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Question 39 of 39

Q.Solve the following linear programming problem by graphical method (Graph sheet is not required): Minimize Z=3x+5yZ = 3x + 5y subject to x+3y≥3x + 3y \ge 3, x+y≥2x + y \ge 2 and x,y≥0x, y \ge 0.

West Bengal WbchseWest Bengal HS (WBCHSE) Board 2026Subjective· 3mImportance★★★★★
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Graphical LPP: unbounded feasible region with corners (3,0), (3/2,1/2), (0,2) from x+3y≥3, x+y≥2, x,y≥0; Z = 3x + 5y is minimised (7) at (3/2, 1/2).
Graphical LPP: unbounded feasible region with corners (3,0), (3/2,1/2), (0,2) from x+3y≥3, x+y≥2, x,y≥0; Z = 3x + 5y is minimised (7) at (3/2, 1/2).

The feasible corners are (3,0)(3,0), (32,12)\left(\tfrac32,\tfrac12\right) and (0,2)(0,2); evaluating Z=3x+5yZ=3x+5y gives 9, 7, 109,\,7,\,10, so the minimum is Z=7Z=7 at (32,12)\left(\tfrac32,\tfrac12\right).

Concept. Solve graphically by identifying the corner points of the feasible region defined by the constraints, then apply the corner-point theorem. For an unbounded region we also confirm the objective cannot go lower than the best corner value. This is the standard NCERT Class 12 mathematics graphical LPP method.

Constraints. x+3y≥3x+3y\ge3, x+y≥2x+y\ge2, x,y≥0x,y\ge0.

Corner points.

  • Line x+3y=3x+3y=3 meets the xx-axis at (3,0)(3,0); it satisfies x+y≥2x+y\ge2 (3≥23\ge2) — feasible.
  • Line x+y=2x+y=2 meets the yy-axis at (0,2)(0,2); it satisfies x+3y≥3x+3y\ge3 (6≥36\ge3) — feasible. …

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