Components of a Linear Programming Problem
Learn to define decision variables, form a profit objective, translate resource limits into constraints, and write non-negativity conditions using a toy factory example.
From a production story to a mathematical model
Decision variables and the objective
Learning objectives
- Identify the three key components of a linear programming problem (LPP).
- Define decision variables with clear meanings.
- Translate per-unit profits into a total-profit objective.
An LPP describes a choice under limited resources. Its three key components are decision variables, an objective function, and constraints. Non-negativity conditions specify that the production quantities cannot be negative.
Choose the decision variables
The toy factory must decide how many trucks and cars to produce. Let x be the number of toy trucks and y be the number of toy cars. These quantities are the decision variables.
Express the profit objective
Each truck earns ₹10 profit and each car earns ₹5 profit. Producing x trucks earns ₹10x; producing y cars earns ₹5y. Total profit is Z = 10x + 5y, measured in rupees. The goal is to maximize Z.
Maximize Z = 10x + 5y
Answer: Number of toy cars produced. The profit from those cars is 5y rupees; y itself is a production quantity.
Constraints and non-negativity
Translate resource limits
Video focus: 2:20–5:03. Constraints describe limits or conditions on production. Write total resource use on the left and the available resource on the right.
The time expression 3x + 2y is measured in hours. The material expression 3x + y is measured in material units. Both limits must hold for the same production plan.
Zero production is allowed by the non-negativity conditions. Negative production is not.
Answer: 3x + y ≤ 100. Each truck uses 3 material units and each car uses 1. The combined use cannot exceed the 100 units available.
Answer: Quantity ≥ 20. The value 20 is allowed, and values above 20 are also allowed. This is an added practice question applying the video’s wording rule.
Assemble the complete model
The complete toy factory model
Video focus: 5:03–6:13. Define x as the number of trucks and y as the number of cars. Combine the goal and all the conditions in one formulation.
Maximize Z = 10x + 5y Subject to: 3x + 2y ≤ 150 (working hours) 3x + y ≤ 100 (material units) x ≥ 0, y ≥ 0 (non-negativity)
Define
Identify the quantities you can choose and state their units.
Express the goal
Multiply each quantity by its profit per unit and add the contributions.
Write the limits
Translate each resource limit into a separate inequality.
Set non-negativity
State that each production quantity is at least zero.
The video demonstrates formulation. It does not calculate the optimal production quantities. The closing preview says assumptions of LPP models will be discussed in the next video.
Practice and review
Apply the model
The following exercises and worked answers are added supporting practice derived from the video’s coefficients. They are not additional examples presented by the speaker. A feasible plan satisfies every constraint, including non-negativity.
Yes. Working time = 3(20) + 2(30) = 120 hours, within 150. Material = 3(20) + 30 = 90 units, within 100. Both quantities are non-negative. Profit = 10(20) + 5(30) = ₹350. Feasibility alone does not establish that the plan maximizes profit.
Time = 3(30) + 2(20) = 130 hours, within 150. Material = 3(30) + 20 = 110 units, exceeding 100 by 10. The plan is not feasible, even though it satisfies the time limit and non-negativity.
Review checklist
- Can you explain what x and y represent?
- Can you justify the coefficients 10 and 5 in the objective?
- Can you distinguish the hours constraint from the material constraint?
- Can you choose ≤ for “at most” and ≥ for “at least”?
- Can you explain why x ≥ 0 and y ≥ 0 are required?