Math Problem Statement

Formulate the lpp

Formulate the linear programming problem (LPP) based on the scenario provided.

Solution

To formulate the linear programming problem (LPP) for the given scenario, we need to establish the objective function and constraints.

Problem Analysis and Data

  1. Objective: Maximize profit.

  2. Decision Variables:

    • Let xx represent the number of units of part A produced.
    • Let yy represent the number of units of part B produced.
  3. Profit Calculation:

    • Selling price for part A: Rs 5
    • Selling price for part B: Rs 6
    • Cost of casting part A: Rs 2
    • Cost of casting part B: Rs 3
    • Profit for part A: 52=35 - 2 = 3 Rs per unit
    • Profit for part B: 63=36 - 3 = 3 Rs per unit

    Therefore, the objective function is: Maximize Z=3x+3y\text{Maximize } Z = 3x + 3y

  4. Constraints:

    Each part requires machining, boring, and polishing, with different capacities for each process. We need to include these capacities in the constraints.

    • Machining Capacity:

      • Part A: 25 units per hour
      • Part B: 40 units per hour
      • Constraint: x25+y401\frac{x}{25} + \frac{y}{40} \leq 1
    • Boring Capacity:

      • Part A: 28 units per hour
      • Part B: 35 units per hour
      • Constraint: x28+y351\frac{x}{28} + \frac{y}{35} \leq 1
    • Polishing Capacity:

      • Part A: 35 units per hour
      • Part B: 25 units per hour
      • Constraint: x35+y251\frac{x}{35} + \frac{y}{25} \leq 1
  5. Non-negativity Constraints:

    • x0x \geq 0
    • y0y \geq 0

Formulated LPP

Thus, the linear programming problem can be formulated as:

Maximize Z=3x+3y\text{Maximize } Z = 3x + 3y

Subject to: x25+y401\frac{x}{25} + \frac{y}{40} \leq 1 x28+y351\frac{x}{28} + \frac{y}{35} \leq 1 x35+y251\frac{x}{35} + \frac{y}{25} \leq 1 x0,y0x \geq 0, \, y \geq 0

This LPP can be solved using the graphical method by plotting the constraints and finding the feasible region, then identifying the point(s) within this region that maximize the objective function ZZ.

Would you like a graphical solution or further details on solving this LPP?


  1. How do you determine the feasibility of points within the graphical method?
  2. What is the significance of each constraint in this LPP?
  3. How would you adjust the LPP if the costs of the machines were variable?
  4. How would changes in production rates affect the objective function?
  5. How would this LPP differ if the goal was to minimize cost rather than maximize profit?

Tip: When plotting constraints for the graphical method, check the feasible region by testing points against each inequality to confirm they satisfy all conditions.

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Math Problem Analysis

Mathematical Concepts

Linear Programming
Optimization
Inequalities
Graphical Method

Formulas

Objective function: Maximize Z = 3x + 3y
Machining capacity constraint: x/25 + y/40 ≤ 1
Boring capacity constraint: x/28 + y/35 ≤ 1
Polishing capacity constraint: x/35 + y/25 ≤ 1

Theorems

Feasible Region
Linear Programming Theorem

Suitable Grade Level

Undergraduate