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Linear programming is a short, predictable chapter in CBSE Class 12 maths, yet students lose marks on the same few steps: shading the wrong side, misreading a corner, or declaring a minimum for an unbounded region without checking. This guide walks through the method, three worked problems with every number checked, and a short practice plan.
What does the CBSE syllabus cover?
The CBSE Mathematics 2026-27 curriculum gives Unit V, Linear Programming, 5 marks out of 80. It covers constraints, objective function, the graphical method for two variables, feasible and infeasible regions (bounded or unbounded) and optimal feasible solutions with up to three non-trivial constraints. The same document lists the graphical solution of linear inequalities in Class XI as assessed only formatively, so you may not have practised graphing inequalities under exam conditions. Check the current curriculum before you plan.
The idea has real reach. Leonid Kantorovich and Tjalling Koopmans received the 1975 economics prize in memory of Alfred Nobel "for their contributions to the theory of optimum allocation of resources".
Why are corner points enough?
The NCERT Linear Programming chapter states two theorems. If the objective Z = ax + by has an optimal value, it occurs at a corner point of the feasible region. If the region is bounded, Z has both a maximum and a minimum, each at a corner point. If the region is unbounded, a maximum or minimum may not exist.
NCERT's corner point method:
Draw the feasible region and find its corner points by solving pairs of boundary equations.
Evaluate Z at each corner. Call the largest value M and the smallest m.
If the region is bounded, M and m are the answers.
If the region is unbounded, M is the maximum only if the open half-plane ax + by > M has no point in common with the region. Likewise, m is the minimum only if ax + by < m has no common point.
Worked example 1: a bounded region
Maximise Z = 2x + 5y subject to x + y ≤ 4, x + 3y ≤ 6, x ≥ 0, y ≥ 0.
Boundary lines. x + y = 4 meets the axes at (4, 0) and (0, 4). x + 3y = 6 meets them at (6, 0) and (0, 2).
Shading. Test (0, 0): 0 ≤ 4 and 0 ≤ 6 are both true, so shade towards the origin for both.
Corners. O(0, 0); A(4, 0), since (6, 0) breaks x + y ≤ 4; C(0, 2), since (0, 4) breaks x + 3y ≤ 6; and B where the two lines meet. Subtracting x + y = 4 from x + 3y = 6 gives 2y = 2, so y = 1 and x = 3. B = (3, 1).
Evaluate. Z(O) = 0. Z(A) = 8. Z(B) = 6 + 5 = 11. Z(C) = 10.
Maximum Z = 11 at (3, 1). Check B: 3 + 1 = 4 and 3 + 3 = 6, so it satisfies both constraints exactly.
Worked example 2: two corners tie
Same region, but maximise Z = x + y. Z(A) = 4, Z(B) = 4, Z(C) = 2, Z(O) = 0.
The maximum 4 occurs at both A and B. NCERT notes that when two corners give the same optimal value, every point on the segment joining them does too. So the answer is every point on the segment from (4, 0) to (3, 1). Write that, not just one point.
Worked example 3: an unbounded region
Minimise Z = 3x + 5y subject to x + 3y ≥ 3, x + y ≥ 2, x ≥ 0, y ≥ 0.
Shading. Test (0, 0): 0 ≥ 3 is false, so shade away from the origin. The region is unbounded.
Corners. On the x-axis, x + 3y ≥ 3 needs x ≥ 3, so the corner is (3, 0). On the y-axis, x + y ≥ 2 needs y ≥ 2, so the corner is (0, 2). Where the lines meet: subtract x + y = 2 from x + 3y = 3 to get 2y = 1, so y = 0.5 and x = 1.5.
Evaluate. Z(3, 0) = 9. Z(1.5, 0.5) = 4.5 + 2.5 = 7. Z(0, 2) = 10. So m = 7.
Half-plane check. Draw 3x + 5y = 7. It passes through (1.5, 0.5) and the region lies entirely on the side where 3x + 5y ≥ 7, so 3x + 5y < 7 has no common point.
Algebraic check: 3x + 5y = (x + 3y) + 2(x + y) ≥ 3 + 2(2) = 7 on the whole region. Minimum Z = 7 at (1.5, 0.5). There is no maximum, because Z grows without limit as x or y grows.
What are the common mistakes?
Shading the wrong side. Always test a point such as (0, 0), unless the line passes through it.
Forgetting x ≥ 0, y ≥ 0. These cut the region to the first quadrant.
Reading corners off the graph. Solve the two equations. A corner at (3, 1) can look like (3, 1.2) on a rough sketch.
Skipping the half-plane check. In an unbounded region, a smallest corner value is not automatically the minimum.
Badly defined variables in word problems. Write "let x be the number of ..." with units before writing any constraint.
A practice routine for the week before your test
Day 1: Five NCERT exercise problems with bounded regions. Draw every graph with a ruler.
Day 2: Three unbounded problems, writing the half-plane check in full each time.
Day 3: Two word problems: define variables, write constraints, then solve.
Day 4: One timed past-paper LP question, marked against the CBSE marking scheme.
For the wider exam plan, see our CBSE Class 12 maths strategy. If a word problem stalls you at the set-up, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Tonight, redo worked example 3 from scratch on graph paper and check each corner by substitution.
