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Class 12 Maths NCERT Solutions

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Linear Programming Class 12 Maths NCERT Solutions

The complete NCERT exercise solutions for Chapter 12, Linear Programming — 18 questions from Ex 12.1 to Ex 12.2, each worked through step by step in the CBSE marking pattern. Formulating and solving linear programming problems, the feasible region, and the graphical method for optimum points.

Class:12Subject:MathsChapter:12
3 Key Formulas
DWritten byDeep Narayan
Updated
Key Concept Summary

How many questions are in NCERT Class 12 Maths Chapter 12?

Chapter 12 carries 2 exercise questions, numbered Ex 12.1 to Ex 12.2. All of them are solved step by step on this page, along with the chapter's key formulas and exam pointers.

01

Chapter Overview

Linear programming is a mathematical technique for optimising (maximising or minimising) a linear objective function subject to linear constraints. It has wide applications in business, economics, industry and military planning. In this chapter we learn to formulate real-world problems as linear programming problems (LPPs), solve them using the graphical method, and identify optimal solutions at corner points of the feasible region.

Board pattern

In board exams, 4-6 mark questions on LPP expect you to (i) define variables, (ii) write the objective function, (iii) list all constraints including non-negativity, (iv) draw the feasible region accurately, (v) find corner points, and (vi) evaluate Z at each corner point to state the optimal value. Always label axes, shade the feasible region, and verify whether it is bounded or unbounded.
02

Exercise 12.1 — Linear Programming: Graphical Method

8Exercise questions

Step-by-step solution

  1. 1Constraints: x + y ≤ 4, x ≥ 0, y ≥ 0.
  2. 2The feasible region is bounded by the lines x + y = 4, x = 0, y = 0.
  3. 3Corner points: O(0,0), A(4,0), B(0,4).
  4. 4Z at O(0,0) = 0; Z at A(4,0) = 12; Z at B(0,4) = 16.
  5. 5Maximum value is 16 at (0, 4).

Final answer

Step-by-step solution

  1. 1Plot the lines x + 2y = 8 and 3x + 2y = 12 along with x = 0, y = 0.
  2. 2Feasible region is bounded. Solve x + 2y = 8 and 3x + 2y = 12 simultaneously: subtracting gives 2x = 4, so x = 2, y = 3. Intersection at (2, 3).
  3. 3Corner points: O(0,0), A(4,0), B(2,3), C(0,4).
  4. 4Z at O = 0; Z at A = -12; Z at B = 9; Z at C = 16.
  5. 5Minimum value is -12 at (4, 0).

Final answer

Step-by-step solution

  1. 1Plot 3x + 5y = 15 and 5x + 2y = 10 in the first quadrant.
  2. 2Find intersection: from 3x + 5y = 15 and 5x + 2y = 10. Multiply first by 2 and second by 5: 6x + 10y = 30 and 25x + 10y = 50. Subtracting: 19x = 20, x = 20/19. Then 5(20/19) + 2y = 10 gives y = 45/19.
  3. 3Corner points: O(0,0), A(2,0), B(20/19, 45/19), C(0,3).
  4. 4Z at O = 0; Z at A = 10; Z at B = 5(20/19) + 3(45/19) = 100/19 + 135/19 = 235/19 ≈ 12.37; Z at C = 9.
  5. 5Maximum value is 235/19 at (20/19, 45/19).

Final answer

Step-by-step solution

  1. 1Plot x + 3y = 3 and x + y = 1 in the first quadrant. Shade regions above both lines.
  2. 2The feasible region is unbounded (extends infinitely upward).
  3. 3Corner points: A(0,1), B(3,0).
  4. 4Z at A(0,1) = 5; Z at B(3,0) = 9.
  5. 5Since the region is unbounded, check: can Z < 5 be achieved? Draw 3x + 5y = 5; it passes through (0,1). For any point in the feasible region, 3x + 5y ≥ 5.
  6. 6Minimum value is 5 at (0, 1).

Final answer

Step-by-step solution

  1. 1Plot x + 3y = 3 and x + y = 1.
  2. 2Feasible region: below both lines in the first quadrant.
  3. 3Corner points: O(0,0), A(1,0), B(0,1).
  4. 4Z at O = 0; Z at A = 3; Z at B = 5.
  5. 5Maximum value is 5 at (0, 1).

Final answer

Step-by-step solution

  1. 1Plot x + 2y = 12 and x + y = 6.
  2. 2Feasible region: bounded by x + 2y ≤ 12 (below), x + y ≥ 6 (above), x ≥ 0, y ≥ 0.
  3. 3On x-axis (y=0): x ≤ 12 and x ≥ 6, so x in [6,12]. On y-axis (x=0): 2y ≤ 12 gives y ≤ 6, and y ≥ 6, so y = 6.
  4. 4Corner points: A(6,0), B(12,0), C(0,6).
  5. 5Z at A = 30; Z at B = 60; Z at C = 60.
  6. 6Minimum value is 30 at (6, 0); maximum value is 60 at (12, 0) and (0, 6) and along the line segment joining them.

Final answer

Step-by-step solution

  1. 1Plot 2x + y = 3 and x + 2y = 6.
  2. 2Find intersection: from 2x + y = 3, y = 3 - 2x. Substitute: x + 2(3 - 2x) = 6, x + 6 - 4x = 6, -3x = 0, x = 0, y = 3. Intersection at (0, 3).
  3. 3Feasible region: 2x + y ≥ 3 (above), x + 2y ≤ 6 (below), x ≥ 0, y ≥ 0.
  4. 4On x-axis: 2x ≥ 3 gives x ≥ 3/2, and x ≤ 6. So A(3/2, 0), B(6, 0).
  5. 5Corner points: A(3/2, 0), B(6, 0), C(0, 3).
  6. 6Z at A = 3/2; Z at B = 6; Z at C = 6.
  7. 7The region is unbounded (x can grow large along the x-axis). As x → ∞, Z = x + 2y → ∞, so no finite maximum.
  8. 8Minimum value is 3/2 at (3/2, 0).

Final answer

Step-by-step solution

  1. 1Plot x₁ + x₂ = 4 and x₁ + 3x₂ = 6.
  2. 2Find intersection: from x₁ + x₂ = 4 and x₁ + 3x₂ = 6, subtracting gives 2x₂ = 2, x₂ = 1, x₁ = 3. Intersection at (3, 1).
  3. 3Feasible region is bounded. Corner points: O(0,0), A(4,0), B(3,1), C(0,2).
  4. 4Z at O = 0; Z at A = -4; Z at B = -3 + 2 = -1; Z at C = 4.
  5. 5Maximum value is 4 at (0, 2).

Final answer

03

Exercise 12.2 — Applications of Linear Programming

10Exercise questions

Step-by-step solution

  1. 1Let x = units of F₁ and y = units of F₂.
  2. 2Objective: Minimize Z = 4x + 6y.
  3. 3Constraints: 3x + 5y ≥ 80 (vitamin A), 5x + 2y ≥ 100 (minerals), x ≥ 0, y ≥ 0.
  4. 4Plot 3x + 5y = 80 and 5x + 2y = 100.
  5. 5Intersection: 3x + 5y = 80 and 5x + 2y = 100. Multiply first by 2 and second by 5: 6x + 10y = 160, 25x + 10y = 500. Subtracting: 19x = 340, x = 340/19. Then y = (80 - 3(340/19))/5 = (1520 - 1020)/95 = 500/95 = 100/19.
  6. 6Corner points of the feasible region (unbounded, above both lines): A(0, 40), B(340/19, 100/19), C(20, 0).
  7. 7Z at A = 240; Z at B = 4(340/19) + 6(100/19) = (1360 + 600)/19 = 1960/19 ≈ 103.16; Z at C = 80.
  8. 8Check unbounded region: draw 4x + 6y = 80. This does not intersect the feasible region in its interior, so Z ≥ 80 throughout.
  9. 9Minimum cost is Rs 80 at (20, 0).

Final answer

Step-by-step solution

  1. 1Let x = number of tables and y = number of chairs.
  2. 2Objective: Maximize Z = 500x + 300y.
  3. 3Constraints: 2x + y ≤ 120 (carpentry), x + y ≤ 80 (finishing), x ≥ 0, y ≥ 0.
  4. 4Plot 2x + y = 120 and x + y = 80.
  5. 5Intersection: subtracting gives x = 40, y = 40.
  6. 6Corner points: O(0,0), A(60,0), B(40,40), C(0,80).
  7. 7Z at O = 0; Z at A = 30000; Z at B = 20000 + 12000 = 32000; Z at C = 24000.
  8. 8Maximum profit is Rs 32000 at (40, 40).

Final answer

Step-by-step solution

  1. 1Let x = units of Food I, y = units of Food II.
  2. 2Objective: Minimize Z = 50x + 70y.
  3. 3Constraints: 2x + y ≥ 10 (vitamin A), x + 2y ≥ 8 (vitamin B), x + 3y ≥ 7 (vitamin C), x ≥ 0, y ≥ 0.
  4. 4Plot the three lines and identify the feasible region (unbounded, above all three lines).
  5. 5Find key intersections: 2x + y = 10 and x + 2y = 8 give x = 4, y = 2. Check third: 4 + 6 = 10 ≥ 7 ✓.
  6. 6x + 2y = 8 and x + 3y = 7 give y = -1 (infeasible).
  7. 72x + y = 10 and x + 3y = 7 give x = 23/5, y = 4/5. Check second: 23/5 + 8/5 = 31/5 = 6.2 < 8, fails.
  8. 8Corner points of feasible region: A(0, 10/2) on y-axis: y ≥ 10 from first constraint (x=0 gives y ≥ 10). Actually (0, 10): 0+20=20≥8, 0+30=30≥7. But also check (0, 7/3): 0+7/3 = 2.33 < 8, fails. So on y-axis, need 2(0)+y≥10 gives y≥10. And (0,8) from second: 0+16≥8, but 0+8≥10 fails. So A(0,10).
  9. 9Also B(4,2) and on x-axis: 2x≥10 gives x≥5. C(5,0): check x+2y=5≥8 fails! So need x≥8 on x-axis. C(8,0): 16+0≥10, 8+0≥8, 8+0≥7 ✓.
  10. 10Corner points: A(0, 10), B(4, 2), C(8, 0).
  11. 11Z at A = 700; Z at B = 200 + 140 = 340; Z at C = 400.
  12. 12Minimum cost is Rs 340 at (4, 2). Check unbounded: 50x + 70y < 340 has no feasible point.

Final answer

Step-by-step solution

  1. 1Let x = units of A, y = units of B.
  2. 2Objective: Maximize Z = 30x + 40y.
  3. 3Constraints: 2x + y ≤ 200 (Machine I), x + 3y ≤ 300 (Machine II), x ≥ 0, y ≥ 0.
  4. 4Plot 2x + y = 200 and x + 3y = 300.
  5. 5Intersection: from first y = 200 - 2x, substitute: x + 600 - 6x = 300, -5x = -300, x = 60, y = 80.
  6. 6Corner points: O(0,0), A(100, 0) [from 2x+y=200 on x-axis], B(60, 80), C(0, 100) [from x+3y=300 on y-axis].
  7. 7Z at O = 0; Z at A = 3000; Z at B = 1800 + 3200 = 5000; Z at C = 4000.
  8. 8Maximum profit is Rs 5000 at (60, 80).

Final answer

Step-by-step solution

  1. 1Let x = units of P, y = units of Q.
  2. 2Objective: Maximize Z = 200x + 300y.
  3. 3Constraints: x + 2y ≤ 8 (Machine A), 2x + y ≤ 10 (Machine B), x + 3y ≤ 12 (Machine C), x ≥ 0, y ≥ 0.
  4. 4Find intersections of constraint pairs.
  5. 5x + 2y = 8 and 2x + y = 10: multiply first by 2: 2x + 4y = 16, subtract second: 3y = 6, y = 2, x = 4. Check Machine C: 4 + 6 = 10 ≤ 12 ✓.
  6. 6x + 2y = 8 and x + 3y = 12: subtracting gives y = 4, x = 0. Check Machine B: 0 + 4 = 4 ≤ 10 ✓.
  7. 72x + y = 10 and x + 3y = 12: from first y = 10 - 2x, x + 30 - 6x = 12, -5x = -18, x = 18/5 = 3.6, y = 2.8. Check Machine A: 3.6 + 5.6 = 9.2 > 8, fails.
  8. 8Corner points: O(0,0), A(5,0) [from 2x+y=10], B(4,2), C(0,4) [from x+2y=8].
  9. 9Z at O = 0; Z at A = 1000; Z at B = 800 + 600 = 1400; Z at C = 1200.
  10. 10Maximum profit is Rs 1400 at (4, 2).

Final answer

Step-by-step solution

  1. 1Let a = fraction of capacity used on Machine A, b = fraction on Machine B (0 ≤ a, b ≤ 1).
  2. 2Item X produced: 120a + 60b ≥ 200. Item Y produced: 80a + 140b ≥ 160.
  3. 3Simplify: 6a + 3b ≥ 10 (from X), 4a + 7b ≥ 8 (from Y).
  4. 4Objective: Minimize Z = 2000a + 1600b.
  5. 5Constraints: 6a + 3b ≥ 10, 4a + 7b ≥ 8, 0 ≤ a ≤ 1, 0 ≤ b ≤ 1.
  6. 6Plot 6a + 3b = 10 and 4a + 7b = 8 in the a-b plane (first quadrant).
  7. 7Intersection: multiply first by 7 and second by 3: 42a + 21b = 70, 12a + 21b = 24. Subtracting: 30a = 46, a = 46/30 = 23/15 > 1. Not feasible.
  8. 8Since intersection is outside [0,1] x [0,1], the binding constraints in the feasible region are different. On b = 1: 6a + 3 ≥ 10 gives a ≥ 7/6 > 1, fails. On a = 1: 6 + 3b ≥ 10 gives b ≥ 4/3 > 1, fails.
  9. 9The problem as stated may be infeasible within the [0,1] box. In practice, the NCERT problem uses different numbers. With the given data, the factory cannot meet both demand constraints with one day of each machine. The minimum cost solution requires running Machine A for 5/3 days and Machine B for 0 days (cost Rs 3333.33), or running both for extended periods.
  10. 10For a one-day allocation: a = 1, b = 1. Check: 6+3=9 < 10, fails X demand. The demand cannot be met in one day.
  11. 11Reinterpret: let a and b be the number of days each machine runs (no upper bound). Then minimize Z = 2000a + 1600b subject to 6a + 3b ≥ 10, 4a + 7b ≥ 8, a ≥ 0, b ≥ 0.
  12. 12Corner points: A(0, 10/3), B(23/15, 8/7)... since intersection at a = 23/15 ≈ 1.53, b = (10 - 6(23/15))/3 = (150 - 138)/45 = 12/45 = 4/15 ≈ 0.27. Check 4a+7b = 92/15 + 28/15 = 120/15 = 8 ✓.
  13. 13Corner points: A(0, 10/3), B(23/15, 4/15), C(5/3, 0).
  14. 14Z at A = 16000/3 ≈ 5333; Z at B = 2000(23/15) + 1600(4/15) = 46000/15 + 6400/15 = 52400/15 ≈ 3493; Z at C = 10000/3 ≈ 3333.
  15. 15Minimum cost is Rs 10000/3 ≈ Rs 3333 at (5/3, 0) — run Machine A for 5/3 days, Machine B for 0 days.

Final answer

Step-by-step solution

  1. 1Let x = pills of Brand X, y = pills of Brand Y.
  2. 2Objective: Minimize Z = 40x + 60y.
  3. 3Constraints: 20x + 30y ≥ 200 (vitamin A), 100x + 150y ≥ 500 (vitamin B), 30x + 20y ≥ 150 (vitamin C), x ≥ 0, y ≥ 0.
  4. 4Simplify: 2x + 3y ≥ 20, 2x + 3y ≥ 10, 3x + 2y ≥ 15. The second constraint is redundant (implied by the first).
  5. 5Effective: 2x + 3y ≥ 20, 3x + 2y ≥ 15.
  6. 6Plot 2x + 3y = 20 and 3x + 2y = 15.
  7. 7Intersection: multiply first by 3, second by 2: 6x + 9y = 60, 6x + 4y = 30. Subtract: 5y = 30, y = 6, x = (20 - 18)/2 = 1.
  8. 8On y-axis: 3y ≥ 20 gives y ≥ 20/3; 2y ≥ 15 gives y ≥ 7.5. So y ≥ 20/3. On x-axis: 2x ≥ 20 gives x ≥ 10; 3x ≥ 15 gives x ≥ 5. So x ≥ 10.
  9. 9Corner points: A(0, 20/3), B(1, 6), C(10, 0).
  10. 10Z at A = 60 × 20/3 = 400; Z at B = 40 + 360 = 400; Z at C = 400.
  11. 11All three corner points give Z = 400. Minimum cost is Rs 400, achieved at every point on the boundary of the feasible region (all give the same value).

Final answer

Step-by-step solution

  1. 1Let x = units of T₁, y = units of T₂.
  2. 2Objective: Maximize Z = 80x + 120y.
  3. 3Constraints: 2x + y ≤ 120 (dept A), x + 3y ≤ 150 (dept B), x ≥ 0, y ≥ 0.
  4. 4Plot 2x + y = 120 and x + 3y = 150.
  5. 5Intersection: y = 120 - 2x, x + 3(120 - 2x) = 150, x + 360 - 6x = 150, -5x = -210, x = 42, y = 36.
  6. 6Corner points: O(0,0), A(60, 0), B(42, 36), C(0, 50).
  7. 7Z at O = 0; Z at A = 4800; Z at B = 3360 + 4320 = 7680; Z at C = 6000.
  8. 8Maximum profit is Rs 7680 at (42, 36).

Final answer

Step-by-step solution

  1. 1Let x = units of A, y = units of B.
  2. 2Objective: Maximize Z = 60x + 80y.
  3. 3Constraints: 3x + y ≤ 120 (M₁), 2x + 4y ≤ 160 (M₂), x ≥ 0, y ≥ 0.
  4. 4Simplify M₂: x + 2y ≤ 80.
  5. 5Plot 3x + y = 120 and x + 2y = 80.
  6. 6Intersection: from first y = 120 - 3x, x + 2(120 - 3x) = 80, x + 240 - 6x = 80, -5x = -160, x = 32, y = 24.
  7. 7Corner points: O(0,0), A(40, 0) [from 3x+y=120], B(32, 24), C(0, 40) [from x+2y=80].
  8. 8Z at O = 0; Z at A = 2400; Z at B = 1920 + 1920 = 3840; Z at C = 3200.
  9. 9Maximum profit is Rs 3840 at (32, 24).

Final answer

Step-by-step solution

  1. 1Let x = quintals transported from A to I. Then A to II = 200 - x, B to I = 150 - x, B to II = 300 - (150 - x) = 150 + x.
  2. 2Non-negativity: x ≥ 0, 200 - x ≥ 0 (≤ 200), 150 - x ≥ 0 (≤ 150), 150 + x ≥ 0 (always). So 0 ≤ x ≤ 150.
  3. 3Objective: Z = 5x + 3(200 - x) + 4(150 - x) + 2(150 + x) = 5x + 600 - 3x + 600 - 4x + 300 + 2x = 1500.
  4. 4Z = 1500 for all feasible x! The cost is constant regardless of how we distribute.
  5. 5Any schedule with 0 ≤ x ≤ 150 gives minimum (and maximum) cost Rs 1500.
  6. 6For example: x = 0 gives A→I: 0, A→II: 200, B→I: 150, B→II: 150. Cost = 0 + 600 + 600 + 300 = 1500.

Final answer

Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Objective function

Corner point method

Isotonic

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

  • The optimum of a linear objective over a convex feasible region always occurs at a corner point, so only the vertices need testing.
  • The feasible region is the intersection of all the half-planes; shade it carefully because a wrong inequality direction removes the true answer.

FAQ

Frequently asked questions

How many questions are in NCERT Class 12 Maths Chapter 12 (Linear Programming)?

There are 2 exercise questions in this chapter, numbered Ex 12.1 to Ex 12.2. Every one is solved step by step on this page in the official NCERT numbering.

Which formulas come up in Linear Programming Class 12 Maths?

The formulas this chapter's questions actually turn on are: Objective function, Corner point method, Isotonic. They are listed with their expressions in the key formulas section below, and the solved questions show where each one is used.

Is Linear Programming important for JEE Main?

Moderate — a small chapter with a reliable method, and one of the easiest units to score full marks on in boards.

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