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The following LLP

$\text{Maximize } z=100x_1 +2x_2+5x_3$

Subject to

$14x_1+x_2-6x_33+3x_4=7$

$32x_1+x_2-12x_3 \leq 10$

$3x_1-x_2-x_3 \leq 0$

$x_1, x_2, x_3, x_4 \geq 0$ has

  1. Solution : $x_1=100, \:  x_2=0, \:  x_3=0$
  2. Unbounded solution
  3. No solution
  4. Solution : $x_1=50, \:  x_2=70, \:  x_3=60$
in Others by Boss (10.8k points)
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2 Answers

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OPTION 4
by (287 points)
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How ??

explain it !!
0
Option 4 is wrong B will be correct because according to the above the unbounded feasible solution can not be determined, instead there are infinite many solutions.
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B will be correct because according to the above the unbounded feasible solution can not be determined, instead there are infinite many solutions.
by (27 points)

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