Linear Programming — Class 12 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 12 Mathematics chapter "Linear Programming" — 6 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 12 Mathematics chapter "Linear Programming" — 6 important questions with detailed answers for CBSE board e…
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NCERT Solutions for Class 12 Mathematics — Linear Programming. Step-by-step answers to all 6 textbook questions from this chapter, written for CBSE board preparation and free to use.
Linear Programming — All 6 Questions Solved
Q1. Solve the linear programming problem: Maximize z = 3x + 4y subject to x + y ≤ 4, x ≥ 0, y ≥ 0.
To solve the LPP:
Step 1: Identify constraints.
(1) x + y ≤ 4
(2) x ≥ 0
(3) y ≥ 0
Objective: Maximize z = 3x + 4y
Step 2: Find corner points of feasible region.
Constraint (1): x + y = 4
Intersection with x = 0: (0, 4)
Intersection with y = 0: (4, 0)
Origin: (0, 0)
Corner points: (0, 0), (0, 4), (4, 0)
Step 3: Evaluate objective function at each corner.
At (0, 0): z = 3(0) + 4(0) = 0
At (0, 4): z = 3(0) + 4(4) = 16
At (4, 0): z = 3(4) + 4(0) = 12
Step 4: Identify optimal solution.
Maximum value of z = 16 at (0, 4)
Final Answer: Maximum z = 16, achieved at x = 0, y = 4
Q2. Minimize z = 2x + 3y subject to x + y ≥ 5, 2x + y ≥ 8, x ≥ 0, y ≥ 0.
To minimize z = 2x + 3y:
Step 1: Identify constraints and find boundary lines.
(1) x + y = 5
(2) 2x + y = 8
(3) x = 0 (y-axis)
(4) y = 0 (x-axis)
Step 2: Find intersection points.
Line 1 ∩ Line 2:
x + y = 5 ... (i)
2x + y = 8 ... (ii)
Subtracting: x = 3, y = 2
Point: (3, 2)
Line 1 ∩ y-axis (x = 0):
0 + y = 5, so y = 5
Point: (0, 5)
Line 2 ∩ x-axis (y = 0):
2x + 0 = 8, so x = 4
Point: (4, 0)
Line 1 ∩ x-axis: (5, 0)
Line 2 ∩ y-axis: (0, 8)
Step 3: Check feasibility and find corner points.
Test (3, 2): 3 + 2 = 5 ✓, 2(3) + 2 = 8 ✓
Test (0, 5): 0 + 5 = 5 ✓, 2(0) + 5 = 5 < 8 ✗
Test (4, 0): 4 + 0 = 4 < 5 ✗, 2(4) + 0 = 8 ✓
Test (0, 8): 0 + 8 = 8 ≥ 5 ✓, 2(0) + 8 = 8 ✓
Feasible corner points: (3, 2) and (0, 8)
Step 4: Evaluate objective function.
At (3, 2): z = 2(3) + 3(2) = 6 + 6 = 12
At (0, 8): z = 2(0) + 3(8) = 24
Step 5: Find minimum.
Minimum z = 12 at (3, 2)
Final Answer: Minimum z = 12, achieved at x = 3, y = 2
Q3. A factory produces chairs and tables. Each chair takes 2 hours and each table takes 3 hours. The factory has 36 hours available. Profit per chair is Rs 100 and per table is Rs 150. Maximize profit.
LPP formulation and solution:
Step 1: Define variables.
Let x = number of chairs
Let y = number of tables
Step 2: Formulate constraints.
Time constraint: 2x + 3y ≤ 36
Non-negativity: x ≥ 0, y ≥ 0
Step 3: Formulate objective function.
Maximize P = 100x + 150y
Step 4: Find corner points of feasible region.
Constraint boundary: 2x + 3y = 36
Intercepts:
When x = 0: 3y = 36, y = 12, Point: (0, 12)
When y = 0: 2x = 36, x = 18, Point: (18, 0)
Origin: (0, 0)
Corner points: (0, 0), (0, 12), (18, 0)
Step 5: Evaluate profit at corner points.
At (0, 0): P = 100(0) + 150(0) = 0
At (0, 12): P = 100(0) + 150(12) = 1800
At (18, 0): P = 100(18) + 150(0) = 1800
Step 6: Find optimal solution.
Maximum profit = Rs 1800
This occurs at both (0, 12) and (18, 0)
Alternative optimal solutions:
Option 1: Produce 0 chairs and 12 tables
Option 2: Produce 18 chairs and 0 tables
Final Answer: Maximum profit = Rs 1800. Optimal points: (0, 12) or (18, 0)
Q4. A diet must contain at least 400 units of calcium and 900 units of protein. Food A contains 100 units calcium and 300 units protein per kg, costing Rs 50/kg. Food B contains 200 units calcium and 150 units protein per kg, costing Rs 60/kg. Find minimum cost diet.
LPP formulation:
Step 1: Define variables.
x = kg of Food A
y = kg of Food B
Step 2: Formulate constraints.
Calcium: 100x + 200y ≥ 400
Protein: 300x + 150y ≥ 900
x ≥ 0, y ≥ 0
Simplify:
Constraint 1: x + 2y ≥ 4
Constraint 2: 2x + y ≥ 6
Step 3: Objective function.
Minimize Cost C = 50x + 60y
Step 4: Find corner points.
Line 1: x + 2y = 4
Line 2: 2x + y = 6
Intersection:
From Line 1: x = 4 - 2y
Substitute in Line 2: 2(4 - 2y) + y = 6
8 - 4y + y = 6
-3y = -2
y = 2/3
x = 4 - 2(2/3) = 4 - 4/3 = 8/3
Point: (8/3, 2/3)
Line 1 ∩ x-axis: x + 0 = 4, Point: (4, 0)
Line 2 ∩ y-axis: 0 + y = 6, Point: (0, 6)
Step 5: Check feasibility.
At (8/3, 2/3):
100(8/3) + 200(2/3) = 800/3 + 400/3 = 400 ✓
300(8/3) + 150(2/3) = 800 + 100 = 900 ✓
At (4, 0):
100(4) + 200(0) = 400 ✓
300(4) + 150(0) = 1200 ≥ 900 ✓
At (0, 6):
100(0) + 200(6) = 1200 ≥ 400 ✓
300(0) + 150(6) = 900 ✓
Step 6: Evaluate cost.
At (8/3, 2/3): C = 50(8/3) + 60(2/3) = 400/3 + 120/3 = 520/3 ≈ Rs 173.33
At (4, 0): C = 50(4) + 60(0) = Rs 200
At (0, 6): C = 50(0) + 60(6) = Rs 360
Step 7: Find minimum.
Minimum cost = Rs 520/3 ≈ Rs 173.33 at x = 8/3 kg Food A and y = 2/3 kg Food B
Final Answer: Minimum cost = Rs 520/3 (≈ Rs 173.33), with 8/3 kg Food A and 2/3 kg Food B
Q5. A company makes two products A and B. Each unit of A requires 3 hours of skilled labor and 2 units of raw material. Each unit of B requires 4 hours and 3 units. Available: 120 hours labor and 90 units material. Profit per A is Rs 40, per B is Rs 50. Maximize profit.
LPP formulation:
Step 1: Define variables.
x = units of product A
y = units of product B
Step 2: Formulate constraints.
Labor: 3x + 4y ≤ 120
Raw material: 2x + 3y ≤ 90
x ≥ 0, y ≥ 0
Step 3: Objective function.
Maximize P = 40x + 50y
Step 4: Find corner points.
Line 1: 3x + 4y = 120
Line 2: 2x + 3y = 90
Intercepts of Line 1:
x = 0: y = 30, Point: (0, 30)
y = 0: x = 40, Point: (40, 0)
Intercepts of Line 2:
x = 0: y = 30, Point: (0, 30)
y = 0: x = 45, Point: (45, 0)
Intersection of Lines 1 and 2:
3x + 4y = 120 ... (i)
2x + 3y = 90 ... (ii)
Multiply (i) by 3: 9x + 12y = 360
Multiply (ii) by 4: 8x + 12y = 360
Subtract: x = 0
From (ii): 0 + 3y = 90, y = 30
Point: (0, 30)
Corner points in feasible region: (0, 0), (0, 30), (40, 0)
Note: Check (40, 0): material = 2(40) = 80 ≤ 90 ✓
Check (45, 0): material = 2(45) = 90, but labor = 3(45) = 135 > 120 ✗
Step 5: Evaluate profit.
At (0, 0): P = 0
At (0, 30): P = 40(0) + 50(30) = 1500
At (40, 0): P = 40(40) + 50(0) = 1600
Step 6: Find maximum.
Maximum profit = Rs 1600 at x = 40, y = 0
Final Answer: Maximum profit = Rs 1600, produce 40 units of A and 0 units of B
Q6. State the Fundamental Theorem of Linear Programming.
Fundamental Theorem of Linear Programming:
Theorem Statement:
If a linear programming problem has an optimal solution, then it occurs at one of the corner (vertex) points of the feasible region.
Key Points:
1. Existence of Optimal Solution:
If the feasible region is non-empty and bounded, an optimal solution exists.
2. Location of Optimal Solution:
The optimal solution lies at a corner point of the feasible region (not in the interior).
3. Multiple Optimal Solutions:
If the objective function is parallel to a constraint, multiple optimal solutions may exist along an edge.
4. Corner Point Method:
To find the optimal solution:
- Find all corner points of the feasible region
- Evaluate the objective function at each corner point
- The point giving the maximum (or minimum) value is the optimal solution
5. Mathematical Reasoning:
The objective function is linear: z = cx + dy
At any interior point, we can move in the direction of increasing (or decreasing) z until we reach a boundary.
On a boundary edge, we can continue moving until a corner is reached.
Therefore, the optimum must occur at a corner.
Importance:
This theorem makes solving LPPs practical by reducing the infinite feasible region to a finite set of corner points.
Final Answer: If an LPP has an optimal solution, it occurs at a corner (vertex) point of the feasible region.
Linear Programming — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. What is the primary goal of an objective function in a Linear Programming Problem (LPP)?
- A. To define the feasible region
- B. To be optimized (maximized or minimized)
- C. To represent the constraints
- D. To provide non-negative restrictions
Answer: B. To be optimized (maximized or minimized) — The objective function in an LPP is the linear function that the decision-maker wants to either maximize (e.g., profit) or minimize (e.g., cost).
MCQ 2. The set of all points satisfying all the constraints and non-negative restrictions of an LPP is called the:
- A. Optimal solution
- B. Objective function
- C. Feasible region
- D. Corner point
Answer: C. Feasible region — The feasible region is the area on the graph that satisfies all the given linear inequalities (constraints) simultaneously, including non-negativity.
MCQ 3. A company produces two types of toys, A and B. Toy A requires 2 hours of assembly and 1 hour of finishing. Toy B requires 1 hour of assembly and 2 hours of finishing. The company has 100 hours of assembly time and 80 hours of finishing time available. If 'x' is the number of toy A and 'y' is the number of toy B, which constraint represents the assembly time?
- A. x + 2y ≤ 100
- B. 2x + y ≤ 100
- C. 2x + y ≤ 80
- D. x + 2y ≤ 80
Answer: B. 2x + y ≤ 100 — Toy A requires 2 hours of assembly (2x) and Toy B requires 1 hour (1y). The total assembly time cannot exceed 100 hours, so the constraint is 2x + y ≤ 100.
MCQ 4. In an LPP, the decision variables (e.g., x, y) must always be:
- A. Negative
- B. Non-negative
- C. Any real number
- D. Integers only
Answer: B. Non-negative — In most practical LPPs, the quantities of items produced or resources used cannot be negative, hence the non-negativity restrictions x ≥ 0, y ≥ 0 are imposed.
MCQ 5. For a maximization problem, if the feasible region is unbounded and the objective function has positive coefficients for all variables, what can be concluded about the optimal solution?
- A. A unique maximum always exists.
- B. No maximum solution exists (unbounded solution).
- C. Multiple maximum solutions exist.
- D. The maximum must be at the origin.
Answer: B. No maximum solution exists (unbounded solution). — If the feasible region is unbounded in a maximization problem and the objective function can be increased indefinitely within this region, then no finite maximum solution exists.
MCQ 6. An optimal solution in an LPP refers to:
- A. Any solution that satisfies the constraints.
- B. The solution that yields the best (maximum or minimum) value of the objective function.
- C. A solution that lies on the boundary of the feasible region.
- D. A solution where all variables are zero.
Answer: B. The solution that yields the best (maximum or minimum) value of the objective function. — The optimal solution is the feasible solution that gives the maximum or minimum value of the objective function, depending on the problem's goal.
MCQ 7. Consider an LPP with constraints: x + y ≤ 4, x + y ≥ 6, x ≥ 0, y ≥ 0. What type of solution does this LPP have?
- A. Unique optimal solution
- B. Multiple optimal solutions
- C. Unbounded solution
- D. No feasible solution
Answer: D. No feasible solution — The constraints x + y ≤ 4 and x + y ≥ 6 are contradictory. There is no point (x, y) that can satisfy both conditions simultaneously, hence there is no feasible region.
MCQ 8. According to the Corner Point Theorem, the optimal solution for a bounded feasible region in an LPP always occurs at:
- A. The origin
- B. The center of the feasible region
- C. One of the corner points of the feasible region
- D. A point on the boundary, but not necessarily a corner point
Answer: C. One of the corner points of the feasible region — The Corner Point Theorem states that if a feasible region is bounded, the optimal value (maximum or minimum) of the objective function occurs at a corner point.
MCQ 9. If the objective function line coincides with one of the edges of the feasible region at the optimal value, what does it imply?
- A. The problem has a unique optimal solution.
- B. The problem has no feasible solution.
- C. The problem has multiple optimal solutions, specifically along that edge.
- D. The problem has an unbounded solution.
Answer: C. The problem has multiple optimal solutions, specifically along that edge. — If the objective function line, when shifted to its optimal value, overlaps with an entire edge of the feasible region, then every point on that edge (including its endpoints) provides the same optimal value, indicating multiple optimal solutions.
MCQ 10. In an LPP, the restrictions or limitations on the availability of resources or other conditions are called:
- A. Objective functions
- B. Decision variables
- C. Constraints
- D. Optimal solutions
Answer: C. Constraints — Constraints are the linear inequalities or equations that impose limitations on the decision variables in an LPP.
MCQ 11. A factory produces two products, P1 and P2. To produce P1, it requires 2 units of raw material A and 3 units of raw material B. To produce P2, it requires 3 units of raw material A and 2 units of raw material B. The factory has 60 units of raw material A and 80 units of raw material B available. If x and y are the number of units of P1 and P2 respectively, what is the constraint for raw material B?
- A. 2x + 3y ≤ 60
- B. 3x + 2y ≤ 60
- C. 3x + 2y ≤ 80
- D. 2x + 3y ≤ 80
Answer: C. 3x + 2y ≤ 80 — For raw material B, P1 requires 3 units (3x) and P2 requires 2 units (2y). The total usage cannot exceed 80 units, so the constraint is 3x + 2y ≤ 80.
MCQ 12. For a minimization problem, if the feasible region is unbounded and the objective function Z = ax + by (where a > 0, b > 0) is minimized, can an optimal solution exist?
- A. No, it always leads to an unbounded solution.
- B. Yes, if the minimum value occurs at a corner point.
- C. Only if the origin is a feasible point.
- D. Only if the objective function coefficients are negative.
Answer: B. Yes, if the minimum value occurs at a corner point. — Even with an unbounded feasible region, a minimization problem can have a finite optimal solution if the objective function's value does not decrease indefinitely within the feasible region and the minimum is attained at a corner point.
MCQ 13. The inequality 2x + 3y ≤ 6 represents a region that includes:
- A. Points above the line 2x + 3y = 6.
- B. Points below or on the line 2x + 3y = 6.
- C. Only points on the line 2x + 3y = 6.
- D. Points on the x-axis only.
Answer: B. Points below or on the line 2x + 3y = 6. — For an inequality of the form Ax + By ≤ C with positive coefficients, the feasible region typically lies on or below the line Ax + By = C (if testing with (0,0) yields 0 ≤ C, which is true).
MCQ 14. Which of the following is NOT a possible outcome for an LPP?
- A. Unique optimal solution
- B. Multiple optimal solutions
- C. Solution with infinitely many variables
- D. No feasible solution
Answer: C. Solution with infinitely many variables — LPPs deal with a finite number of decision variables. Infinitely many variables are not part of the standard LPP framework. The other options are standard outcomes.
MCQ 15. In an LPP, a constraint whose removal would not change the feasible region is called a:
- A. Binding constraint
- B. Non-binding constraint
- C. Redundant constraint
- D. Trivial constraint
Answer: C. Redundant constraint — A redundant constraint is one that does not restrict the feasible region any further than what is already restricted by other constraints. Its removal would not alter the feasible region.
MCQ 16. To maximize Z = 5x + 3y subject to x + y ≤ 2, x ≥ 0, y ≥ 0, what are the corner points of the feasible region?
- A. (0,0), (2,0), (0,2)
- B. (0,0), (5,0), (0,3)
- C. (0,0), (2,0), (1,1)
- D. (0,0), (0,2), (2,2)
Answer: A. (0,0), (2,0), (0,2) — The feasible region is bounded by the axes (x=0, y=0) and the line x+y=2. The corner points are the origin (0,0), the x-intercept of x+y=2 which is (2,0), and the y-intercept which is (0,2).
MCQ 17. If the objective function is Z = 3x + 2y, and the corner points of the feasible region are (0,0), (5,0), (3,4), (0,5), what is the minimum value of Z?
- A. 0
- B. 15
- C. 17
- D. 10
Answer: A. 0 — Evaluate Z at each corner point: Z(0,0) = 0; Z(5,0) = 3(5)+2(0) = 15; Z(3,4) = 3(3)+2(4) = 9+8 = 17; Z(0,5) = 3(0)+2(5) = 10. The minimum value is 0.
MCQ 18. Consider the objective function Z = -x - y subject to x ≥ 0, y ≥ 0. If the feasible region is the entire first quadrant, which of the following is true?
- A. Z has a maximum value but no minimum value.
- B. Z has a minimum value but no maximum value.
- C. Z has both maximum and minimum values.
- D. Z has neither maximum nor minimum values.
Answer: B. Z has a minimum value but no maximum value. — The feasible region is unbounded. For Z = -x - y, as x and y increase, Z becomes more negative (approaches -∞), so there's no maximum. The minimum value occurs at the corner point (0,0) where Z=0.
MCQ 19. In an LPP, the quantities whose values need to be determined are called:
- A. Parameters
- B. Decision variables
- C. Constants
- D. Coefficients
Answer: B. Decision variables — Decision variables are the unknown quantities in an LPP whose optimal values are sought to maximize or minimize the objective function.
MCQ 20. A baker makes two types of cakes, C1 and C2. C1 requires 200g of flour and 25g of sugar. C2 requires 100g of flour and 50g of sugar. The baker has 5kg of flour and 1kg of sugar. Let x be the number of C1 cakes and y be the number of C2 cakes. Which pair of inequalities correctly represents the resource constraints?
- A. 200x + 100y ≤ 5000; 25x + 50y ≤ 1000
- B. 200x + 100y ≥ 5000; 25x + 50y ≥ 1000
- C. 200x + 25y ≤ 5000; 100x + 50y ≤ 1000
- D. 200x + 100y ≤ 5; 25x + 50y ≤ 1
Answer: A. 200x + 100y ≤ 5000; 25x + 50y ≤ 1000 — First, convert kg to grams: 5kg = 5000g, 1kg = 1000g. The flour constraint is 200x + 100y ≤ 5000. The sugar constraint is 25x + 50y ≤ 1000.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.