Class 12 Maths · Original EZT practice
Linear Programming (Class 12 Maths, NCERT, Chapter 12)
Linear Programming is Chapter 12 of the NCERT Class 12 Mathematics Part 2 textbook. A furniture maker with limited machine hours and a profit per table and chair faces the same shape as a diet plan on a budget: 'such type of problems which seek to maximise (or, minimise) profit (or, cost) form a general class of problems called optimisation problems'. This chapter turns Class 11's linear inequalities into a solving machine: formulate - decision variables, objective function, constraints - shade the feasible region, and read the optimum off the corner points. (Note: this is original practice, not a textbook answer key.)
6 original worked examples, 6 fresh self-check questions and 4 mixed questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 26 October 2026:
1. What the chapter covers
12.1 Introduction: from systems of linear inequalities (Class 11) to optimisation; the Koopmans/Kantorovich lineage. 12.2 Linear Programming Problem and its Mathematical Formulation: 12.2.1 formulating the problem - decision variables, linear objective function, linear constraints and non-negativity restrictions; 12.2.2 the graphical method - feasible region (bounded and unbounded), feasible and optimal solutions, the corner-point (vertex) theorem. EXERCISE 12.1 closes the chapter in this printing.
2. Key ideas at a glance
Every LPP has three parts: decision variables, a linear objective function to maximise or minimise, and linear constraints (plus x, y >= 0). The feasible region is the common shaded area of all constraints; points inside it are feasible solutions. Corner-point theorem: if an optimal value exists over a bounded feasible region, it occurs at a vertex - so evaluate Z at every corner. An unbounded feasible region can still have a maximum or minimum - the open half-plane test decides. No feasible region means no solution: the constraints contradict each other.
3. Traps this chapter sets
Declaring the line-intersection point the optimum by habit: every vertex competes, including the axis intercepts, and the maximum can sit on an axis. Forgetting non-negativity: x, y >= 0 restricts the picture to the first quadrant and adds the origin and axis intercepts as candidate vertices. Skipping the half-plane test on an unbounded region: the best corner value is not yet the optimum until the open half-plane below (or above) it misses the region. Misreading which constraint binds on an axis: the vertex is where the region first meets the axis, which can come from either constraint.
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Formulate first
Choose the method: Variables, objective, constraints.
EZT practice question
A home baker takes daily orders for cakes (profit Rs 400 each, 2 oven-hours) and batches of cookies (profit Rs 150 each, 1 oven-hour). The oven is available 16 hours a day and she accepts at most 10 orders a day. Formulate the LPP for maximum daily profit.
Show the worked solution
- Let x = cakes, y = cookie batches.
- Maximise Z = 400x + 150y.
- Subject to 2x + y <= 16 (oven-hours), x + y <= 10 (order count), x >= 0, y >= 0.
Answer: Maximise Z = 400x + 150y subject to 2x + y <= 16, x + y <= 10, x >= 0, y >= 0 - decision variables, objective, constraints and non-negativity, the formulation pattern of section 12.2.1.
Check: Each sentence of the word problem matched to one constraint; profits in the objective, resources in the constraints.
Watch for this: Do not put the profits in the constraints or the oven-hours in the objective - money measures the goal, hours measure the limits; swapping them is the most common formulation error.
Now try it yourself
Why do x >= 0 and y >= 0 appear even when nobody says so?
Check your answer
Negative cakes and negative cookie batches are meaningless in the story, so non-negativity restrictions are part of every formulation - stated or not.
Formulation is translation: the real world's silent assumptions still have to be written down.
Check every corner
Choose the method: Vertices, values, verdict.
EZT practice question
Maximise Z = 5x + 4y subject to 2x + 5y <= 16, x + y <= 5, x >= 0, y >= 0. Find the corner points and the maximum - and note which vertex wins.
Show the worked solution
- Vertices: (0, 0), (5, 0), (0, 16/5), and the intersection of 2x + 5y = 16 with x + y = 5 at (3, 2).
- Z values are 0, 25, 12.8, 23.
- The maximum 25 is at the axis vertex (5, 0), not at the intersection.
Answer: Maximum 25 at (5, 0). Vertices (0, 0), (5, 0), (0, 16/5), (3, 2) give Z = 0, 25, 12.8, 23 - the interior intersection (3, 2) is feasible (it meets both constraints with equality) but it is not the optimum.
Check: All four vertices listed including the origin and both axis intercepts, and each Z value recomputed.
Watch for this: Do not assume the intersection of the constraint lines is automatically the optimum - here it loses to an axis vertex, because the objective's slope decides which corner wins, not the drawing's centre.
Now try it yourself
When does the intersection point win?
Check your answer
When the objective's gradient points between the outward normals of the two binding constraints - geometrically, when moving along either edge from the intersection lowers Z. Nothing about being an intersection guarantees it.
The corner-point theorem promises the optimum is a vertex; it never says which one.
Run the extra check
Choose the method: Corners, then half-plane.
EZT practice question
Minimise Z = 4x + 6y subject to x + 2y >= 8, 2x + y >= 7, x, y >= 0. Find the vertices of the feasible region and the minimum, stating every check the unbounded region demands.
Show the worked solution
- On the y-axis the binding constraint is 2x + y >= 7, giving vertex (0, 7) - note x + 2y >= 8 only asks y >= 4 there, so (0, 8) is NOT a vertex.
- On the x-axis the binding one is x + 2y >= 8, giving (8, 0); the lines cross at (2, 3).
- Z values are 42, 32, 26. Region is unbounded, so apply the open-half-plane check on 4x + 6y < 26, i.e. 2x + 3y < 13, shares no point with the feasible region (2x + 3y takes its smallest value 13 at (2, 3) itself) - so the minimum is 26 at (2, 3).
Answer: Minimum 26 at (2, 3), vertices (0, 7), (8, 0), (2, 3) with Z = 42, 32, 26 - confirmed by the open-half-plane test: 4x + 6y < 26 has no point in common with the feasible region, since 2x + 3y bottoms out at 13 exactly at (2, 3).
Check: The binding constraint on each axis identified (not assumed), the intersection recomputed, and the half-plane test written out for the unbounded region.
Watch for this: Do not stop at the smallest corner value on an unbounded region - and do not read (0, 8) off the first constraint alone: on the y-axis the binding constraint is 2x + y >= 7, so the true vertex is (0, 7).
Now try it yourself
What would it mean if the half-plane 4x + 6y < 26 DID meet the region?
Check your answer
Some feasible point would beat 26, so 26 would not be the minimum - and since the region is unbounded, Z might fall without limit, meaning no minimum exists at all.
The half-plane test is the difference between the best corner seen and the true optimum.
Why corners suffice
Choose the method: Linearity flattens the interior.
EZT practice question
Explain in one sentence why checking only the corner points of a bounded feasible region is enough.
Show the worked solution
- A linear function on a bounded polygon attains its maximum and minimum at a vertex.
- So evaluating Z at the corners finds every possible optimum.
- This is the corner-point theorem.
Answer: A linear function on a bounded polygon attains its maximum and minimum at a vertex - the corner-point theorem - so the corner values exhaust the possibilities.
Check: The sentence mentions linearity, the bounded region, and attainment at a vertex.
Watch for this: Do not stretch the theorem to nonlinear objectives or unbounded regions without its extra test - linearity and boundedness are the hypotheses, not decoration.
Now try it yourself
Why does linearity force the optimum to a vertex?
Check your answer
A linear function has no interior peaks or valleys - along any edge it rises, falls or stays flat, so the best value on each edge sits at an endpoint, and the region is the union of such behaviour.
Flat objectives cannot hide their best point inside.
Spot the empty region
Choose the method: Shade and find nothing.
EZT practice question
A problem has constraints x + y >= 8 and x + y <= 3 with x, y >= 0. What can you say about solutions?
Show the worked solution
- No point can satisfy x + y >= 8 and x + y <= 3 at once.
- So the feasible region is empty.
- The LPP has no solution - the constraints are inconsistent.
Answer: No solution exists: x + y cannot be at least 8 and at most 3 simultaneously, so the feasible region is empty - inconsistent constraints.
Check: The two half-planes shown disjoint; conclusion stated as no feasible region, hence no solution.
Watch for this: Do not go hunting for vertices or an optimum once the region is empty - formulation checks come before the graphical method, and an empty region ends the problem.
Now try it yourself
How is an empty region different from an unbounded one?
Check your answer
Empty means no feasible point exists at all; unbounded means feasible points exist but the objective may improve forever - one kills the problem, the other only warns it.
Both are failure modes, but of different stages: feasibility and optimality.
Name the safeguard
Choose the method: Best corner, then test.
EZT practice question
For an unbounded feasible region with Z = 3x + 2y to minimise, the smallest corner value is 10. State the extra check before declaring 10 the minimum.
Show the worked solution
- Draw the open half-plane 3x + 2y < 10.
- Check whether it shares any point with the feasible region.
- Only if it has no point in common with the region is 10 truly the minimum.
Answer: Test the open half-plane 3x + 2y < 10 against the feasible region: if they share no point, 10 is the minimum; if they do, 10 is beaten and may not be attainable at all - the open-half-plane test for unbounded regions.
Check: The strict inequality (open half-plane), the comparison with the region, and the conditional conclusion.
Watch for this: Do not use 3x + 2y <= 10 in the test - the half-plane must be strictly below the candidate value; points with Z = 10 exactly are allowed, points below it are not.
Now try it yourself
Why does the bounded case not need this check?
Check your answer
On a bounded region the corner-point theorem guarantees a vertex attains the optimum - there is nowhere for the objective to escape to; unboundedness opens that escape route, and the test closes it.
Every extra hypothesis removed costs an extra check.
Try a mixed set
Choose a method or explanation before opening each answer.
1. Three parts of every LPP?
Show answer and reason
Decision variables, a linear objective function, and linear constraints - plus non-negativity restrictions.
Formulation is the first half of every mark.
2. Feasible region?
Show answer and reason
The common shaded region of all constraints; every point in it is a feasible solution.
No feasible region, no solution.
3. Corner-point theorem?
Show answer and reason
Over a bounded feasible region, the optimum (if it exists) occurs at a vertex - so evaluate Z at every corner.
Corners only; never assume which corner.
4. Unbounded region, minimum?
Show answer and reason
Best corner value, then the open-half-plane test: the half-plane below that value must miss the region.
The smallest corner seen is not the minimum until the test passes.
Try it yourself
Work these on paper before checking any notes. Answers are deliberately not shown here.
- Explain in your own words why the optimum of a linear objective over a bounded feasible region cannot hide strictly inside the region.
- Take the baker's LPP (2x + y <= 16, x + y <= 10) and list all four vertices of its feasible region with the value of Z = 400x + 150y at each.
- State in one sentence the difference between an LPP with no feasible region and one whose feasible region is unbounded.
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Questions about this practice page
- Which book is Class 12 Maths Chapter 12 Linear Programming from?
- It is Chapter 12 of the NCERT Class 12 Mathematics Part 2 textbook (official NCERT chapter PDF lemh206.pdf). No printing-year footer was observed in the fetched chapter text; the front matter lists Edition January 2023 with reprints through 'Reprint 2026-27' - check your own copy, as printings can differ.
- What is a linear programming problem?
- An optimisation problem - per the chapter, problems that 'seek to maximise (or, minimise) profit (or, cost)' - where the objective function and every constraint are linear in the decision variables, with non-negativity restrictions.
- What is a feasible region?
- The common region satisfying all the constraints at once, found by shading each inequality on the graph. Every point of it is a feasible solution; the optimal solution is the feasible point giving the best objective value.
- What is the corner-point method?
- Evaluate the objective function Z at each vertex of the feasible region: the largest value is the maximum and the smallest the minimum, because a linear objective over a bounded feasible region attains its optimum at a vertex.
- Can an LPP have no solution?
- Yes, two ways: the constraints can contradict each other (empty feasible region), or the feasible region can be unbounded in the direction that improves Z without limit - no finite optimum exists then.