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Find the Max and Mini values of the objective function Z=3x+4y on the region bounded by 2x+y>5x+5y>162x+y<14 -x+4y<20Needs to be solved with the simplex method

Question

Find the Max and Mini values of the objective function Z=3x+4y on the region bounded by 

2x+y>5

x+5y>16

2x+y<14 

-x+4y<20

Needs to be solved with the simplex method 

check_circleAnswer
Step 1

First find the minimum value: so, the problem is,

 

Min Z = 3 x1 + 4 x2

 

Subject to

 

2x1 + x2 > 5

x1 + 5x2 > 16

2x1 + x2 < 14

-x1 + 4x2 < 20

 

 And x1, x2 0

 

Now, Max Z = -3 x1 – 4 x2

 

The problem is converted to canonical form by adding slack, surplus and artificial variable:

 

Max Z = -3 x1 - 4 x2 + 0 S1 + 0 S2 + 0 S3 +0 S4 -MA1-MA2

 

Subject to

 

2x1 + x2 - S1                     + A1            = 5

x1 + 5x2        - S2                      + A2 =16

2x1 + x2                      + S3                                    =14

-x1 + 4x2                      + S4                       =20

 

 And x1, x2, S1, S2, S3, S4, A1, A2 0

 

 

Iteration 1

.

0000-M
C,
3
-M
Min
S S SA
х,
В
С,
A
X
ratio
2
-1 0 0 0
0-1 0 0
001
001
м м 0 0 | -м| -М
A
— м 5
1
2
1
0
5
-M
16
1
0
3.2
А,
1
14
2
0
14
0
1
0
20
-1
4
0
0
5
0
0
Z =-21M
Z
-3М
-6M
-3 | -6М-4 | М | м| 0
-C, | -3М
0
0
Z
help_outline

Image Transcriptionclose

0000-M C, 3 -M Min S S SA х, В С, A X ratio 2 -1 0 0 0 0-1 0 0 001 001 м м 0 0 | -м| -М A — м 5 1 2 1 0 5 -M 16 1 0 3.2 А, 1 14 2 0 14 0 1 0 20 -1 4 0 0 5 0 0 Z =-21M Z -3М -6M -3 | -6М-4 | М | м| 0 -C, | -3М 0 0 Z

fullscreen
Step 2

Pivot element is 5 and leaving basis variable is A2 and entering variable is x2.

 

Iteration table 2:

00 M
C
4
0
х,
Min ratio
B
с,
х,
xS
х,
1.8
00
01
-M
1.8
1.8
0.2
1
1.8
3.2
-16
0.2
00
3.2
4
0.2
-0.2
1
х,
10.8
=6
1.8
00
10.8
1.8
0.2
0
0
1
0 0
0
7.2
-1.8
0.8
0
0
1
-1.8M+0.8 4 M-0.2M-0.8
Z1.8M+12.8
Z
0
0
-M
-3М -3
М -0.2М -0.8
Z -с
0
0
0
0
en
help_outline

Image Transcriptionclose

00 M C 4 0 х, Min ratio B с, х, xS х, 1.8 00 01 -M 1.8 1.8 0.2 1 1.8 3.2 -16 0.2 00 3.2 4 0.2 -0.2 1 х, 10.8 =6 1.8 00 10.8 1.8 0.2 0 0 1 0 0 0 7.2 -1.8 0.8 0 0 1 -1.8M+0.8 4 M-0.2M-0.8 Z1.8M+12.8 Z 0 0 -M -3М -3 М -0.2М -0.8 Z -с 0 0 0 0 en

fullscreen
Step 3

Pivot element is 1.8, entering = x1 and leaving basis is A1

 

Iteration ...

C
3
4
0
0
0
0
S
C,
S S
0 0
-0.2222 0 0
10
01
41.22220.5556 0 0
-1.2222-05556 0 0
B
х,
Min ratio
х, | х,
00.5556
01
3
0.1111
1
1
3
0.1111
27
4
х,
S
0
0
0
0
1
S
0
0
0
-1
1
Z-15
3
Z
Z-C
0
0
help_outline

Image Transcriptionclose

C 3 4 0 0 0 0 S C, S S 0 0 -0.2222 0 0 10 01 41.22220.5556 0 0 -1.2222-05556 0 0 B х, Min ratio х, | х, 00.5556 01 3 0.1111 1 1 3 0.1111 27 4 х, S 0 0 0 0 1 S 0 0 0 -1 1 Z-15 3 Z Z-C 0 0

fullscreen

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