Previous year questions: Linear Programming

Previous year questions: linear programming

Seven questions here are printed board questions from 2023 to 2026 and three are from official sample papers. The one-mark questions test the vocabulary and the corner value check; the longer ones ask for a full graphical solution. Read the question, list the corner points, and only then evaluate the objective.

Q1 (CBSE Board 2023, 1 mark)

The objective function Z = ax + by of an LPP has maximum value 42 at (4, 6) and minimum value 19 at (3, 2). Which is true? (A) a = 9, b = 1, (B) a = 5, b = 2, (C) a = 3, b = 5, (D) a = 5, b = 3.

Answer: Option (C). At (4, 6) the value 4a + 6b equals 42, and at (3, 2) the value 3a + 2b equals 19. Solving the two equations together gives a = 3 and b = 5, which satisfies both. The other options fail one of the two equations, so the pair can be checked in one line by substitution.

Q2 (CBSE Board 2024, 5 marks)

A linear programming problem has Z = 2x + 3y to be made as large as possible under x + y at most 6, x at least 2, y at most 3 and non-negativity. Solve it graphically.

Answer: The boundary lines are x + y = 6, x = 2 and y = 3. The feasible region is bounded by these three lines and the axes, with corner points (2, 0), (6, 0), (3, 3) and (2, 3). The values of Z are 4, 12, 15 and 13, so the maximum is 15 at (3, 3). The point (2, 3) is a corner because it is where x = 2 meets y = 3, and (3, 3) is where y = 3 meets x + y = 6. A common mistake here is to miss (6, 0), where x + y = 6 meets the x-axis; without it the table would still give the right maximum, but the list of corners would be incomplete and the answer would lose a mark.

Q3 (CBSE Board 2024, 5 marks)

A linear programming problem has Z = 6x + 7y to be made as small as possible under x + 2y at least 240, 3x + 4y at most 620, 2x + y at least 180 and non-negativity. Solve it graphically.

Answer: The boundary lines meet at three points that form the feasible region: x + 2y = 240 with 2x + y = 180 gives (40, 100); x + 2y = 240 with 3x + 4y = 620 gives (140, 50); and 3x + 4y = 620 with 2x + y = 180 gives (20, 140). The values of Z = 6x + 7y are 940, 1,190 and 1,100. The minimum is 940 at (40, 100). Note that the point (40, 100) is where the two greater-than constraints meet, so it lies on the lower boundary of the region; the other two corners come from pairing each of those lines with the less-than constraint. This is the pattern to look for whenever the feasible region is a triangle formed by three boundary lines.

Q4 (CBSE Board 2025, 1 mark)

In an LPP the corner points of the feasible region are (1, 1), (3, 0) and (0, 3). If Z = ax + by with a, b greater than zero is to be minimised, and the minimum occurs at (3, 0) and (1, 1), the condition on a and b is: (A) a = 2b, (B) a = b/2, (C) a = 3b, (D) a = b.

Answer: Option (B). A minimum at both corners means the two values are equal: 3a equals a + b, so b = 2a and therefore a = b/2. The other options give unequal values at the two corners and would make only one of them optimal.

Q5 (CBSE Board 2025, 1 mark)

The maximum value of Z = 3x + 4y subject to x + y at most 1, x, y at least zero is: (A) 3, (B) 4, (C) 7, (D) 0.

Answer: Option (B). The feasible region is the triangle with corners (0, 0), (1, 0) and (0, 1). The values of Z are 0, 3 and 4, so the largest is 4 at (0, 1).

Q6 (CBSE Board 2026, 1 mark)

The corner points of the feasible region are (0, 0), (0, 40), (20, 40), (60, 20) and (60, 0). If the objective function is Z = 4x + 3y, the maximum value is: (A) 200, (B) 300, (C) 240, (D) 120.

Answer: Option (B). Evaluating Z at the corners gives 0, 120, 200, 300 and 240. The largest value is 300 at (60, 20).

Q7 (CBSE Board 2026, 1 mark)

Assertion (A): The LPP that minimises Z = x + 2y subject to 2x + y at least 3, x + 2y at least 6, x, y at least zero has its minimum at infinitely many points, and the corner points of the feasible region include (0, 3) and (6, 0). Reason (R): When two corners of the feasible region carry equal minimum values, all points of the edge between them carry that value as well. Choose: (A) both true, R explains A; (B) both true, R does not explain A; (C) A true, R false; (D) A false, R true.

Answer: Option (A). At (0, 3) and at (6, 0) the value of Z is 6 in both cases, and the segment joining them lies on the line x + 2y = 6, along which Z keeps the value 6. So the minimum occurs at infinitely many points, and the reason states exactly the property that produces this. Both statements are true and the reason explains the assertion.

Q8 (CBSE SQP 2022-23, 1 mark)

The corner points of the shaded unbounded feasible region of an LPP are (0, 4), (0.6, 1.6) and (3, 0). The minimum value of Z = 4x + 6y occurs: (a) at (0.6, 1.6) only, (b) at (3, 0) only, (c) at (0.6, 1.6) and (3, 0) only, (d) at every point of the segment joining (0.6, 1.6) and (3, 0).

Answer: Option (d). At (0, 4) the value is 24, while at (0.6, 1.6) and at (3, 0) the value is 12 in both cases. Because the objective is linear, every point on the segment joining those two corners also gives 12, so the minimum is attained at every point of that segment.

Q9 (CBSE SQP 2023-24, 1 mark)

The corner points of the bounded feasible region are (0, 3), (1, 1) and (3, 0). If Z = px + qy with p, q greater than zero, the condition on p and q so that the minimum occurs at (3, 0) and (1, 1) is what?

Answer: A minimum at both corners means the two values are equal: 3p equals p + q, so q = 2p. This is the same multiple-optimum idea as in Q4 and Q7: equal corner values put the whole edge between the corners at the optimum.

Q10 (CBSE SQP 2025-26, 3 marks)

A linear programming problem has Z = 2x + y to be maximised under x + y at most 1,200, x + y at least 600, y at most x/2 and non-negativity. Solve it graphically.

Answer: The boundaries are the parallel lines x + y = 600 and x + y = 1,200, the line y = x/2 and the axes. The corner points are (600, 0), (1,200, 0), (800, 400) and (400, 200). The values of Z are 1,200, 2,400, 2,000 and 1,000. The maximum is 2,400 at (1,200, 0). The region is bounded by two parallel lines and the line y = x/2, and the corner (1,200, 0) is where y = 0 meets x + y = 1,200. A quick check with y = x/2 at x = 1,200 gives y = 600, which violates x + y at most 1,200, so the top corner is (800, 400) and not the point on the slanted line.

Seven of these are board questions from four board exam years. The pattern is the same in every graphical question: draw, list the corners, tabulate the objective, then name the optimum point and value. The one-mark questions are short checks of the same idea. Keep the graph neat enough that the shaded region is unambiguous, write each corner coordinate as a fraction in lowest terms, and finish every answer with a sentence such as the maximum value is 15 at the point (3, 3). Where a question has an OR option, both alternatives test the same corner point routine, so the practice carries over. In the assertion-reason items, decide first whether the two statements are true, and only then ask whether the reason supplies the explanation; that second step is where most marks are lost.