Operations Management and Strategic Management · Optimum Allocation of Resources - LPP

Graphical Method of Solving LPP: Corner Point and Iso-Profit Line

Updated 10 October 2026 · Fact-checked

The graphical method solves a linear programming problem with two decision variables. You plot each constraint as a line, shade the feasible region where all constraints hold, then find the corner point that gives the best objective value. You can also slide an iso-profit line to the last point it touches.

Understand Graphical Method of Solving LPP

A linear programming problem (LPP) asks you to maximise or minimise an objective, such as profit or cost, subject to limits on resources. These limits are the constraints. The graphical method works when there are only two decision variables, say x and y, because each can be shown on one axis.

Each constraint like 2x + 3y ≤ 12 is a half-plane. First draw the boundary line 2x + 3y = 12. Then decide which side satisfies the inequality. The non-negativity conditions x ≥ 0 and y ≥ 0 restrict you to the first quadrant.

The feasible region is the area where all constraints are satisfied together. Any point inside it is a feasible solution. The region is a convex polygon (or an unbounded region). The key theorem is that if an optimal solution exists, it occurs at a corner point (extreme point) of the feasible region.

So you have two ways to find it. In the corner point method, you find the coordinates of every corner, compute the objective value at each, and pick the best. In the iso-profit (or iso-cost) line method, you draw the objective line Z = c for some value c, then move it parallel to itself. For maximisation, you move it away from the origin until it touches the last point of the feasible region. For minimisation, you move it towards the origin until it touches the first point of the region.

Key rules to remember

General LPP form: Max or Min Z = ax + by, subject to constraints px + qy ≤, =, ≥ r, and x, y ≥ 0; Z is the objective function. x and y are decision variables.
Boundary line of a constraint: px + qy = r; Intercepts: x = r ÷ p (put y = 0) and y = r ÷ q (put x = 0).
Test point rule: Put (0, 0) in the constraint; if it holds, shade the side containing the origin; Use only when the line does not pass through the origin. Otherwise test another point such as (1, 0).
Corner point theorem: Optimal value of Z, if it exists, occurs at a corner point of the feasible region; Evaluate Z at every corner and compare.
Iso-profit line: ax + by = c; slope = −a ÷ b; Lines for different c are parallel. Maximise: move away from the origin. Minimise: move towards the origin.
Intersection of two lines: Solve the two boundary equations simultaneously; Use elimination to get the coordinates of a corner that is not on an axis.

How to solve Graphical Method of Solving LPP questions

Use this sequence for any two-variable graphical LPP. Write each step in the answer to earn step marks.

1Define the decision variables clearly, for example x = units of product A and y = units of product B.
2Write the objective function and all constraints in mathematical form, including x, y ≥ 0.
3Convert each inequality to an equation and find its two intercepts. Plot each line on the graph.
4Use the test point (0, 0) to decide the side of each constraint. Shade the common area as the feasible region.
5Mark the corner points of the region. Find the coordinates of non-axis corners by solving the two intersecting equations.
6Compute Z at every corner point and tabulate the values.
7Choose the highest Z for maximisation or the lowest Z for minimisation. As a check, draw an iso-profit line and confirm it leaves the region at that corner.
8State the answer in words with units: how many of each product and the resulting profit or cost.

Quickest way: Corner points without drawing a neat graph

When to use it: Use when constraints are all of the ≤ type with positive terms, and time is short. Still show a rough sketch for marks.

Find the axis intercepts of each constraint and note which are the tightest on each axis.
Sketch roughly and identify which two lines meet at the corner away from the axes.
Solve only those two equations for the intersection point.
List the corners: origin, the binding intercepts on each axis, and the intersection.
Calculate Z at each and pick the best. Verify the intersection point satisfies all other constraints.

Common mistakes in Graphical Method of Solving LPP

Shading the wrong side of a constraint.
Students assume ≤ always means below the line, which fails with negative coefficients or ≥ constraints.
Fix: Always test (0, 0) in the inequality. If the line passes through the origin, test another point.
Taking an intersection point that is not in the feasible region as a corner.
Two lines always intersect somewhere, but the point may violate another constraint.
Fix: Substitute every candidate corner in all constraints before using it.
Forgetting x ≥ 0 and y ≥ 0.
The non-negativity conditions are not given as a separate line in the problem.
Fix: Write them in the formulation and restrict the graph to the first quadrant.
Errors in solving simultaneous equations.
Rushing the elimination under time pressure.
Fix: Substitute the answer back into both equations to verify.
Moving the iso-profit line in the wrong direction for minimisation.
Students memorise the maximisation rule only.
Fix: For maximisation, move away from the origin to the farthest point. For minimisation, move towards the origin to the nearest point of the region.
Reporting only Z and not the values of x and y.
The question seems to ask only for maximum profit.
Fix: Always give the optimal quantities and the optimal Z with units.

Worked examples

Example 1

A firm makes products A and B. Profit is ₹40 per unit of A and ₹30 per unit of B. Each unit of A needs 2 hours of machine time and each unit of B needs 1 hour; 100 machine hours are available. Each unit of A needs 1 hour of labour and each unit of B needs 2 hours; 80 labour hours are available. Find the production plan that maximises profit.

Show the solution

Let x = units of A and y = units of B. Maximise Z = 40x + 30y.
Constraints: 2x + y ≤ 100 (machine), x + 2y ≤ 80 (labour), x, y ≥ 0.
Machine line 2x + y = 100 has intercepts (50, 0) and (0, 100). Labour line x + 2y = 80 has intercepts (80, 0) and (0, 40).
The origin satisfies both constraints, so the feasible region lies on the origin side of both lines.
Intersection: from 2x + y = 100 and x + 2y = 80, multiply the second by 2 to get 2x + 4y = 160. Subtract the first: 3y = 60, so y = 20. Then x = 80 − 40 = 40. The point is (40, 20).
Corner points of the region: (0, 0), (50, 0), (40, 20), (0, 40).
Z at (0, 0) = 0. Z at (50, 0) = 2,000. Z at (40, 20) = 1,600 + 600 = 2,200. Z at (0, 40) = 1,200.
The maximum is 2,200 at (40, 20). Check: machine 80 + 20 = 100 and labour 40 + 40 = 80, both fully used.

Answer: Produce 40 units of A and 20 units of B for a maximum profit of ₹2,200.

Example 2

A feed mix must contain at least 12 units of nutrient P and at least 8 units of nutrient Q. Ingredient X costs ₹6 per kg and gives 3 units of P and 1 unit of Q per kg. Ingredient Y costs ₹4 per kg and gives 2 units of P and 2 units of Q per kg. Find the mix of minimum cost.

Show the solution

Let x = kg of X and y = kg of Y. Minimise Z = 6x + 4y.
Constraints: 3x + 2y ≥ 12, x + 2y ≥ 8, x, y ≥ 0.
Line 3x + 2y = 12 has intercepts (4, 0) and (0, 6). Line x + 2y = 8 has intercepts (8, 0) and (0, 4).
The origin fails both ≥ constraints, so the feasible region lies on the side away from the origin. It is unbounded.
Intersection: subtract x + 2y = 8 from 3x + 2y = 12 to get 2x = 4, so x = 2. Then 2y = 6, so y = 3. The point is (2, 3).
Corner points: (8, 0), (2, 3), (0, 6). Check (8, 0): 3(8) = 24 ≥ 12 and 8 ≥ 8, feasible. Check (0, 6): 12 ≥ 12 and 12 ≥ 8, feasible.
Z at (8, 0) = 48. Z at (2, 3) = 12 + 12 = 24. Z at (0, 6) = 24.
Z is 24 at both (2, 3) and (0, 6). The iso-cost line 6x + 4y = 24 is parallel to the edge 3x + 2y = 12, which joins these two corners, so every point on that edge gives ₹24. This is a case of multiple optimal solutions. Since the region is unbounded and the objective has positive coefficients, the minimum exists.

Answer: The minimum cost is ₹24. It is obtained at (2, 3), or at (0, 6), or at any point on the line segment joining them.

Exam tips

Show the formulation, the plotted constraint lines, a shaded feasible region and a corner point table. Each earns marks even if the final arithmetic slips.
Label every line with its equation and mark the corner coordinates on the graph.
In MCQs, you often need only the corner points. Find the intersection and compare Z values rather than drawing a full graph.
Check whether the objective line is parallel to a constraint. If two corners give the same Z, state that multiple optimal solutions exist.
Write a one-line conclusion in business terms, such as units to produce and the profit earned.

Practice questions from Optimum Allocation of Resources - LPP

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 FinalGraphical Method of Solving LPP

Graphical Method of Solving LPP: frequently asked questions

How do I find the feasible region in the graphical method?

Plot each constraint as a line and test (0, 0) to find the correct side. Shade that side for every constraint. The area common to all shaded sides, within the first quadrant, is the feasible region.

Which is better, the corner point method or the iso-profit line method?

The corner point method is safer because you compare actual values. The iso-profit line method is quicker for a clear picture, but it needs an accurate graph. Many students use the iso-profit line as a cross-check.

Can the graphical method be used for three variables?

Not in practice. It works for two decision variables because the solution is plotted on a plane. For more variables, you use the simplex method.

What if the feasible region is unbounded?

Compute Z at the corner points as usual. For maximisation with an unbounded region the objective may increase without limit, so no finite optimum exists. For minimisation with positive cost coefficients, a minimum usually exists at a corner.