For the LP max x = 3x1 + 5* 8.ti. anata 211 #1, 20 which has optimal tableau So B-¹ = a11 921 ≤ bi ≤ ba 012 €22 -188---88---:00 and B and and using B-¹ and B1¹b formulas (in reverse) fill in the missing values in original LP and optimal tableau 1 0 0 b₁ =; 6₂: and the optimal solution for z is z = 01 Xq 0 0 1 0 0 1 ܤ ܗܝ ܕ ܥܝܪ 82 rhs ◄ TA RESE Z

Elementary Algebra
17th Edition
ISBN:9780998625713
Author:Lynn Marecek, MaryAnne Anthony-Smith
Publisher:Lynn Marecek, MaryAnne Anthony-Smith
Chapter10: Quadratic Equations
Section10.5: Graphing Quadratic Equations
Problem 211E: For the revenue model in Exercise 10.205 and Exercise 10.209, explain what the x-intercepts mean to...
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For the LP
max z = 3x1 +5€
8.t.
which has optimal tableau
So B-1
11+12 bi
211 + a2 ≤ba
X1, X1 ≥ 0
011
21 =
and
-188-189-41:00
and using Band B-¹ b formulas (in reverse) fill in the missing values in original LP and optimal tableau
- 12:
€32
and B
b₁
and the optimal solution for z is z =
b₂ =
Z
1
0
0
01 Dz
0
0
1
0
0
1
81
82
rhs
Z
200
Transcribed Image Text:For the LP max z = 3x1 +5€ 8.t. which has optimal tableau So B-1 11+12 bi 211 + a2 ≤ba X1, X1 ≥ 0 011 21 = and -188-189-41:00 and using Band B-¹ b formulas (in reverse) fill in the missing values in original LP and optimal tableau - 12: €32 and B b₁ and the optimal solution for z is z = b₂ = Z 1 0 0 01 Dz 0 0 1 0 0 1 81 82 rhs Z 200
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