For the constraint of a linear optimizing function z=x1+x2 given by x1+x2≤1, 3x1+x2≥3 and x1, x2≥0 There are infinite feasible regions There are two feasible regions There is no feasible region None of these
Answers
Answer:
It has no feasible region.
Step-by-step explanation:
The given function is z = x₁ + x₂
Given constraints = x₁ + x₂ ≤ 1
3x₁ + x₂ ≥ 3
x₁ ≥ 0
x₂ ≥ 0
Plotting the graph of these equations to find the feasible region(if any)
for x1 + x₂ = 1
x₁ -1 0 1 2
x₂ 2 1 0 -1
for 3x₁ + x₂ ≥ 3
x₁ -1 0 1 2
x₂ 6 3 0 3
On putting (0,0) in x₁ + x₂ ≤ 1 , we get 0 ≤ 1
which is true
this means this graph has the area containing the origin
On putting (0,0) in 3x₁ + x₂ ≥ 3 , we get 0 ≥ 3
which is False
this means this graph has the area not containing the origin.
Graph attached below.
Therefore, we have no common region.
=> It has no feasible region.
