The constraints –x + y ≤ 1, −x + 3y ≤ 9 and x, y ≥ 0 defines on

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Navik GD Mathematics 20 March 2021 (All Shifts) Questions
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  1. Bounded feasible space
  2. Unbounded feasible space
  3. Both unbounded and bounded feasible space
  4. None of the above

Answer (Detailed Solution Below)

Option 2 : Unbounded feasible space
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Detailed Solution

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Concept:

First, graph the "equals" line, then shade in the correct area.

There are three steps:

  • Rearrange the equation so "y" is on the left and everything else on the right.
  • Plot the "y = " line (make it a solid line for y ≤ or y ≥, and a dashed line for y < or y >)
  • Shade above the line for a "greater than" (y > or y ≥) or below the line for a "less than" (y < or y ≤).

 

Calculation:

Given: The constraints –x + y ≤ 1, −x + 3y ≤ 9 and x, y ≥ 0

–x + y ≤ 1

⇒ y ≤ 1 + x

Hence shade below the line.

−x + 3y ≤ 9

3y ≤ 9 + x

⇒ y ≤ 3 + x/3

Hence shade below the line.

F1 A.K 24.4.2 Pallavi D1

We can see through above graph region is unbounded feasible space

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