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Byju's Answer
Standard XII
Mathematics
Application of inequalities and absolute values
Maximize z=15...
Question
Maximize
z
=
15
x
1
+
20
x
2
Subject to :
12
x
1
+
4
x
2
≥
36
12
x
1
+
6
x
2
≤
24
x
1
,
x
2
≥
0
The above linear programming problem has
A
alternative optimum solutions
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B
degenerate solution
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C
infeasible solution
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D
unbounded solution
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Solution
The correct option is
D
unbounded solution
Maximize
z
=
15
x
1
+
20
x
2
Subjected to,
12
x
1
+
4
x
2
≥
36
o
r
3
x
1
+
x
2
≥
9
12
x
1
−
6
x
2
≤
24
o
r
2
x
1
−
x
2
≤
4
Let
3
x
1
+
x
2
=
9
o
r
2
x
1
−
x
2
=
4
x
1
=
0
,
x
2
=
9
o
r
x
1
=
0
,
x
2
=
−
4
x
1
=
3
,
x
2
=
0
o
r
x
1
=
2
,
x
2
=
0
Since maximum value of objective function lies at infinity.
So, the problem has an unbounded solution.
Suggest Corrections
0
Similar questions
Q.
For the linear programming problem:
Maximize
z
=
3
x
1
+
2
x
2
Subject to:
−
2
x
1
+
3
x
2
≤
9
x
1
−
5
x
2
≥
−
20
x
1
,
x
2
≥
0
The above problem has
Q.
Consider the following linear programming problem (LPP):
Maximize
z
=
5
x
1
+
3
x
2
Subjected to the following constraints
x
1
−
x
2
≤
2
x
1
+
x
2
≥
3
x
1
,
x
2
≥
0
The above LPP has
Q.
Consider the following Linear Programming Problem (LPP):
Maximize
z
=
3
x
1
+
2
x
2
Subject to:
x
1
≤
4
x
2
≤
6
3
x
1
+
2
x
2
≤
18
x
1
≥
0
,
x
2
≥
0
Q.
By graphical method, the solution of linear programming problem
Maximize
Z
=
3
x
1
+
5
x
2
Subject
to
3
x
1
+
2
x
2
≤
18
x
1
≤
4
x
2
≤
6
x
1
≥
0
,
x
2
≥
0
,
is
(a) x
1
= 2, x
2
= 0, Z = 6
(b) x
1
= 2, x
2
= 6, Z = 36
(c) x
1
= 4, x
2
= 3, Z = 27
(d) x
1
= 4, x
2
= 6, Z = 42