Solve this LP max z=3x+2x2 +2.5x3 8.t. 1+2+2x3 ≤4 201 +23 <6 1, 2, 30 has optimal tableau Z 21 22 23 81 82 rhs 1 0 0 C3 C 1/2 0 0 1 ā13 1 -1/2 51 5. או 0 1 0 ودة 0 1/2 52 = and Cgy -[00] -188)--1881- So B = and and using B-1 and B-1 cay formulas the missing values in optimal tableau are C = 0; 2 = 0; ā13 შევ 5₁ = 0; 52 = and the optimal solution for zisz: Now find the maximum amount A = that the optimal BV remains optimal. that the value of C3 can be increased so
Solve this LP max z=3x+2x2 +2.5x3 8.t. 1+2+2x3 ≤4 201 +23 <6 1, 2, 30 has optimal tableau Z 21 22 23 81 82 rhs 1 0 0 C3 C 1/2 0 0 1 ā13 1 -1/2 51 5. או 0 1 0 ودة 0 1/2 52 = and Cgy -[00] -188)--1881- So B = and and using B-1 and B-1 cay formulas the missing values in optimal tableau are C = 0; 2 = 0; ā13 შევ 5₁ = 0; 52 = and the optimal solution for zisz: Now find the maximum amount A = that the optimal BV remains optimal. that the value of C3 can be increased so
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter11: Differential Equations
Section11.6: Applications Of Differential Equations
Problem 15E: Solve Exercise 14 if a 25 solution of the same mixture is added instead of pure alcohol.
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![Solve this LP
max z = 3x1 + 2x2 +2.5x3
8.t.
x1 + x2 + 2x3
201
+203
1, 2, 3 > 0
has optimal tableau
So B=
<4
<6
=
Z 21 X2
1 0 0
0
0
0
1
T3
C3
1 ā13
0
ā23
82
₂ 1/2
1
0
ā 13
ā23
5₁ ; 5₂ = [
and the optimal solution for zis z =
Now find the maximum amount A
that the optimal BV remains optimal.
81
-188--18:3
and
=
and using B-¹ and Bcy formulas the missing values in optimal tableau are
€3=0; €₂=0;
rhs
IN L
-1/2 by
1/2
ba
and cgy
-[00]
that the value of c3 can be increased so](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc80b4d1f-cc91-4668-ace5-1850d4a765ec%2F5ee8f47f-96f1-4c77-a6a6-31c2c9236f7f%2Fp7broi8_processed.png&w=3840&q=75)
Transcribed Image Text:Solve this LP
max z = 3x1 + 2x2 +2.5x3
8.t.
x1 + x2 + 2x3
201
+203
1, 2, 3 > 0
has optimal tableau
So B=
<4
<6
=
Z 21 X2
1 0 0
0
0
0
1
T3
C3
1 ā13
0
ā23
82
₂ 1/2
1
0
ā 13
ā23
5₁ ; 5₂ = [
and the optimal solution for zis z =
Now find the maximum amount A
that the optimal BV remains optimal.
81
-188--18:3
and
=
and using B-¹ and Bcy formulas the missing values in optimal tableau are
€3=0; €₂=0;
rhs
IN L
-1/2 by
1/2
ba
and cgy
-[00]
that the value of c3 can be increased so
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