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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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Question

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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Solution

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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Chapter 1.7: Linear Programming Problems - Long Answers II

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SCERT Maharashtra Mathematics and Statistics (Arts and Science) [English] 12 Standard HSC
Chapter 1.7 Linear Programming Problems
Long Answers II | Q 10

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