Linear programming problem formulation

A shopkeeper has a fixed budget, a factory has limited machine hours, a dietician needs a minimum daily intake of vitamins — each of these is really the same mathematical question wearing a different costume. Linear programming problem (LPP) formulation is the skill of reading a word problem about limited resources and translating it, sentence by sentence, into a precise mathematical model: an objective function to be maximised or minimised, sitting on top of a set of linear constraints. This is the opening idea of the Linear Programming chapter in CBSE Class 12 Mathematics, and it is entirely a translation exercise — no graphing yet. The CBSE syllabus keeps this deliberately narrow: two decision variables and at most three non-trivial constraints, so that the resulting model can be solved graphically in the next topic. Get the formulation right and the rest of the chapter is almost mechanical; get it wrong and every later step inherits the error.

The core insight

Formulating an LPP means naming the unknown quantities as decision variables, writing the goal to be optimised as a linear objective function $Z = ax + by$ , and turning every resource limit or requirement into a linear inequality of the form $px + qy \le r$ or $px + qy \ge r$ , usually alongside the non-negativity conditions $x \ge 0,\ y \ge 0$ .

The core idea

Every LPP formulation has exactly three ingredients, and the entire skill is extracting them from prose in the right order.

1. Decision variables. These are the quantities whose values we are free to choose — almost always written $x$ and $y$ in a two-variable CBSE problem. They must be defined in a full sentence, e.g. "let $x$ = number of tables and $y$ = number of chairs stocked," because every constraint and the objective function will be written in terms of them.

2. Objective function. This is the linear expression in $x$ and $y$ that the problem wants maximised (profit, output) or minimised (cost, time, distance). It is written

Z = ax + by

where: $Z$ is the quantity being optimised, and $a, b$ are the per-unit contributions of $x$ and $y$ to it (e.g. profit per table, profit per chair).

3. Constraints. These are the linear inequalities that come from every limited resource or minimum requirement stated in the problem, of the general form

px + qy \le r \qquad \text{or} \qquad px + qy \ge r

where: $p, q$ are the per-unit consumption or contribution of $x$ and $y$ toward that resource, and $r$ is the total amount available (for $\le$ ) or required (for $\ge$ ). Non-negativity restrictions $x \ge 0,\ y \ge 0$ are added almost automatically whenever $x, y$ represent physical counts, since a factory cannot make $-3$ chairs. The figure below is the map every formulation follows, from raw sentence to finished model.

Figure

A flow diagram showing a word problem broken into decision variables then an objective function then constraints then non-negativity then the assembled linear programming problem

Derivation ATurning a word problem into a complete LPP

The goal is to show, with a single worked scenario, exactly how each sentence of a word problem maps onto one piece of the mathematical model — decision variables from the question being asked, the objective function from the quantity to be optimised, and one inequality per resource sentence, so that the finished model is ready to hand to the graphical method without any further translation.

Step 1. Read the question being asked, not the numbers, first. A dealer sells only tables and chairs; he wants to know how many of each to stock. That single phrase fixes the decision variables: let $x$ = number of tables and $y$ = number of chairs to be stocked. Naming them in a full sentence, with units, is non-negotiable — an unlabelled $x, y$ earns no credit even if every inequality afterward is correct.

Step 2. Find the sentence describing what is to be optimised, and convert its per-unit figures directly into coefficients. The dealer earns profits of ₹250 per table and ₹75 per chair, and wants to maximise total profit, so $Z = 250x + 75y$ , to be maximised.

Step 3. Convert every remaining resource sentence into one inequality each, matching the direction word to the correct inequality sign. The dealer has ₹50,000 to invest, and a table costs ₹2,500 while a chair costs ₹500, so the investment used is $2500x + 500y$ , and it "cannot exceed" ₹50,000, giving $2500x + 500y \le 50000$ , which simplifies (dividing by 500) to $5x + y \le 100$ . He also has storage for "at most" 60 pieces, giving $x + y \le 60$ directly.

Step 4. Add non-negativity and assemble. Since $x, y$ count physical pieces of furniture, $x \ge 0,\ y \ge 0$ . The complete model is then written as one labelled block: maximise $Z = 250x + 75y$ subject to $5x + y \le 100$ , $x + y \le 60$ , $x \ge 0,\ y \ge 0$ — exactly two decision variables and two non-trivial constraints, within the CBSE limit of three.

Worked example

A dealer sells only tables and chairs. He has ₹50,000 to invest and storage for at most 60 pieces. A table costs ₹2,500 and a chair costs ₹500, giving profits of ₹250 and ₹75 respectively. He wants to know how many of each to stock to maximise profit, assuming he sells everything he buys. Formulate this as an LPP.

Step 1 — Decision variables. Let $x$ = number of tables and $y$ = number of chairs to be stocked.

Step 2 — Objective function. Profit from tables is $250x$ and from chairs is $75y$ , so the total profit to be maximised is

Z = 250x + 75y

Step 3 — Constraints. The investment constraint (total cost cannot exceed ₹50,000) gives $2500x + 500y \le 50000$ , which simplifies to $5x + y \le 100$ . The storage constraint (at most 60 pieces) gives $x + y \le 60$ . Since $x, y$ are counts of furniture, $x \ge 0,\ y \ge 0$ .

Step 4 — Assemble the model. The finished LPP is

\text{Maximise } Z = 250x + 75y

\text{subject to } 5x + y \le 100,\quad x + y \le 60,\quad x \ge 0,\ y \ge 0

This model — with exactly two constraints besides non-negativity — is now ready to be solved graphically in the next topic.

Competitive stretch

The same reading skill flips direction for a minimisation problem. A diet must contain at least 80 units of vitamin A and 100 units of minerals, using foods $F_1$ (₹4/unit, 3 units vitamin A, 4 units minerals) and $F_2$ (₹6/unit, 5 units vitamin A, 2 units minerals). With $x, y$ = units of $F_1, F_2$ , the cost to minimise is $Z = 4x + 6y$ , and "at least" turns both requirements into $\ge$ inequalities: $3x + 5y \ge 80$ (vitamin A) and $4x + 2y \ge 100$ , i.e. $2x + y \ge 50$ (minerals), with $x \ge 0,\ y \ge 0$ . Notice the sign flip from the tables-and-chairs example: resource limits give $\le$ , but requirement minimums give $\ge$ — the same word-reading discipline, applied to opposite language.

Common traps

Flipping the inequality direction. "At least," "not less than," "minimum" mean $\ge$ ; "at most," "not more than," "cannot exceed" mean $\le$ . This single reading error invalidates the whole model even when every coefficient is correct.
Forgetting non-negativity constraints. Since $x, y$ almost always represent physical quantities, $x \ge 0,\ y \ge 0$ must be stated explicitly unless the problem says otherwise.
Leaving decision variables undefined. Writing " $x$ = tables" without the full sentence ("let $x$ = number of tables stocked") is marked as incomplete, since the objective function and constraints are meaningless without it.
Writing extra or redundant constraints. The CBSE syllabus caps formulation at three non-trivial constraints; splitting one resource limit into two separate inequalities invites arithmetic slips later.

Ask the tutor: if a word problem's phrasing has you unsure whether it means $\le$ or $\ge$ , or how to split its numbers into constraints, ask the AI tutor to walk through it sentence by sentence with you.

How do I decide which quantity becomes x and which becomes y?

Look at the question being asked, not the order the numbers appear in the paragraph. Whatever two quantities the problem wants you to find "how many of each" — tables and chairs, or units of two foods — become $x$ and $y$ . The assignment order does not usually matter for correctness, but once chosen it must stay fixed and be defined in a full sentence, since every later inequality depends on that definition.

Why do we simplify an inequality like 2500x + 500y le 50000 to 5x + y le 100?

Simplifying to the smallest integer coefficients makes the constraint easier to graph later and reduces arithmetic errors, but it is not compulsory — the unsimplified inequality is equally correct in the formulation stage. Always simplify by dividing through by a common factor (here 500) rather than by guessing smaller numbers, and check the direction of the inequality does not flip, which it would if you divided by a negative number.

Do I always need to write x ge 0 and y ge 0, even if the problem does not mention it?

Yes, whenever $x$ and $y$ represent counts or amounts of something physical — units produced, items stocked, quantities consumed — because negative values would be meaningless. The only time you would omit them is if a variable is explicitly allowed to be negative (rare at this level) or if the problem states its own bounds instead. When in doubt for a CBSE-style problem, include them.

Linear programming problem formulation

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads