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CBSE Important Questions Class 12 Maths Chapter 12 – Linear Programming

Class 12 Maths Chapter 12 Linear Programming is part of the CBSE Class 12 Maths syllabus. Linear programming is a method of finding the best result — the largest profit or the smallest cost — when the quantities involved are limited by a set of linear conditions. Students learn the terms objective function, constraints, feasible region and corner points, and solve problems by the graphical method.

These CBSE important questions for Class 12 Maths Chapter 12 cover the basic terms of a linear programming problem, finding the feasible region, the corner point method, maximising profit and minimising cost, and word problems on manufacturing, diet and transportation. Every question has a clear step-by-step solution in simple English with figures, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. Define the following terms used in linear programming: objective function, constraints, feasible region and optimal solution.

Answer: • Objective function: The linear function Z = ax + by that we want to maximise or minimise.
• Constraints: The linear inequalities or equations that restrict the variables. The conditions x ≥ 0 and y ≥ 0 are called non-negative constraints.
• Feasible region: The common region determined by all the constraints together. Points in it are called feasible solutions.
• Optimal solution: The feasible solution that gives the maximum or minimum value of the objective function.

Q2. State the corner point method for solving a linear programming problem.

Answer: The corner point method works like this:
1. Draw the graph of all the constraints and find the feasible region.
2. Find the coordinates of all the corner points (vertices) of the feasible region.
3. Evaluate the objective function Z at each corner point.
4. The largest value is the maximum and the smallest is the minimum.
This works because the optimal value of a linear objective function always occurs at a corner point of the feasible region.

Q3. Maximise Z = 3x + 4y subject to x + y ≤ 4, x ≥ 0 and y ≥ 0.

Answer: Corner points: The feasible region is the triangle with vertices O(0, 0), A(4, 0) and B(0, 4).
Evaluate Z = 3x + 4y at each:
At O(0, 0): Z = 0
At A(4, 0): Z = 12
At B(0, 4): Z = 16
The largest value is 16.
So the maximum value of Z is 16 at the point (0, 4).

Q4. Minimise Z = 3x + 5y subject to x + 3y ≥ 3, x + y ≥ 2, x ≥ 0 and y ≥ 0.

Answer: Corner points: Solving the constraint lines in pairs gives A(3, 0), B(3/2, 1/2) and C(0, 2).
Evaluate Z = 3x + 5y:
At A(3, 0): Z = 9
At B(3/2, 1/2): Z = 4.5 + 2.5 = 7
At C(0, 2): Z = 10
The smallest value is 7.
Since the feasible region is unbounded, we must check whether 3x + 5y < 7 has any point in common with the feasible region. It does not.
So the minimum value of Z is 7 at (3/2, 1/2).

Q5. What is a bounded and an unbounded feasible region? What extra check is needed for an unbounded region?

Answer: A feasible region is bounded if it can be enclosed within a circle, that is it does not extend indefinitely. Otherwise it is unbounded.
For a bounded region: The objective function always has both a maximum and a minimum, and both occur at corner points.
For an unbounded region: A maximum or minimum may not exist. After finding the best corner point value M, we must check:
• For a maximum: draw ax + by > M. If it has no point in common with the feasible region, M is the maximum.
• For a minimum: draw ax + by < m. If it has no common point, m is the minimum.

Q6. A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts. It takes 3 hours on machine A and 1 hour on machine B to produce a package of bolts. He earns a profit of ₹17.50 per package of nuts and ₹7 per package of bolts. How many packages of each should be produced each day to maximise his profit, if he operates his machines for at most 12 hours a day?

Answer: Let x be the packages of nuts and y the packages of bolts.
Objective function: Maximise Z = 17.5x + 7y
Constraints:
Machine A: x + 3y ≤ 12
Machine B: 3x + y ≤ 12
x ≥ 0, y ≥ 0
Corner points: (0, 0), (4, 0), (3, 3) and (0, 4)
Evaluate Z:
At (0, 0): 0
At (4, 0): 70
At (3, 3): 52.5 + 21 = 73.5
At (0, 4): 28
Maximum profit is ₹73.50, obtained by producing 3 packages of nuts and 3 packages of bolts.

Q7. Maximise Z = 5x + 3y subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0 and y ≥ 0.

Answer: Corner points: Solving the two lines together gives the point (20/19, 45/19). The other vertices are (0, 0), (2, 0) and (0, 3).
Evaluate Z = 5x + 3y:
At (0, 0): 0
At (2, 0): 10
At (20/19, 45/19): 100/19 + 135/19 = 235/19 ≈ 12.37
At (0, 3): 9
Maximum value of Z is 235/19 at the point (20/19, 45/19).

Q8. A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F₁ and F₂ are available, costing ₹4 and ₹6 per unit. One unit of F₁ contains 3 units of vitamin A and 4 units of minerals, and one unit of F₂ contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem.

Answer: Let x units of F₁ and y units of F₂ be used.
Objective function: Minimise the cost
Z = 4x + 6y
Constraints:
Vitamin A: 3x + 6y ≥ 80
Minerals: 4x + 3y ≥ 100
Non-negative: x ≥ 0, y ≥ 0
This is the complete formulation. The feasible region here is unbounded, so after finding the smallest corner value we would need the extra check described earlier.

Q9. Why does the optimal value of the objective function always occur at a corner point?

Answer: The objective function Z = ax + by is a linear function. If we draw the lines Z = constant, they form a family of parallel straight lines.
As we slide this family of parallel lines across the feasible region, the value of Z steadily increases in one direction and decreases in the other.
The last point of the feasible region that the sliding line touches before leaving the region is always a vertex (corner point), or sometimes a whole edge if the line is parallel to that edge.
That is why checking only the corner points is enough.

Q10. Can a linear programming problem have more than one optimal solution? Explain.

Answer: Yes. If the objective function line happens to be parallel to one of the edges of the feasible region, then every point on that edge gives the same optimal value.
In that case the problem has infinitely many optimal solutions, all lying along that edge, including its two end corner points.
For example, if Z = 2x + 2y is maximised over a region with an edge along x + y = 10, then every point on that edge gives Z = 20.

Q11. Minimise and maximise Z = x + 2y subject to x + 2y ≥ 100, 2x − y ≤ 0, 2x + y ≤ 200, x ≥ 0 and y ≥ 0.

Answer: Corner points: (0, 50), (0, 200), (20, 40) and (50, 100)
Evaluate Z = x + 2y:
At (0, 50): 100
At (0, 200): 400
At (20, 40): 20 + 80 = 100
At (50, 100): 50 + 200 = 250
Minimum value is 100, and it occurs at two corner points (0, 50) and (20, 40). So every point on the line segment joining them also gives 100 — this is the case of multiple optimal solutions.
Maximum value is 400 at (0, 200).

Q12. A company manufactures two types of toys A and B. Type A requires 5 minutes of cutting and 10 minutes of assembling. Type B requires 8 minutes of cutting and 8 minutes of assembling. There are 3 hours available for cutting and 4 hours for assembling. The profit is ₹50 on each toy of type A and ₹60 on each of type B. Formulate the problem to maximise profit.

Answer: Let x toys of type A and y toys of type B be made.
Objective function: Maximise Z = 50x + 60y
Constraints:
Cutting: 3 hours = 180 minutes, so 5x + 8y ≤ 180
Assembling: 4 hours = 240 minutes, so 10x + 8y ≤ 240
Non-negative: x ≥ 0, y ≥ 0
Solving graphically, the corner points are (0, 0), (24, 0), (12, 15) and (0, 22.5). Evaluating Z at these gives a maximum of ₹1500 at (12, 15), that is 12 toys of type A and 15 of type B.

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Q1.

Solve the following linear programming problem graphically:
Maximise Z = 60x+15y
Subject to constraints
x + y ? 50
3x + y ? 90
x, y ? 0

Opt.

We have,
Maximise Z = 60x + 15y
Subject to constraints,
x + y ? 50 ? (i)
3x + y ? 90 ? (ii)
x ? 0, y ? 0 ? (iii)
Plotting the inequations (i), (ii) and (iii) on the graph,

From the graph, it is clear that the shaded portion is the optimum solution. Possible points
Ans.

We have,
Maximise Z = 60x + 15y
Subject to constraints,
x + y 50 (i)
3x + y 90 (ii)
x 0, y 0 (iii)
Plotting the inequations (i), (ii) and (iii) on the graph,

Q2.

A man has Rs. 1,500 for purchase of rice and wheat. A bag of rice and a bag of wheat cost Rs. 180 and Rs. 120 respectively. He has a storage capacity of 10 bags only. He earns a profit of Rs. 11 and Rs. 9 respectively per bag of rice and wheat. Formulate it as a linear programming problem and solve it graphically for maximum profit.

Opt.

Let the man purchases x bags of rice and y bags of wheat. So, the objective function is to minimise Z = 11x + 9y
Subject to constraints,
x + y ? 10 ? (i)
180x + 120y ? 1500
i.e., 3x + 2y ? 25 ? (ii)
x ? 0, y ? 0? (iii)
Plotting the inequations (i), (ii) and (iii) on the graph,

Ans.

Let the man purchases x bags of rice and y bags of wheat. So, the objective function is to minimise Z = 11x + 9y
Subject to constraints,
x + y 10 (i)
180x + 120y 1500
i.e., 3x + 2y 25 (ii)
x 0, y 0 (iii)
Plotting the inequations (i), (ii) and (iii) on the graph,

The feasible region of L.P.P. is the shaded region.
Maximum profit can be at the points A(25/2,0), P(5,5) and B(0,10).
The value of the objective function at these points is given:

$\begin{array}{l} Points & Value of the objective function Z = 11x + 9y \\ A(25/3, 0) & 275 / 3 + 0 = 91 .66 \\ P(5, 5) & 55 + 45 = 100 [Maximum] \\ P(0, 10) & 0 + 90 = 90 \end{array}$

Z is maximum at x = 5 and y = 5.
Hence, for the maximum profit, he will purchase 5 bags of rice and 5 bags of wheat.

Q3.

A manufacturer manufactures two types of steel trunks. He has two machines ? A and B. For completing the first type of trunk, it requires 3 hours on machine A and 1 hour on machine B. Machine A can work for 18 hours and B for 8 hours only per day. There is a profit of Rs.30 on the first type of trunk and Rs. 48 on the second type of trunk. How many trunks of each type should be manufactured every day to earn maximum profit? Solve graphically.

Opt.

Let the number of A type steel trunk be x per day and number of B type steel trunk be y per day. According to the given conditions, L.P.P. is
x ? 0, y ? 0 ? (1)
3x + 3y ? 48 ? x + y ? 6 ? (ii)
x + 2y ? 8 ? (iii)
and to maximise Z = 30x + 48y
Plotting these inequations on the graph, we get

We get a shaded portion as the optimum solution. Maximum pr
Ans.

Let the number of A type steel trunk be x per day and number of B type steel trunk be y per day. According to the given conditions, L.P.P. is
x 0, y 0 (1)
3x + 3y 48 x + y 6 (ii)
x + 2y 8 (iii)
and to maximise Z = 30x + 48y
Plotting these inequations on the graph, we get

We get a shaded portion as the optimum solution. Maximum profit can be at the points A(6, 0), B(4, 2) and C(0, 4).

$\begin{array}{l} Points & Z = 30x + 48y & Value \\ A(6, 0) & 180 + 0 & 180 \\ B(4, 2) & 120 + 96 & 216 [Maximum] \\ C(0, 4) & 0 + 192 & 192 \end{array}$

We notice maximum profit is at B (4, 2), i.e., x = 4, y = 2.
Hence, 4 trunks of type I and 2 trunks of type II must be manufactured each day to get a maximum profit of Rs. 216.

CBSE Important Questions Class 12 Maths Chapter 12 – Linear Programming

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