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X − y ≤ 1, x − y ≥ 0, x ≥ 0, y ≥ 0 are the constant for the objective function z = x + y. It is solvable for finding optimum value of z? Justify? - Mathematics and Statistics

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प्रश्न

x − y ≤ 1, x − y ≥ 0, x ≥ 0, y ≥ 0 are the constant for the objective function z = x + y. It is solvable for finding optimum value of z? Justify?

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उत्तर

To draw the feasible region, construct table as follows:

Inequality x − y ≥ 1 x − y ≥ 0
Corresponding equation (of line) x − y = 1 x − y = 0
Intersection of line with X-axis (1, 0) (0, 0)
Intersection of line with Y-axis (0, −1) (0, 0)
Region Origin Side Test point: (1, 0)
1 − 0 ≥ 0, which is true.
The side containing (1, 0)

x ≥ 0, y ≥ 0 represent 1st quadrant.

Here, the feasible region is unbounded.

So, the objective function does not have finite maximum value, i.e. the value of objective function, increases indefinitely and hence the L.P.P has unbounded solution.

∴ The optimal solution does not exist.

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Linear Programming Problem (L.P.P.)
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अध्याय 1.7: Linear Programming Problems - Long Answers II

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