Question 431 markSection E

Write the constraints that relate the total cost of both types of bags.

Linear Programming
LPP Formulation — Cost Constraint

Official answer

The cost of the bags bought cannot exceed the Rs 15,000 available.


Cost (budget) constraint:


  • Cost of rice = Rs 1,800 per bag ×x=1800x\times x = 1800x.
  • Cost of wheat = Rs 1,200 per bag ×y=1200y\times y = 1200y.
  • Total cost ≤ available money: 1800x+1200y≤150001800x + 1200y \le 15000.

In simplest form (dividing by 600):


  • 3x+2y≤253x + 2y \le 25, with x≥0x \ge 0, y≥0y \ge 0.
Answer keywords:cost constraintbudget constraint1800x + 1200y ≤ 150003x + 2y ≤ 25less than or equallinear programming constraintnon-negativity

Marking scheme

  1. 1.1 mark: writing the cost constraint 1800x+1200y≤150001800x + 1200y \le 15000 (or the reduced form 3x+2y≤253x + 2y \le 25).
  2. 2.Accept either the full or simplified inequality; deduct if written as an equality or with the wrong inequality direction.

Hint

Multiply each variable by its cost per bag, keep the total within Rs 15,000, and use '≤' (not '=').

Quick oral answer

The money spent — Rs 1,800 per rice bag and Rs 1,200 per wheat bag — must stay within Rs 15,000, so 1800x+1200y≤150001800x + 1200y \le 15000, which simplifies to 3x+2y≤253x + 2y \le 25.

Common mistakes

  1. 1.Writing the constraint as an equation 1800x+1200y=150001800x + 1200y = 15000 instead of an inequality with '≤'.
  2. 2.Reversing the inequality sign to '≥', which would wrongly force spending of at least Rs 15,000.
  3. 3.Mixing in the storage limit (x+y≤10x + y \le 10) when only the cost relation is asked.

Analysis and explanation

This part supplies the budget (money) constraint of the LPP.


Concept:


  • A constraint is an inequality that a feasible solution must satisfy. The spending on rice plus wheat must stay within the Rs 15,000 budget, giving a '≤' inequality.

Exam trap:


  • Do not write it as an equation (= 15000); the man need not spend every rupee, so it is '≤'.
  • The storage limit x+y≤10x + y \le 10 is a separate constraint — the question here asks only for the cost relation, so keep them distinct.

Simplification:


  • Reducing 1800x+1200y≤150001800x + 1200y \le 15000 to 3x+2y≤253x + 2y \le 25 keeps the arithmetic light when you later find corner points, and either form earns full marks.

Interesting facts

In LPP terminology this is a 'resource constraint' — money is the limited resource being allocated between two competing goods.

Dividing 1800x+1200y≤150001800x + 1200y \le 15000 by 600 is valid because dividing an inequality by a positive number preserves its direction — a small but frequently misapplied algebra rule.

The feasible region of this problem is bounded by the budget line, the storage line and the two axes, forming a convex polygon whose corners hold the optimum.

Question details

Q43 of 44 questions

Board
CBSE
Class
Class 12
Subject
Applied Mathematics
Year
2026
Section
Section E
Marks
1
Chapter
Linear Programming
Topic
LPP Formulation — Cost Constraint
Bloom's level
understand
Syllabus topic
Unit 8: Linear Programming — Mathematical Formulation of an LPP: Writing the Constraints (Budget/Cost Constraint)
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

Why is the cost constraint an inequality and not an equation?

The man has up to Rs 15,000; he is not required to spend all of it. His total spending 1800x+1200y1800x + 1200y must be less than or equal to 15000, so the correct relation is 1800x+1200y≤150001800x + 1200y \le 15000.

Is 3x+2y≤253x + 2y \le 25 an acceptable answer?

Yes. Dividing 1800x+1200y≤150001800x + 1200y \le 15000 by 600 gives 3x+2y≤253x + 2y \le 25. Since 600 is positive, the inequality direction is unchanged, so both forms are equivalent and earn full marks.