Class 12 Maths Chapter 12: Linear Programming MCQs with answers

Answer: (b) the objective function

A linear function Z = ax + by that is to be maximised or minimised is the linear objective function. The inequalities are the constraints, and x and y are the decision variables.

Answer: (d) y ≥ 0

The conditions x ≥ 0, y ≥ 0 are the non-negative restrictions. x + y ≤ 12 and x − 2y ≥ 0 are other constraints (x − 2y ≥ 0 allows negative values of y), and Z = 5x + 3y is an objective function.

Answer: (c) Z = x² + 4y

In a linear programming problem every relation is linear, so the objective function must be of the form ax + by. Z = x² + 4y contains x² and is not linear; the other three are linear, whatever the signs or decimals of the coefficients.

Answer: (a) (5, 3)

At (5, 3): 2(5) + 3 = 13 > 12, so the first constraint fails and the point lies outside the region. (3, 4) gives 10 ≤ 12 and 15 ≤ 15; (6, 0) gives 12 ≤ 12 and 6 ≤ 15; (0, 5) gives 5 ≤ 12 and 15 ≤ 15. Points on the boundary are feasible.

Answer: (c) 25

Z at the corners: (0, 0) → 0, (6, 0) → 12, (4, 3) → 8 + 15 = 23, (0, 5) → 25. The region is bounded, so the largest corner value, 25 at (0, 5), is the maximum. 37 = 2(6) + 5(5) combines coordinates of two different corners.

Answer: (a) 11

Z at the corners: (2, 1) → 11, (5, 1) → 23, (5, 4) → 32, (1, 3) → 13. For a bounded region the smallest corner value is the minimum: 11 at (2, 1). 13 is the value at (1, 3), the corner with the smallest x, and 32 is the maximum.

Answer: (d) Z has both a maximum and a minimum value on R, each occurring at a corner point of R

Theorem 2: if R is bounded, Z has both a maximum and a minimum value on R and each of these occurs at a corner point (vertex) of R. The possibility of no maximum or no minimum arises only for an unbounded region.

Answer: (a) it can be enclosed within a circle

By definition a feasible region is bounded if it can be enclosed within a circle; otherwise it is unbounded. A region in the first quadrant, or one containing the origin, can still extend indefinitely, and a region with only two corner points can be unbounded.

Answer: (b) k > 3

On the triangle 2x + 3y ≤ 6, x, y ≥ 0, with corners (0, 0), (3, 0) and (0, 2), the largest value of x + y is 3, at (3, 0). So some point satisfies x + y ≥ k exactly when k ≤ 3, and there is no feasible region when k > 3. k > 2 uses the corner (0, 2) only; k > 6 uses the right side of 2x + 3y ≤ 6.

Answer: (c) Z also takes the value 48 at (4, 6)

When two corner points produce the same maximum value, every point on the segment joining them gives the same maximum. (4, 6) is the midpoint of PQ, so Z = 48 there. (From 2a + 8b = 6a + 4b = 48, a = b = 4.8, and 4.8(4 + 6) = 48 confirms it.)