/150 ≥ –6 ∆/30 ≥ –20
∆ ≤ 900 ∆ ≥ –600
Therefore, –600 ≤ ∆ ≤ 900. Since
q2 = 800 + ∆∆ = q2 – 800
–600 ≤ q2 – 800 ≤ 900200 ≤ q2 ≤ 1,700
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 300
301
f) The marginal value of 1 lb of chili beans is$4.13. The sensitivity range for q1, chili beans,is 235.3 ≤ q1q ≤ 2,000. Thus, the companywould purchase up to 2,000 lb at the marginalvalue price.
g) Groundbeef
h) No effect
44. a) Minimize Zd = 60y1 + 40y2subject to
12y1 + 4y2 ≥ 94y1 + 8y2 ≥ 7
y1, y2 ≥ 0
b) y1 = the marginal value of one additional hr ofprocess 1; y2 = the marginal value of oneadditional hr of process 2
For the s1 column, the cj – zj value of $.55 is the marginal value of 1 hr of process 1production time. For the s2 column, the cj – zjvalue of $0.60 is the marginal value of 1 hr ofprocess 2 production time.
cj Basic 9 + ∆ 7 0 0
Variables Quantity x1 x2 s1 s2
9 + ∆ x1 4 1 0 1/10 –1/20
7 x2 3 0 1 –1/20 3/20
zj 57 + 4∆ 9 + ∆ 7 11/20 – ∆/10 12/20 + ∆/20
cj – zj 0 0 –11/20 – ∆/10 –12/20 + ∆/20
c) cj, basic:
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 301
302
Solving for the cj – zj inequalities:
–11/20 – ∆/10 ≤ 0–∆/10 ≤ 11/20
–∆ ≤ 11/2∆ ≥ –11/2
Since c1 = 9 + ∆ , ∆ = c1 – 9. Thus
c1 – 9 ≥ 11/2c1 ≥ 7/2
–12/20 + ∆/20 ≤ 0∆/20 ≤ 12/20
∆ ≤ 12
Since c1 = 9 + ∆ , ∆ = c1 – 9. Thus
c1 – 9 ≤ 12c1 ≤ 12
Summarizing, 7/2 ≤ c1 ≤ 12.
c2, basic:
Since c2 = 7 + ∆; ∆ = c2 – 7. Thus
c2 – 7 ≥ –4c2 ≥ 3
Summarizing, 3 ≤ c2 ≤ 18.
d) q1:
x1: 4 + ∆/10 ≥ 0 x2: 3 – ∆/20 ≥ 0∆/10 ≥ –4 –∆/20 ≥ –3
–∆ ≥ –80 –∆ ≥ –60∆ ≥ –40 ∆ ≤ 60
Therefore, –40 ≤ ∆ ≤60. Since
q1 = 60 + ∆∆ = q1 – 60
–40 ≤ q1 – 60 ≤ 6020 ≤ q1 ≤ 120
cj Basic 9 7 + ∆ 0 0
Variables Quantity x1 x2 s1 s2
9 x1 4 1 0 1/10 –1/20
7 + ∆ x2 3 0 1 –1/20 3/20
zj 57 + 3∆ 9 7 + ∆ 11/20 – ∆/20 12/20 + 3∆/20
cj – zj 0 0 –11/20 + ∆/20 –12/20 – 3∆/20
Solving for the cj – zj inequalities:
–11/20 + ∆/20 ≤ 0∆/20 ≤ 11/20
∆ ≤ 11
Since c2 = 7 + ∆; ∆ = c2 – 7. Thus
c2 – 7 ≤ 11c2 ≤ 18
–12/20 – 3∆/20 ≤ 0–3∆/20 ≤ 12/20
∆ ≥ –4
q2:
x1: 4 – ∆/20 ≥ 0 x2: 3 +3 ∆/20 ≥ 0–∆/20 ≥ –4 3∆/20 ≥ –3
–∆ ≥ –80 ∆ ≥ –20∆ ≤ 80
Therefore, –20 ≤ ∆ ≤ 80. Since
q2 = 40 + ∆∆ = q2 – 40
–20 ≤ q2 – 40 ≤ 8020 ≤ q2 ≤ 120
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 302
303
e) The marginal value of 1 hr of process 1production time is $.55. The sensitivity rangefor q1, production hours, is 20 ≤ q1 ≤ 120.Thus, the company would purchase up to 120hr at the marginal value price.
45. a) Minimize Zd = 180y1 + 135y2subject to
2y1 + 3y2 ≥ 2005y1 + 3y2 ≥ 300
y1, y2 ≥ 0
b) y1 = $33.33 = the marginal value of an additional hr of labor; y2 = $44.44 = themarginal value of an additional bd. ft. of wood
c)
Point A must become the optimal solution forx1 = 0; therefore the slope of the objectivefunction must be less than the slope of theconstraint for labor, –2/5. Solving the followingfor the profit, p, of coffee tables gives
–200/p = –2/5p = $500
Thus, if the profit for coffee tables is greaterthan $500, no end tables will be produced. Thenew optimal solution will be x1 = 0 and x2 = 36.
d)
The constraint line for wood moves outward,creating a new solution space, and the optimalsolution point shifts from point B to point B´where x1 = 31.67 and x2 = 23.33.0 10 20 30 40 50 60 70
x
x
1C
BA
10
20
30
40
50
60
2
80 90 100
0 10 20 30 40 50 60 70x
x
1C C´
B B´
A
10
20
30
40
50
60
2
80 90 100
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 303
304
cj Basic 200 + ∆ 300 0 0
Variables Quantity x1 x2 s1 s2
300 x2 30 0 1 1/3 –2/9
200 + ∆ x1 15 1 0 –1/3 5/9
zj 12,000 + 15∆ 200 + ∆ 300 100/3 – ∆/3 400/9 + 5∆/9
cj – zj 0 0 –100/3 + ∆/3 –400/9 – 5∆/9
e) c1, basic:
Solving for the cj – zj inequalities:
–100/3 + ∆/3 ≤ 0∆/3 ≤ 100/3
∆ ≤ 100
Since c1 = 200 + ∆; ∆ = c1 – 200. Thus
c1 – 200 ≤ 100c1 ≤ 300
–400/9 – 5∆/9 ≤ 0–5∆/9 ≤ 400/9
–∆ ≤ 80∆ ≥ –80
Since c1 = 200 + ∆; ∆ = c1 – 200. Thus
c1 – 200 ≥ –80c1 ≥ 120
Summarizing, 120 ≤ c1 ≤ 300.
c2, basic:
Since c2 = 300 + ∆; ∆ = c2 – 300. Thus
c2 – 300 ≤ 200c2 ≤ 500
Summarizing, 200 ≤ c2 ≤ 500.
f) q1:
x2: 30 + ∆/3 ≥ 0 x1: 15 – ∆/3 ≥ 0∆/3 ≥ –30 –∆/3 ≥ –15
∆ ≥ –90 –∆ ≤ –45∆ ≤ 45
Therefore, –90 ≤ ∆ ≤ 45. Since
q1 = 180 + ∆∆ = q1 – 180
–90 ≤ q1 – 180 ≤ 4590 ≤ q1 ≤ 225
cj Basic 200 300 + ∆ 0 0
Variables Quantity x1 x2 s1 s2
300 + ∆ x2 30 0 1 1/3 –2/9
200 x1 15 1 0 –1/3 5/9
zj 12,000 + 300∆ 200 300 + ∆ 100/3 + ∆/3 400/9 – 2∆ /9
cj – zj 0 0 –100/3 – ∆/3 –400/9 + 2∆ /9
Solving for the cj – zj inequalities:
–100/3 – ∆/3 ≤ 0–∆/3 ≤ 100/3
–∆ ≤ 100∆ ≥ –100
Since c2 = 300 + ∆; ∆ = c2 – 300. Thus
c2 – 300 ≥ –100c2 ≥ 200
–400/9 + 2∆/9 ≤ 02∆/9 ≤ 400/9
∆ ≤ 200
q2:
x2: 30 – 2∆/9 ≥ 0 x1: 15 + 5∆/9 ≥ 0–2∆/9 ≥ –30 5∆/9 ≥ –15
–∆ ≥ –135 ∆ ≥ –27∆ ≤ 135
Therefore, –27 ≤ ∆ ≤ 135. Since
q2 = 135 + ∆∆ = q2 – 135
–27 ≤ q2 – 135 ≤ 135108 ≤ q2 ≤ 270
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 304
305
g) The marginal value of 1 lb of wood is $44.44.From part f, the sensitivity range for q2, wood,is 108 ≤ q2 ≤ 270. Thus, the company wouldpurchase up to 270 bd. ft. of wood at themarginal value price.
g) The marginal value of labor is $33.33 and themarginal value of wood is $44.44; thus, woodshould be purchased.
46. a) Minimize Zd = 19y1 + 14y2 + 20y3subject to
2y1 + y2 + y3 ≥ 70y1 + y2 + 2y3 ≥ 80
y1, y2, y3 ≥ 0
b) y1 =$20 = the marginal value of an additionalhr of production time; y2 =$0 = the marginalvalue of an additional lb of steel; y3 = $30 = themarginal value of an additional ft of wire
cj Basic 70 + ∆ 80 0 0 0
Variables Quantity x1 x2 s1 s2 s3
70 + ∆ x1 6 1 0 2/3 0 –1/3
0 s2 1 0 0 –1/3 1 –1/3
80 x2 7 0 1 –1/3 0 2/3
zj 980 + 6∆ 70 + ∆ 80 20 + 2∆/3 0 30 – ∆/3
cj – zj 0 0 –20 – 2∆/3 0 –30 + ∆/3
c) c1, basic:
Solving for the cj – zj inequalities:
–20 – 2∆/3 ≤ 0–2∆/3 ≤ 20
–∆ ≤ 30∆ ≥ –30
Since c1 = 70 + ∆; ∆ = c1 – 70. Thus
c1 – 70 ≥ –30c1 ≥ 40
–30 + ∆/3 ≤ 0∆/3 ≤ 30
∆ ≤ 90
Since c1 = 70 + ∆; ∆ = c1 – 70. Thus
c1 – 70 ≤ 90c1 ≤ 160
Summarizing, 40 ≤ c1 ≤ 160.
c2, basic:
Solving for the cj – zj inequalities:
–20 + ∆/3 ≤ 0∆/3 ≤ 20
∆ ≥ 60
Since c2 = 80 + ∆; ∆ = c2 – 80. Thus
c2 – 80 ≤ 60c2 ≤ 140
–30 – 2∆/3 ≤ 0–2∆/3 ≤ 30
–∆ ≤ 45∆ ≥ –45
Since c2 = 80 + ∆; ∆ = c2 – 80. Thus
c2 – 80 ≥ –45c2 ≥ 35
Summarizing, 35 ≤ c2 ≤ 140.
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 305
306
cj Basic 70 80 + ∆ 0 0 0
Variables Quantity x1 x2 s1 s2 s3
70 x1 6 1 0 2/3 0 –1/3
0 s2 1 0 0 –1/3 1 –1/3
80 + ∆ x2 7 0 1 –1/3 0 2/3
zj 980 + 7∆ 70 80 + ∆ 20 – ∆/3 0 30 + 2∆/3
cj – zj 0 0 –20 + ∆/3 0 –30 – 2∆/3
d) q1:
x1: 6 + 2∆/3 ≥ 0 x2: 7 – ∆/3 ≥ 0 s2: 1 – ∆/3 ≥ 02∆/3 ≥ –6 –∆/3 ≥ –7 –∆/3 ≥ –1
∆ ≥ –9 –∆ ≥ –21 –∆ ≥ –3∆ ≤ 21 ∆ ≤ 3
Therefore, –9 ≤ ∆ ≤ 3. Since
q1 = 19 + ∆∆ = q1 – 19
–9 ≤ q1 – 19 ≤ 310 ≤ q1 ≤ 22
q2:
x1: 6 + 0∆ ≥ 0 x2: 7 + 0∆ ≥ 0 s2: 1 + ∆ ≥ 0∆ ≥ –1
Therefore, ∆ ≥ –1. Since
q2 = 14 + ∆∆ = q2 – 14
q2 – 14 ≥ –1q2 ≥ 13
q3:
x1: 6 – ∆/3 ≥ 0 x2: 7 + 2∆/3 ≥ 0 s2: 1 – ∆/3 ≥ 0–∆/3 ≥ –6 2∆/3 ≥ –7 –∆/3 ≥ –1
–∆ ≥ –18 ∆ ≥ –21/2 –∆ ≥ –3∆ ≤ 18 ∆ ≤ 3
Therefore, –21/2 ≤ ∆ ≤ 3. Since
q3 = 20 + ∆ –21/2 ≤ q3 – 20 ≤ 3∆ = q3 – 20 19/2 ≤ q3 ≤ 23
e) The sensitivity range for production hours is 10 ≤ q1 ≤ 22. Since 25 hr exceeds the upper limit of the range, it would change the optimal solution.
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 306
307
47. a) Minimize Zd = 64y1 + 50y2 + 120y3 + 7y4 + 7y5subject to
4y1 + 5y2 + 15y3 + y4 ≥ 98y1 + 5y2 + 8y3 + y5 ≥ 12
y1, y2, y3, y4, y5 ≥ 0
b) y1 = $.75 = the marginal value of one additionalhr of labor for process 1; y2 = $1.20 = the marginal value of one additional hr of labor for process 2; y3, y4, y5 = $0; these resources have no value since there were units available which were not used.
c) c1, basic:
Solving for the cj – zj inequalities:
–3/4 + ∆/4 ≤ 0 –6/5 – 2∆/5 ≤ 0∆/4 ≤ 3/4 –2∆/5 ≤ 6/5
∆ ≤ 3 ∆ ≥ –3
Since c1 = 9 + ∆; ∆ = c1 – 9. Thus
–3 ≤ ∆ ≤ 3–3 ≤ c1 – 9 ≤ 36 ≤ c1 ≤ 12
c2, basic:
Solving for the cj – zj inequalities:
–3/4 + ∆/4 ≤ 0 –6/5 + ∆/5 ≤ 0–∆/4 ≤ 3/4 ∆/5 ≤ 6/5
∆ ≥ –3 ∆ ≤ 6
Since c2 = 12 + ∆; ∆ = c1 – 12. Thus
–3 ≤ c2 – 12 ≤ 69 ≤ c2 ≤ 18
d) q1:
x1: 4 – ∆/4 ≥ 0 s5: 1 – ∆/4 ≥ 0 s3: 12 + 7∆/4 ≥ 0–∆/4 ≥ –4 –∆/4 ≥ –1 7∆/4 ≥ –12
∆ ≤ 16 ∆ ≤ 4 ∆ ≥ –6.86
s4: 3 + ∆/4 ≥ 0 x2: 6 + ∆/4 ≥ 0∆/4 ≥ –3 ∆/4 ≥ –6
∆ ≥ –12 ∆ ≥ –24
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 307
308
Summarizing,
–24 < –12 < –6.86 ≤ ∆ ≤ 4 ≤ 16
and, therefore,
–6.86 ≤ ∆ ≤ 4
Since q1 = 64 + ∆; ∆ = q1 – 64. Therefore,
–6.86 ≤ ∆ q1 – 64 ≤ 457.14 ≤ q1 ≤ 68
e) q3:
x1: 4 + 0∆ ≥ 0 s5: 1 + 0∆ ≥ 0 s3: 12 + ∆ ≥ 00∆ ≥ –4 ∆ ≤ ∞ ∆ ≥ –12∆ ≤ ∞
s4: 3 + 0∆ ≥ 0 x2: 6 + 0∆ ≥ 0∆ ≤ ∞ ∆ ≤ ∞
–12 ≤ ∆ ≤ ∞
Since q3 = 120 + ∆; ∆ = q3 – 120. Therefore,
–12 ≤ ∆ ≤ ∞–12 ≤ q3 – 120 ≤ ∞108 ≤ q3 ≤ ∞
Since 100 pounds is less than the lower limitof the range, the optimal solution mix will change. s2 enters the solution and s3 leaves. The new solution is,
x1 = 3.27s2 = 1.82s5 = 0.64s4 = 3.73x2 = 6.36Z = 105.82
48. a) Minimize Zd = 120y1 + 160y2 + 100y3 + 40y4subject to
2y1 + 4y2 + 3y3 + y4 ≥ 403y1 + 3y2 + 2y3 + y4 ≥ 352y1 + y2 + 4y3 + y4 ≥ 45
y1, y2, y3, y4 ≥ 0
b) y1, y2 = 0; y3 = $5 = the marginal value of 1 hrof operation 3 time; y4 = $25 = the marginal value of 1 ft2 of storage space
c) It does not have an effect. In the alternatesolution the dual values remain the same, i.e.,y3 = $5 and y4 = $25.
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 308
309
cj Basic 40 35 + ∆ 45 0 0 0 0
Variables Quantity x1 x2 x3 s1 s2 s3 s4
0 s1 10 –1/2 0 0 1 0 1/2 –4
0 s2 60 2 0 0 0 1 1 –5
45 x3 10 1/2 0 1 0 0 1/2 –1
35 + ∆ x2 30 1/2 1 0 0 0 –1/2 2
zj 1,500 + 35∆ 40 + ∆/2 35 + ∆ 45 0 0 5 – ∆/2 25 + 2∆
cj – zj –∆/2 0 0 0 0 –5 + ∆/2 –25 – 2∆
d) c2, basic:
Solving for the cj – zj inequalities:
–∆/2 ≤ 0∆ ≥ 0
Since c2 = 35 + ∆; ∆ = c2 – 35. Thus
c2 – 35 ≥ 0c2 ≥ 35
–5 + ∆/2 ≤ 0∆/2 ≤ 5
∆ ≤ 10
Since c2 = 35 + ∆; ∆ = c2 – 35. Thus
c2 – 35 ≤ 10c2 ≤ 45
–25 – 2∆ ≤ 0–2∆ ≤ 25
∆ ≥ –12.5
Since c2 = 35 + ∆; ∆ = c2 – 35. Thus
c2 – 35 ≥ –12.5c2 ≥ 22.5
Summarizing, 35 ≤ c2 ≤ 45.
e) q4:
s1: 10 – 4∆ ≥ 0 s2: 60 – 5∆ ≥ 0–4∆ ≥ –10 –5∆ ≥ –60
∆ ≤ 5/2 ∆ ≤ 12
x3: 10 – ∆ ≥ 10 x2: 30 + 2∆ ≥ 0–∆ ≥ –10 2∆ ≥ –30∆ ≤ 10 ∆ ≥ –15
Therefore, –15 ≤ ∆ ≤ 5/2. Since
q4 = 40 + ∆∆ = q4 – 40
–15 ≤ q4 – 40 ≤ 5/225 ≤ q4 ≤ 42.5
f) The marginal value of 1 ft2 of storage is $25.From part e, the sensitivity range for q4 is 25 ≤ q4 ≤ 42.5. Thus, the company wouldpurchase up to 42.5 ft2 of storage space at themarginal value price.
49. a) Maximize Zd = 20y1 + 30y2 + 12y3subject to
4y1 + 12y2 + 3y3 ≤ .035y1 + 3y2 + 2y3 ≤ .02
y1, y2, y3 ≥ 0
b) y1 = $0 = marginal value of 1 mg of protein;y2 = $0 = marginal value of 1 mg of iron; y3 = $.01 = marginal value of 1 mg ofcarbohydrate (i.e., if one less mg ofcarbohydrate was required, it would be worth$.01 to the dietitian)
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 309
310
cj Basic .03 + ∆ .02 0 0 0
Variables Quantity x1 x2 s1 s2 s3
.02 x2 3.6 0 1 0 .20 –.80
.03 + ∆ x1 1.6 1 0 0 –.13 .20
0 s1 4.4 0 0 1 .47 –3.2
zj .12 + 1.6∆ .03 + ∆ .02 0 0 – .133∆ –.01 + .2∆
zj – cj 0 0 0 0 – .133∆ –.01 + .2∆
c) c1, basic:
Solving for the zj – cj inequalities:
0 – .133∆ ≤ 0–.133∆ ≤ 0
–∆ ≤ 0∆ ≥ 0
Since c1 = .03 + ∆; ∆ = c1 – .03. Thus
c1 – .03 ≥ 0c1 ≥ .03
–.01 + .2∆ ≤ 0.2∆ ≤ .01
∆ ≤ .05
Since c1 = .03 + ∆; ∆ = c1 – .03. Thus
c1 – .03 ≤ .05c1 ≤ .08
Summarizing, .03 ≤ c1 ≤ .08.
c2, basic:
Since c2 = .02 + ∆; ∆ = c1 – .02. Thus
c2 – .02 ≤ 0c2 ≤ .02
–.01 + .8∆ ≤ 0.8∆ ≤ .01
∆ ≤ .0125
Since c2 = .02 + ∆; ∆ = c1 – .02. Thus
c2 – .02 ≤ .0125c2 ≥ .0075
Summarizing, c2 ≤ .0375.
d) When determining sensitivity ranges for qivalues in a minimization problem, sinceartificial variables are eliminated, the surplusvariable column coefficients must be used. Thiscorresponds to a qi – ∆ change.
cj Basic .03 .02 + ∆ 0 0 0
Variables Quantity x1 x2 s1 s2 s3
.02 + ∆ x2 3.6 0 1 0 .20 –.80
.03 x1 1.6 1 0 0 –.13 .20
0 s1 4.4 0 0 1 .47 –3.2
zj .12 + 3.6∆ .03 .02 + ∆ 0 0 + .2∆ –.01 + .8∆
zj – cj 0 0 0 0 + .2∆ –.01 + .8∆
Solving for the zj – cj inequalities:
0 + .2∆ ≤ 0.2∆ ≤ 0
∆ ≤ 0
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 310
311
q1:
x2: 3.6 + 0∆ ≥ 0 x1: 1.6 + 0∆ ≥ 0 s1: 4.4 + ∆ ≥ 0∆ ≥ –4.4
Therefore, ∆ ≥ –4.4. Since
q1 = 20 – ∆∆ = 20 – q1
20 – q1 ≥ –4.4q1 ≤ 24.4
q2:
x2: 3.6 + .2∆ ≥ 0 x1: 1.6 – .133∆ ≥ 0.2∆ ≥ –3.6 –.133∆ ≥ –1.6
∆ ≥ –18 –∆ ≥ –12∆ ≤ 12
s1: 4.4 + .47∆ ≥ 0.47∆ ≥ –4.4
∆ ≥ –9.36
Therefore, –9.36 ≤ ∆ ≤ 12. Since
q2 = 30 – ∆∆ = 30 – q2
–9.36 ≤ 30 – q2 ≤ 1218 ≤ q2 ≤ 39.36
q3:
x2: 3.6 – .8∆ ≥ 0 x1: 1.6 + .2∆ ≥ 0–.8∆ ≥ –3.6 .2∆ ≥ –1.6
–∆ ≥ –4.5 ∆ ≥ –8∆ ≤ 4.5
s1: 4.4 – 3.2∆ ≥ 0–3.2∆ ≥ –4.4
–∆ ≥ –1.375∆ ≤ –1.375
Therefore, –8 ≤ ∆ ≤ 1.375. Since
q3 = 12 – ∆∆ = 12 – q3
–8 ≤ 12 – q3 ≤ 1.37510.625 ≤ q3 ≤ 20
e) The marginal value of 1 mg of carbohydrates is $.01. From part d, the sensitivity range for q3, carbohydrates, is 10.625 ≤ q3 ≤ 20. Thus, the dietitian could lower the requirements for carbohydrates to10.625 at the marginal value without the solution becoming infeasible.
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 311
312
50. a) Minimize Zd = 1,200y1 + 500y3subject to
.50y1 + y2 ≥ 1.251.2y1 – y2 + y3 ≥ 2.00
.80y1 + y2 ≥ 1.75y1, y2, y3 ≥ 0
y1 = the marginal value of an additional hour ofproduction time
= $0y2 = the marginal value of increasing the
combined demand for cheese sandwichesby one sandwich
= $1.75y3 = the marginal value of producing an
additional ham salad sandwich= $3.75
b) c1, non-basic:
–5 + ∆ ≤ 0∆ ≤ .50
Since c1 = 1.25 + ∆; ∆ = c1 – 1.25. Therefore,
c1 – 1.25 ≤ .50c1 ≤ 1.75
c2, basic:
–3.75 – ∆ ≤ 0–∆ ≤ 3.75∆ ≥ –3.75
Since c2 = 2 + ∆; ∆ = c2 – 2. Therefore,
c2 –2 ≥ –3.75c2 ≥ –1.75
c3, basic:
–.5 – ∆ ≤ 0 –3.75 – ∆ ≤ 0–∆ ≤ .5 –∆ ≤ 3.75∆ ≥ –.5 ∆ ≥ –3.75
Summarizing,
–3.75 ≤ –.5 ≤ ∆
and, therefore,
–.5 ≤ ∆
Since c3 = 1.75 + ∆; ∆ = c3 – 1.75. Therefore,
–.5 ≤ ∆–.5 ≤ c3 – 1.75
1.25 ≤ c3
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 312
313
c) q3:
s1: 200 – 2∆ ≥ 0 x3: 500 + ∆ ≥ 0 x2: 500 + ∆ ≥ 0–2∆ ≥ –200 ∆ ≥ –500 ∆ ≥ –500
∆ ≤ 100
Summarizing,
–500 ≤ ∆ ≤ 100
Since q3 = 500 + ∆, ∆ = q3 – 500. Therefore,
–500 ≤ q3 – 500 ≤ 1000 ≤ q3 ≤ 600
d) The marginal value for demand for cheese sandwiches is $1.75. The range for q2 iscomputed as follows.
s1: 200 – .8∆ ≥ 0 x3: 500 + ∆ ≥ 0 x2: 500 + 0∆ ≥ 0–.8∆ ≥ –200 ∆ ≥ –500 0∆ ≥ –500
∆ ≤ 250 ∆ ≤ ∞
Summarizing,
–500 ≤ ∆ ≤ 250
Since q2 = 0 + ∆, ∆ = q2. Therefore,
–500 ≤ q2 ≤ 250
Thus, the demand for cheese sandwiches canbe increased up to a maximum of 250sandwiches.
The additional profit for 200 more cheesesandwiches would be,
($1.75) (200) = $350
Since the cost of advertising is $100, a $250profit would result, therefore the company should advertise.
51. a) c3, nonbasic:
–13 – ∆ ≤ 0–∆ ≤ 13
∆ ≥ –13
Since c3 = 2 + ∆, ∆ = c3 – 2. And
c3 – 2 ≥ –13c3 ≥ –11
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 313
314
c1, basic:
cj Basic 3 + ∆ 5 2 0 0 0
Variables Quantity x1 x2 x3 s1 s2 s3
0 s2 15 0 0 –4 –3/2 1 –1/2
3 + ∆ x1 5 1 0 –2 –1/2 0 1/2
5 x2 30 0 1 –1 –1/2 0 –1/2
zj 165 + 5∆ 3 + ∆ 5 –11 – 2∆ –4 – ∆/2 0 –1 + ∆/2
zj – cj 0 0 –13 – 2∆ –4 – ∆/2 0 –1 + ∆/2
Solving for the zj – cj inequalities:
–13 – 2∆ ≤ 0–2∆ ≤ 13
∆ ≥ –13/2
Since c1 = 3 + ∆, ∆ = c1 – 3. Thus
c1 – 3 ≥ –13/2c1 ≥ –7/2
–4 – 2∆ ≤ 0–2∆ ≤ 4
∆ ≥ –2
Since c1 = 3 + ∆, ∆ = c1 – 3. Thus
c1 – 3 ≥ –2c1 ≥ 1
–1 + ∆/2 ≤ 0∆/2 ≤ 1
∆ ≤ 2
Since c1 = 3 + ∆, ∆ = c1 – 3. Thus
c1 – 3 ≤ 2c1 ≤ 5
Summarizing, 1 ≤ c1 ≤ 5.
c2: basic:
Solving for the zj – cj inequalities:
–13 – 2∆ ≤ 0–2∆ ≤ 13
∆ ≥ –13/2
Since c2 = 5 + ∆, ∆ = c2 – 5. Thus
c2 – 5 ≥ –13/2c2 ≥ –3/2
–4 – ∆/2 ≤ 0–∆/2 ≤ 4
∆ ≥ –8
Since c2 = 5 + ∆, ∆ = c2 – 5. Thus
c2 – 3 ≥ –8c2 ≥ –3
–1 – ∆/2 ≤ 0–∆/2 ≤ 1
∆ ≥ –2
Since c2 = 5 + ∆, ∆ = c2 – 5. Thus
c2 – 5 ≥ –2c2 ≥ 3
Summarizing, c2 > 3.
cj Basic 3 5 + ∆ 2 0 0 0
Variables Quantity x1 x2 x3 s1 s2 s3
0 s2 15 0 0 –4 –3/2 1 –1/2
3 x1 5 1 0 –2 –1/2 0 1/2
5 + ∆ x2 30 0 1 –1 –1/2 0 –1/2
zj 165 + 30∆ 3 5 + ∆ –11 – 2∆ –4 – ∆/2 0 –1 – ∆/2
zj – cj 0 0 –13 – 2∆ –4 – ∆/2 0 –1 – ∆/2
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 314
315
q1:
s2: 15 – 3∆/2 ≥ 0 x1: 5 – ∆/2 ≥ 0 x2: 30 – ∆/2 ≥ 0–3∆/2 ≥ –15 –∆/2 ≥ –5 –∆/2 ≥ –30
∆ ≤ 10 ∆ ≤ 10 ∆ ≤ 60
Therefore, ∆ ≤ 10. Since
q1 = 35 – ∆∆ = 35 – q1
35 – q1 ≤ 10q1 ≥ 25
q2:
s2: 15 + ∆ ≥ 0 x1: 5 – 0∆ ≥ 0 x2: 30 – 0∆ ≥ 0∆ ≥ –15
Therefore, ∆ ≤ –15. Since
q2 = 50 – ∆∆ = 50 – q2
50 – q2 ≤ –15q2 ≤ 65
q3:
s2: 15 – ∆/2 ≥ 0 x1: 5 + ∆/2 ≥ 0 x2: 30 – ∆/2 ≥ 0–∆/2 ≥ –15 ∆/2 ≥ –5 –∆/2 ≥ –30
∆ ≤ 30 ∆ ≥ –10 ∆ ≤ 60
Therefore, –10 ≤ ∆ ≤ 30. Since q3 = 25 – ∆ and ∆ = 25 – q3
–10 ≤ 25 – q3 ≤ 30–5 ≤ q3 ≤ 35
b) When determining sensitivity ranges for qivalues, since artificial values are eliminated,the surplus variable column coefficients mustbe used. This corresponds to a qi – ∆ change.
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 315
316
52. a) y1 = 7/15 = $.467
Range for q1:
Since q1 = 320 + ∆,
∆ = q1 – 320–120 ≥ q1 – 320 ≤ 80200 ≤ q1 ≤ 400
As many as 400 pears can be purchased.
b) y2 = 1/15 = $.067
Range for q2:
x1: 8 + ∆/30 ≥ 0 x2: 16 + ∆/15 ≥ 0∆/30 ≥ –8 ∆/15 ≥ –16
∆ ≥ –240 ∆ ≥ –240
x3: 3 – ∆/10 ≥ 0 s4: 36 – ∆/30 ≥ 0–∆/10 ≥ –3 –∆/30 ≥ –36
∆ ≤ 30 ∆ ≤ 1,080
–240 ≤ ∆ ≤ 30
Since q2 = 400 + ∆,
∆ = q2 – 400–240 ≤ q2 – 400 ≤ 30160 ≤ q2 ≤ 430
Range over which the value of peaches is valid
c) Range for q3:
s3: 3 + ∆ ≥ 0∆ ≥ –3
Since q3 = 43 + ∆,
∆ = q3 – 43q3 – 43 ≥ –3
q3 ≥ 40
No, increasing q3 from 43 to 60 will not affectthe optimal solution.
x1: 8 + ∆/15 ≥ 0 x2: 16 – ∆/30 ≥ 0∆/15 ≥ –8 –∆/30 ≥ –16
∆ ≥ –120 ∆ ≤ 480
x3: 3 – ∆/40 ≥ 0∆ ≥ 120
s4: 36 – ∆/30 ≥ 0 s5: 8 – ∆/10 ≥ 0–∆/30 ≥ – 36 –∆/10 ≥ –8
∆ ≤ 1,080 ∆ ≤ 80–120 ≤ ∆ ≤ 80
d) Range for q5:
s5: 8 + ∆ ≥ 0∆ ≥ –8
Since q5 = 40 + ∆,
∆ = q5 – 40q5 – 40 ≥ –8
q5 ≥ 32
No, increasing q5 from 40 to 50 will have noaffect on the optimal solution.
e) Since y1 = 7/15 = $.467, pears should be secured.
53. a) y2 = $0, spruce has no marginal value
q2:
s2: 70 + ∆ ≥ 0∆ ≥ –70
Since q2 = 160 + ∆,
∆ = q2 – 160q2 – 160 ≥ –70
q2 ≥ 90
b) y3 = $2, marginal value of cutting hours
q3:
s1: 80 – 4∆/3 ≥ 0 s2: 70 + ∆/3 ≥ 0–4∆/3 ≥ –80 ∆/3 ≥ –70
∆ ≤ 60 ∆ ≥ –210
x3: 20 + 2∆/3 ≥ 0 x2: 10 – ∆/3 ≥ 02∆/3 ≥ –20 –∆/3 ≥ –10
∆ ≥ –30 ∆ ≤ 30
Therefore, –30 ≤ ∆ ≤ 30. Since q3 = 50 + ∆,
∆ = q3 – 5020 ≤ q3 ≤ 80
c) y3 = $2, cutting hours; y4 = $2, pressing hours.Since they both have the same marginal value,management could choose either.
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 316
317
d) From part a, q2 ≥ 90; thus, a decrease from 160to 100 lb of spruce will not affect the solution.
e) Compute the range for c1, a nonbasic cj value.
c1 = 4 + ∆–2 + ∆ ≤ 0
∆ ≤ 2c1 – 4 ≤ 2
c1 ≤ 6
The unit profit from Western paneling would have to be $6 or more before it would beproduced.
f) Compute the range for c3.
c3 = 8 + ∆
–2 – ∆/3 ≤ 0 –2 – 2∆/3 ≤ 0 –2 +∆/6 ≤ 0–∆/3 ≤ 2 –2∆/3 ≤ 2 ∆/6 ≤ 2
∆ ≥ –6 ∆ ≥ –3 ∆ ≤ 12
Since ∆ = c3 – 8,
c3 – 8 ≥ –6 c3 – 8 ≥ –3 c3 – 8 ≤ 12c3 ≥ 2 c3 ≥ 5 c3 ≤ 20
Summarizing, 5 ≤ c3 ≤ 20. If the unit profit ofColonial paneling is increased to $13, thepercent solution would not be affected.
54. y1 = $1.33
q1:
x4: 80 + 2∆/3 ≥ 0 x2: 40 – ∆/3 ≥ 02∆/3 ≥ –80 –∆/3 ≥ –40
∆ ≥ –120 ∆ ≤ 120
–120 ≤ ∆ ≤ 120
Since q1 = 200 + ∆,
∆ = q1 – 200–120 ≤ q1 – 200 ≤ 120
80 ≤ q1 ≤ 320
TaylMod-Aff.qxd 4/21/06 8:50 PM Page 317
