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Question 28 of 28

Q.Solve the following L.P.P. graphically:
Minimize: Z=3x+2yZ = 3x + 2y,
Subject to the constraints
x−y≤1x - y \leq 1,
x+y≥3x + y \geq 3,
x≥0x \geq 0, y≥0y \geq 0.

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2020Subjective· 4mImportance★★★★★
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The feasible region's relevant corners are (0,3)(0, 3) and (2,1)(2, 1); Z(0,3)=6Z(0,3) = 6 and Z(2,1)=8Z(2,1) = 8, so the minimum is Z=6Z = 6 at (0,3)(0, 3).

Constraints: x−y≤1x - y \le 1, x+y≥3x + y \ge 3, x≥0x \ge 0, y≥0y \ge 0.

Boundary lines:

  • x−y=1x - y = 1 passes through (1,0)(1, 0) and (0,−1)(0, -1).
  • x+y=3x + y = 3 passes through (3,0)(3, 0) and (0,3)(0, 3).

Corner points of the feasible region:

  • Intersection of x−y=1x - y = 1 and x+y=3x + y = 3: adding gives 2x=4⇒x=2, y=12x = 4 \Rightarrow x = 2,\ y = 1, i.e. (2,1)(2, 1).
  • Intersection of x+y=3x + y = 3 with the yy-axis (x=0x = 0): (0,3)(0, 3) — it satisfies x−y=−3≤1x - y = -3 \le 1. ✓ …

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