Linear Programming: Simplex MethodI attached 2nd photo so you could use structure of that tableau… it’s easier for me to understand. First convert to standard form though 3. Solve:Dobivor Minimize 2x₁ + 2x2 — 5x3offrwsubject to3x1 + 2x2 + 4x3 = 7and to eletx1 - x2 + 3x3 = 21850Of burebird 1, 2, 3 ≥ 0.(10 552 (28bao1T ZOITOUNTE985s vlasplo otisw 920hinja sil #3 Cont+ R₂10QR₁-5R2R3+10R₂5R₂R₁ +422R3 + 2 R₂1217(2 + ² R₁min -10X₁ - 6x2 +8x3St.bvjX₂X45510X 4XI1-1051Pela|--101bloO5 X ₁-2 X2 +6X3 + xy =110 X₁ + 4x₂ - 6x3 + x5X₁1 X21 X 31X4 X5 20X41X1 5/20X4 10X25/25X4 10/3X2 5/25X3 10/3X21525/3X₂-2니-6-2NOWYJNIN51-4Dto-2-X3 X4 X56O01O1O4-20109/209180OOOOOOO0-31209123.1- 23/201-1130 to 3305-3/1/20-bo-o01/12321-2/22= 20=30121/121ib2030O203Oi/c515/23/2451/3 16 35/31/4 15/23035)O3/2/4543 16 35/325½/225/3 170/3

Answered: Image /qna-images/answer/e6be64e4-324f-4c9b-a9a3-72b48b054bab.jpg

Answered: bivor Minimize 2x1 + 2x25x3 JITVI da Be…

bivor Minimize 2x1 + 2x25x3 JITVI da Be HIB subject to 3x1 + 2x2 + 4x3 = 7 ang jo plost x1 - x2 + 3x3 = 2 TREDISO TO X1, X2, X3 ≥ 0. oldo

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Linear Programming: Simplex Method I attached 2nd photo so you could use structure of that tableau… it’s easier for me to understand. First convert to standard form though

3. Solve:
Dobivor Minimize 2x₁ + 2x2 — 5x3
offrw
subject to
3x1 + 2x2 + 4x3 = 7
and to elet
x1 - x2 + 3x3 = 2
1850
Of burebird 1, 2, 3 ≥ 0.
(10 552 (28
bao1T ZOITOUNTE
985
s vlasplo otisw 920
hinja sil

#3 Cont
+ R₂
10
Q
R₁-5R2
R3+10R₂
5R₂
R₁ +422
R3 + 2 R₂
1217
(2 + ² R₁
min -10X₁ - 6x2 +8x3
St.
bv
jX₂
X45
510
X 4
XI
1-10
5
1
Pela
|--10
1
blo
O
5 X ₁-2 X2 +6X3 + xy =
110 X₁ + 4x₂ - 6x3 + x5
X₁1 X21 X 31
X4 X5 20
X41
X1 5/2
0
X4 10
X25/2
5
X4 10/3
X2 5/2
5
X3 10/3
X215
25/3
X₂
-2
니
-6
-2
NOWYJNIN
5
1-4
D
to
-2
-
X3 X4 X5
6
O
0
1
O
1
O
4
-2
0
1
0
9/20
9
1
80
OOOOOOO
0
-3
1
20
91
2
3.
1
- 23/20
1-1
1
30 to 3
30
5
-3/1/20
-bo-o
01/12
32
1
-2/22
= 20
=30
1
21/12
1
ib
20
30
O
20
3
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15/2
3/2
45
1/3 16 35/3
1/4 15/2
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5/3 170/3

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You didn't follow the same structure of the tableau i gave you in 2nd photo. Please refer back to using that structure... I don't understand this

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