Strategic Cost Management · Linear Programming

Graphical Method of Solving LPP for CMA Final

Updated 11 October 2026 · Fact-checked

The graphical method solves a two-variable linear programming problem by drawing each constraint as a line, shading the feasible region where all constraints hold together, and listing its corner points. You then calculate the objective function at each corner. The highest value is the maximum and the lowest is the minimum.

Understand Graphical Method of Solving LPP

A linear programming problem (LPP) asks you to maximise something, such as contribution, or minimise something, such as cost, while staying within limits on resources. With only two decision variables, you can solve it on a graph. Let x be units of product A and y be units of product B.

Each constraint is a straight line when you treat the inequality as an equation. The inequality sign tells you which side of the line is allowed. A "≤" constraint (resource limit) usually allows the side towards the origin. A "≥" constraint (minimum requirement) usually allows the side away from the origin. Non-negativity (x ≥ 0, y ≥ 0) keeps you in the first quadrant.

The area that satisfies every constraint at once is the feasible region. Every point inside it is a possible production plan. The points where boundary lines meet are the corner points (extreme points).

The key idea is the corner point theorem: if an LPP has an optimal solution, it occurs at a corner point of the feasible region. The objective function is a straight line, so as you slide it across the region, the last point it touches is a corner. This means you need not test every point. Test only the corners.

Two special cases appear. If constraints contradict each other, there is no feasible region (infeasible). If the region is open-ended, a maximisation may have no finite answer (unbounded), though a cost-minimisation with positive cost coefficients usually still has a minimum.

Key rules to remember

General two-variable LPP: Maximise or minimise Z = ax + by, subject to a₁x + b₁y (≤ or ≥) c₁, a₂x + b₂y (≤ or ≥) c₂, x ≥ 0, y ≥ 0; Z is the objective function. Write it clearly before you draw anything.
Axis intercepts of a constraint line: Put y = 0 to get the x-intercept: x = c ÷ a. Put x = 0 to get the y-intercept: y = c ÷ b.; Two intercepts are enough to draw each line.
Corner point theorem: Optimal value of Z, if it exists, occurs at a corner point of the feasible region; Evaluate Z at every corner. Pick the largest for maximisation and the smallest for minimisation.
Slope of the objective (iso-line): Slope of ax + by = Z is −a ÷ b; Useful to cross-check which corner is optimal. Lines of equal profit or cost are parallel.
Corner at intersection of two lines: Solve the two constraint equations simultaneously; Then check the point satisfies all other constraints.
Test for the side to shade: Substitute (0, 0) into the constraint. If true, shade the side containing the origin.; Do not use this test if the line passes through the origin.

How to solve Graphical Method of Solving LPP questions

Use this order for any two-variable graphical LPP. It works for both maximisation and minimisation.

1Define x and y in words with units, and write the objective function Z and all constraints, including x ≥ 0 and y ≥ 0.
2Convert each inequality into an equation and find its two intercepts (x-intercept with y = 0, y-intercept with x = 0).
3Draw each line on the graph. Decide the allowed side by testing (0, 0), and mark it.
4Identify the feasible region: the area allowed by all constraints together.
5List the corner points. Read intercept corners from the axes and find intersection corners by solving two equations. Check each against all constraints.
6Calculate Z at every corner point in a small table.
7Choose the largest Z for maximisation or the smallest Z for minimisation.
8State the answer in business terms: how many units of each product, and the resulting profit or cost. Mention any unused resource if asked.

Quickest way: Corner table method without perfect drawing

When to use it: Use when time is short and the question needs only the optimal mix and value, not a neat graph.

Find the intercepts of each constraint and sketch a rough graph to see which lines form the boundary.
Write the corner points directly: (0, 0) if allowed, the binding intercepts on each axis, and the intersection of the two lines that bound the region.
Solve only the intersection equations that are really needed, by elimination.
Put all corners in a two-column table of point and Z, then choose the best.
Verify the winning point in every constraint. This takes seconds and protects your marks.

Common mistakes in Graphical Method of Solving LPP

Shading the wrong side of a constraint line
Students assume "≤" always means below the line and "≥" always means above, without testing.
Fix: Substitute (0, 0) into the inequality. If it holds, shade the origin side. Otherwise shade the other side.
Taking an intercept as a corner point when it is outside another constraint
Students list all intercepts of all lines without checking them against the other constraints.
Fix: Test every candidate corner in all constraints. Only points satisfying all of them belong to the feasible region.
Errors in solving the simultaneous equations
Rushing the elimination, or using the wrong constants.
Fix: Substitute the answer back into both equations. If either fails, redo the working.
Choosing the maximum corner in a minimisation problem, or the reverse
Students stop at the first corner that looks good or forget the direction of the objective.
Fix: Write "Maximise" or "Minimise" beside Z and circle it. Compare all corners before choosing.
Forgetting non-negativity or leaving the answer as a bare number
Students focus on the graph and ignore x ≥ 0, y ≥ 0, and the business meaning of the result.
Fix: Always include x ≥ 0 and y ≥ 0 in the formulation and finish with a sentence giving the units of each product and the value of Z in rupees.

Worked examples

Example 1

A firm makes products A and B. Contribution is ₹40 per unit of A and ₹50 per unit of B. Each unit of A needs 1 machine hour and each unit of B needs 2 machine hours; 40 machine hours are available. Each unit of A needs 3 labour hours and each unit of B needs 2 labour hours; 60 labour hours are available. Find the production mix that maximises total contribution.

Show the solution

Let x = units of A and y = units of B. Maximise Z = 40x + 50y.
Constraints: x + 2y ≤ 40 (machine hours); 3x + 2y ≤ 60 (labour hours); x ≥ 0, y ≥ 0.
Intercepts of x + 2y = 40: (40, 0) and (0, 20). Intercepts of 3x + 2y = 60: (20, 0) and (0, 30).
Since both constraints are "≤", the feasible region lies towards the origin and under both lines. On the x-axis the binding limit is x = 20. On the y-axis it is y = 20.
Intersection: subtract the first equation from the second: 2x = 20, so x = 10. Then 10 + 2y = 40 gives y = 15. Check labour: 30 + 30 = 60.
Corner points: (0, 0), (20, 0), (10, 15), (0, 20).
Z at (0, 0) = 0. Z at (20, 0) = 800. Z at (10, 15) = 400 + 750 = 1,150. Z at (0, 20) = 1,000.
The highest Z is 1,150 at (10, 15).

Answer: Produce 10 units of A and 15 units of B. Maximum contribution is ₹1,150. Both machine hours and labour hours are fully used.

Example 2

A canteen mixes two food items, X and Y, to meet nutrition needs. X costs ₹20 per kg and Y costs ₹30 per kg. Each kg of X gives 2 units of nutrient P and 1 unit of nutrient Q. Each kg of Y gives 1 unit of P and 2 units of Q. The mix must supply at least 8 units of P and at least 10 units of Q. Find the least-cost mix.

Show the solution

Let x = kg of X and y = kg of Y. Minimise Z = 20x + 30y.
Constraints: 2x + y ≥ 8 (nutrient P); x + 2y ≥ 10 (nutrient Q); x ≥ 0, y ≥ 0.
Intercepts of 2x + y = 8: (4, 0) and (0, 8). Intercepts of x + 2y = 10: (10, 0) and (0, 5).
Both constraints are "≥", so (0, 0) fails. The feasible region lies away from the origin, above both lines, and is open-ended.
On the y-axis the binding point is (0, 8). On the x-axis the binding point is (10, 0).
Intersection: from 2x + y = 8, y = 8 − 2x. Put in x + 2y = 10: x + 16 − 4x = 10, so x = 2 and y = 4. Check: 4 + 4 = 8 and 2 + 8 = 10.
Corner points: (0, 8), (2, 4), (10, 0).
Z at (0, 8) = 240. Z at (2, 4) = 40 + 120 = 160. Z at (10, 0) = 200.
The region is unbounded, but costs are positive, so Z cannot fall without limit. The smallest value is 160.

Answer: Mix 2 kg of X and 4 kg of Y. The minimum cost is ₹160.

Exam tips

Write the formulation first: define variables, objective and constraints. Marks are usually given for correct formulation even if arithmetic slips later.
Label each line with its equation and mark the feasible region clearly. A clean graph earns presentation marks and helps you find corners.
Show a corner point table with Z values. It makes the examiner's job easy and shows you compared all corners.
In MCQs, you can often skip the graph: find the corners by intercepts and one intersection, then evaluate Z.
End with a recommendation in business language: units to produce, resulting profit or cost, and which resources are fully used.

Practice questions from Linear Programming

Graphical Method of Solving LPP in other exams

The same ground in other exams, if you are preparing for more than one or want another angle on it.

CMA IntermediateGraphical Method of Solving LPP

Graphical Method of Solving LPP: frequently asked questions

What is the feasible region in LPP?

It is the area on the graph where every constraint, including non-negativity, is satisfied at the same time. Every point inside it is a possible plan. The optimal solution lies on a corner of this region.

Why do we test only corner points?

The objective function is linear, so its best value over a feasible region occurs at an extreme point, which is a corner. Points inside the region or along an edge never beat the best corner, except when an edge ties with it.

Can the graphical method handle three variables?

Not in a practical exam setting. It works for two decision variables. For more variables you use the simplex method.

What if there is no feasible region?

Then the constraints contradict each other and the problem is infeasible. No plan can satisfy all the limits, so there is no optimal solution.