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Class 12 Linear Programming · Long Answer3 Marksgraphical_optimization

Class 12 Maths · Linear Programming · Long Answer #61

A bounded feasible polygon has successive vertices P(3/13,24/13),Q(3/2,15/4),R(7/2,3/4),S(18/7,2/7). Find min and max of Z=x+2y.
Coordinate-derived feasible-region graph for cbse12-math-lp-p1-practice_short-012
Coordinate-derived feasible-region graph for cbse12-math-lp-p1-practice_short-012

✅ Step-by-Step Solution

Final Answer: Maximum 9 at Q; minimum 22/7 at S.
  1. Evaluate the objective at all four vertices.
  2. Z(P)=51/13, Z(Q)=9.
  3. Z(R)=5, Z(S)=22/7.
  4. Hence maximum 9 at (3/2,15/4) and minimum 22/7 at (18/7,2/7).

📋 CBSE Stepwise Marking Scheme

  • [0.5 marks] Evaluate the objective at all four vertices.
  • [1 mark] Z(P)=51/13, Z(Q)=9.
  • [1 mark] Z(R)=5, Z(S)=22/7.
  • [0.5 marks] Hence maximum 9 at (3/2,15/4) and minimum 22/7 at (18/7,2/7).

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