To maintain his health, a person must fulfill certain minimum daily requirements for several kinds of nutrients. Assuming that there are only three kinds of nutrients — calcium, protein and Calories — and the person's diet consist of only two food items 1 and 2, whose price and nutrient contents are shown in the table below:
| Nutrients | Food I (per lb) | Food II (per lb) | Minimum daily requirement |
|---|---|---|---|
| Calcium | 10 | 4 | 20 |
| Protein | 5 | 5 | 20 |
| Calories | 2 | 6 | 13 |
| Price (in Rs.) | 0.60 | 1.00 |
What combination of two food items will satisfy the daily requirement and entail the least cost? Formulate this problem as a LPP.
Concept understanding — Linear Programming Formulation
Linear Programming Formulation: From Intuition to Precision
Imagine you run a small factory that makes two products: chairs and tables. Each chair gives you ₹200 profit, each table gives you ₹300 profit. You have limited wood (240 units) and limited labour hours (100 hours). A chair needs 2 units of wood and 1 hour of labour; a table needs 4 units of wood and 3 hours of labour. You can't make negative chairs or tables. How many of each should you produce to maximise your profit?
This is the kind of problem Linear Programming (LP) is built to solve. The word "programming" here doesn't mean computer programming — it's an old term for "planning". So linear programming is about planning with linear relationships.
The Core Intuition
Every LP problem has three ingredients:
- Decision variables — the quantities you control (how many chairs, how many tables)
- An objective — what you want to maximise (profit) or minimise (cost)
- Constraints — the limits you must respect (wood, labour, non-negativity)
The "linear" part means everything — the profit, the resource usage — adds up in straight-line proportions. Double the chairs, double the wood needed. No fancy curves, no discounts for bulk.
The Precise Statement
A Linear Programming problem is an optimisation problem of the following form:
Maximise (or Minimise) Z=c1x1+c2x2+⋯+cnxn
subject to:
a11x1+a12x2+⋯+a1nxn≤b1
a21x1+a22x2+⋯+a2nxn≤b2
⋮
am1x1+am2x2+⋯+amnxn≤bm
x1,x2,…,xn≥0
Let's decode this piece by piece.
Decision Variables
x1,x2,…,xn are the variables you control. In our factory: let x1 = number of chairs, x2 = number of tables.
Objective Function
Z=c1x1+c2x2+⋯+cnxn is what you want to optimise. The cj are coefficients — profit per unit or cost per unit. For our factory: Z=200x1+300x2 (profit to maximise).
Constraints
Each constraint is a linear inequality (or equality) that limits the variables. The aij are the resource usage per unit of product j for resource i. The bi are the available amounts of each resource.
For our factory:
- Wood: 2x1+4x2≤240
- Labour: 1x1+3x2≤100
Non-negativity
xj≥0 for all j. You can't produce negative chairs. This is almost always present in real problems.
A common mistake is forgetting the non-negativity constraints. Without them, the solver might "produce" negative quantities — which is nonsense in the real world.
The Complete Formulation for Our Example
Maximise Z=200x1+300x2
subject to:
2x1+4x2≤240
x1+3x2≤100
x1,x2≥0
That's it. This is a complete LP formulation. Every LP problem, no matter how complex, follows this same skeleton.
Why "Linear"?
The objective and every constraint are linear functions — they involve only first powers of variables, no x2, no sinx, no x1x2. This linearity is what makes LP problems solvable efficiently (using the Simplex method or interior-point methods). If you have products of variables or powers, it's not linear programming anymore — it becomes nonlinear programming, which is much harder.
A Quick Checklist for Formulating Any LP
When you see a word problem, ask yourself in order:
- What are the decisions? → Define variables with units.
- What is the goal? → Maximise or minimise? Write the objective.
- What are the limits? → Each limit becomes a constraint. Check units match.
- Are there hidden constraints? → Non-negativity is standard. Sometimes you need integer constraints (but that's Integer Programming, not LP).
Always write the units in your head. If a constraint says 2x1+4x2≤240, and x1 is chairs, then 2 must be "units of wood per chair" and 240 must be "total units of wood available". Unit consistency catches most formulation errors.
What LP Cannot Do
LP assumes everything is continuous — you can produce 3.7 chairs. If you need whole numbers (you can't sell half a chair), you need Integer Programming. LP also assumes certainty — if wood supply is random, you need stochastic programming. But for a first meeting, LP is the foundation that all these advanced topics build upon.
Linear Programming Formulation is the opening skill taught in the NCERT Class 12 Mathematics chapter on Linear Programming, matching searches like "linear programming problem formulation examples" or "LPP important questions class 12 maths". Correctly translating a word problem into an objective function and constraints is a heavily weighted, scoring-friendly topic in CBSE Class 12 board exams.
Concept: Linear Programming Formulation — We minimise cost subject to nutrient constraints.
Step 1: Define decision variables
Let x1 = pounds of Food I, x2 = pounds of Food II.
Step 2: Write the objective function
Minimise total cost:
Z=0.60x1+1.00x2
Step 3: Write the constraints from the table
- Calcium: 10x1+4x2≥20
- Protein: 5x1+5x2≥20
- Calories: 2x1+6x2≥13
- Non-negativity: x1≥0,x2≥0
The LPP is: Minimise Z=0.60x1+1.00x2 subject to 10x1+4x2≥20, 5x1+5x2≥20, 2x1+6x2≥13, x1,x2≥0.
LPP: minimise Z=0.60x+1.00y subject to the three nutrient constraints. The least-cost diet is 2.75 lb of Food I and 1.25 lb of Food II, costing Rs. 2.90.
Formulation
Let x = pounds of Food I and y = pounds of Food II, with x≥0, y≥0.
Minimise (cost):
Z=0.60x+1.00y
Subject to (each nutrient must meet its minimum daily requirement):
10x+4y5x+5y2x+6yx,y≥20≥20≥13≥0(calcium)(protein)(calories)
Solving graphically
The feasible region is unbounded (above all three lines). Its lower-boundary corner points and their costs:
| Corner point | How obtained | Cost Z=0.60x+1.00y |
|---|---|---|
| (0,5) | calcium ∩ y-axis | 5.00 |
| (32,310) | calcium ∩ protein | 3.73 |
| (2.75,1.25) | protein ∩ calories | 2.90 |
| (6.5,0) | calories ∩ x-axis | 3.90 |
(The calcium ∩ calories point (1.31,1.73) fails protein, since x+y=3.04<4, so it is not a feasible vertex.)
The point (2.75,1.25) satisfies all constraints — calcium 32.5≥20, protein 20≥20, calories 13≥13 — at the lowest cost:
Z=0.60(2.75)+1.00(1.25)=1.65+1.25=Rs. 2.90.
Minimise Z=0.60x+1.00y subject to 10x+4y≥20, 5x+5y≥20, 2x+6y≥13, x,y≥0. The least-cost combination is 2.75 lb of Food I and 1.25 lb of Food II, at a minimum cost of Rs. 2.90.
- CBSE 2025Set 465/W1XZY/41 markMCQQ.In a LPP, the maximum value of z=3x+4y subject to the constraints x+y≤40, x+2y≤60, x,y≥0 is (A) 120 (B) 140 (C) 150 (D) 130
›Reveal solutionSolution
Corner points are (0,0),(40,0),(0,30),(20,20); z is largest at (20,20) with z=140.
Corner-point method: the optimum of z=3x+4y over a bounded feasible region occurs at a vertex of that region.
- Find the vertices of the region defined by x+y≤40, x+2y≤60, x,y≥0.
- Axis vertices: (0,0); x+y=40 meets the x-axis at (40,0); x+2y=60 meets the y-axis at (0,30).
- Intersection of x+y=40 and x+2y=60: subtracting gives y=20, then x=20, so (20,20) (it satisfies both constraints).
- Evaluate z=3x+4y: at (0,0):0; at (40,0):120; at (0,30):120; at (20,20):60+80=140.
- The maximum is 140 at (20,20).
✓Final answerMaximum z=140 — option (B).
- CBSE 2025Set 465/S/WXYZ/41 markMCQQ.The graph of the inequality 3x+2y>6 is the : (A) entire XOY plane (B) whole XOY plane excluding the points on the line 3x+2y=6 (C) half plane that contains the origin (D) half plane that neither contains the origin nor the points on the line 3x+2y=6
›Reveal solutionSolution
3x+2y>6 is the open half-plane on the far side of the line from the origin, excluding the line itself.
A strict linear inequality ax+by>c is a half-plane not including the boundary line ax+by=c; use a test point to pick the correct side.
- Boundary: 3x+2y=6. Because the inequality is strict (>), points on this line do not satisfy it, so the line is excluded.
- Test the origin (0,0): 3(0)+2(0)=0, and 0>6 is false.
- So the origin lies in the region that is not a solution.
- Hence the solution is the half-plane on the opposite side of the line from the origin, excluding the boundary — it contains neither the origin nor the line's points.
✓Final answer(D) half plane that neither contains the origin nor the points on the line 3x+2y=6
- CBSE 2024Set 465/RQPS/41 markMCQQ.The graph of the inequation 2x+3y>6 is the : (A) entire XOY-plane (B) half-plane that contains the origin (C) half-plane that neither contains the origin nor the points on the line 2x+3y=6 (D) whole XOY-plane excluding the points on the line 2x+3y=6
›Reveal solutionSolution
2x+3y>6 is the open half-plane on the far side of the line from the origin, with the line itself excluded.
For a linear inequality, substitute a test point: if it satisfies the inequality, the region is on that side; a strict (>) inequality excludes the boundary line.
- Test the origin (0,0): 2(0)+3(0)=0, and 0>6 is false, so the origin is NOT in the region.
- Hence the solution is the half-plane on the opposite side of the line from the origin.
- Because the inequality is strict (>, not ≥), points on the boundary 2x+3y=6 are excluded.
- So the graph is the half-plane containing neither the origin nor the line's points.
✓Final answer(C) half-plane that neither contains the origin nor the points on the line 2x+3y=6
- CBSE 2024Set 465/RQPS/41 markMCQQ.In an LPP, if the objective function Z=ax+by has same maximum value on two corner points of the feasible region, then the number of points at which maximum value of Z occurs is : (A) 0 (B) 2 (C) finite (D) infinite
›Reveal solutionSolution
Equal maxima at two corner points means the whole connecting edge is optimal — infinitely many solutions.
In an LPP, if Z=ax+by has the same maximum at two distinct corner points, it has that same value at every point of the line segment joining them (alternative/multiple optimal solutions).
- Let the two corner points be P1 and P2 with Z(P1)=Z(P2)=Zmax.
- Any point on segment P1P2 can be written as λP1+(1−λ)P2, 0≤λ≤1.
- By linearity, Z(λP1+(1−λ)P2)=λZ(P1)+(1−λ)Z(P2)=Zmax.
- Thus every point of the segment is optimal, giving infinitely many optimal points.
✓Final answer(D) infinite
- CBSE 2024Set 465/S/RQPS/41 markMCQQ.The number of solutions of an L.P.P. to minimize z=3x+2y under the constraints x+y≥8, 3x+5y≤15 and x,y≥0, is : (A) 2 (B) 5 (C) infinitely many (D) zero
›Reveal solutionSolution
The two constraints are contradictory — the feasible region is empty — so the LPP has zero solutions.
An LPP has a solution only if its feasible region (intersection of all constraints) is non-empty. Check consistency of x+y≥8 and 3x+5y≤15 with x,y≥0.
- For x,y≥0, 3x+5y≥3x+3y=3(x+y).
- The first constraint forces x+y≥8, so 3(x+y)≥24, hence 3x+5y≥24.
- But the second constraint demands 3x+5y≤15. Since 24>15, both cannot be satisfied simultaneously.
- Therefore no feasible point exists — the number of solutions is zero.
✓Final answer(D) zero
- CBSE 2023Set 465/EF1GH/41 markMCQQ.Region represented by x≥0,y≥0 lies in(a) I quadrant(b) II quadrant(c) III quadrant(d) IV quadrant
›Reveal solutionSolution
x≥0, y≥0 is the non-negative region — the first quadrant.
Sign conventions: I quadrant x>0,y>0; II x<0,y>0; III x<0,y<0; IV x>0,y<0.
- x≥0 restricts to points on or to the right of the y-axis.
- y≥0 restricts to points on or above the x-axis.
- Their intersection is the region where both coordinates are non-negative — the first quadrant.
- (This is why LPP feasible regions with non-negativity constraints lie in quadrant I.)
✓Final answer(a) I quadrant
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