Question

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

Answer 1

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.