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Question

Consider the primal problem in LP.

Maximize Z=4X1+3X2

Subject to

X1+X282X1+X210X1;X20

together with its dual (LD) Then

A
LP and LD both are infeasible
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B
LP is feasible but LD is infeasible
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C
LP and LD both are feasible
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D
LP is infeasible but LD is feasible
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Solution

The correct option is C LP and LD both are feasible
Zmax=4X1+3X2

Statements :

X1+X28

2X1+X210

X1;X20

X18+X281

X15+X2101


Primal problem has feasible solution space. Hence an optimal solution exists for primal. If primal is feasible, dual is also feasible.

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