ISC Class 12 • 2022 • 5 Marks

Linear Programming: LPP Production Optimization Model

Official examination question with verified M1/A1 mark scheme and step-by-step mathematical reasoning.

Problem Statement

A company manufactures two types of products, $A$ and $B$. Each unit of $A$ requires $3$ hours of machining and $1$ hour of polishing. Each unit of $B$ requires $1$ hour of machining and $2$ hours of polishing. The factory has at most $30$ hours of machining and at most $20$ hours of polishing available per week. If profit on $A$ is $₹300$ and on $B$ is $₹400$, formulate and solve the LPP to maximize profit.

Verified Solution & Marking Scheme

Formulate Objective Function and Constraints
Maximize $Z = 300x + 400y$. Subject to: $3x + y \le 30 \quad (\text{Machining})$ $x + 2y \le 20 \quad (\text{Polishing})$ $x \ge 0, \quad y \ge 0$
Determine Corner Points of Feasible Region
1. Origin: $(0, 0)$. 2. On $x$-axis: $3x + 0 = 30 \implies (10, 0)$. 3. On $y$-axis: $0 + 2y = 20 \implies (0, 10)$. 4. Intersection of $3x + y = 30$ and $x + 2y = 20$: From (1): $y = 30 - 3x$. Substitute into (2): $x + 2(30 - 3x) = 20 \implies -5x + 60 = 20 \implies 5x = 40 \implies x = 8$. $y = 30 - 3(8) = 6$. Intersection point is $(8, 6)$.
Evaluate Profit Z at All Corner Points
- At $(0, 0)$: $Z = 0$ - At $(10, 0)$: $Z = 300(10) + 0 = ₹3,000$ - At $(0, 10)$: $Z = 0 + 400(10) = ₹4,000$ - At $(8, 6)$: $Z = 300(8) + 400(6) = 2400 + 2400 = ₹4,800$
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