Linear Programming | Class 12 Mathematics Notes

12.2 Linear Programming Problem and its Mathematical Formulation

This section elaborates on the furniture dealer example to demonstrate how to formulate a linear programming problem mathematically. The dealer can invest in tables (x) and chairs (y), both non-negative quantities. The constraints include a maximum investment of Rs 50,000 and a storage limit of 60 pieces. The dealer's profit depends on the number of tables and chairs bought. The section explains how to translate these real-world constraints into linear inequalities. For example, if the dealer buys only tables, he can buy 20 tables (50000 ÷ 2500), yielding a profit of Rs 5000 (250 × 20). If only chairs are bought, storage limits restrict to 60 chairs, profit Rs 4500 (60 × 75). Various combinations are possible, such as 10 tables and 50 chairs, yielding Rs 6250 profit. The problem is to find the combination maximizing profit. The variables x and y represent the number of tables and chairs respectively, with non-negative constraints x ≥ 0 and y ≥ 0. The investment constraint is 2500x + 500y ≤ 50000, and storage constraint x + y ≤ 60. The objective function to maximize is Z = 250x + 75y. This mathematical formulation converts the real problem into a linear programming problem: maximize Z subject to linear inequality constraints and non-negativity conditions. The section also defines key terms: objective function (Z = ax + by), constraints (linear inequalities), and optimization problem (maximizing or minimizing a linear function under constraints).

🧪 Activity: No specific activity in this subsection.

🔗 Connection: Prepares for graphical solution of the formulated linear programming problem in Section 12.2.2.