Class 12 Maths Chapter 12: Linear Programming worksheet with answer key

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.)

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) 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) 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.

Z(A) = 6, Z(B) = 18, Z(C) = 15 − 16 = −1, Z(D) = 3 − 12 = −9. The region is bounded, so the maximum is 18 at B(6, 0) and the minimum is −9 at D(1, 3).

• Z at all four corners: 6, 18, −1, −9 (1)
• maximum 18 at (6, 0); minimum −9 at (1, 3) (1)

Equal values at the two corners: 6p = 4p + 12, so p = 6. Then Z = 36 at (6, 0) and at (4, 4), while Z(0, 5) = 15 and Z(0, 0) = 0, so the maximum value is 36 (attained at every point of the segment joining (6, 0) and (4, 4)).

• equates 6p = 4p + 12, p = 6 (1)
• maximum 36, checked against the other corners (1)

Since the region is unbounded, check whether the open half plane 4x + 3y < 17 has a point in common with it. For any feasible point, 4x + 3y = (5/2)(x + y) + (1/2)(3x + y) ≥ (5/2)(5) + (1/2)(9) = 17. So no feasible point has 4x + 3y < 17; the half plane and the region have no common point, and 17 is the minimum, at (2, 3).

• states the test: the open half plane 4x + 3y < 17 must have no point in common with the region (1)
• shows 4x + 3y ≥ 17 on the region (or by the graph) and concludes (1)

(a) The boundary lines are x + y = 3 (through (3, 0), (0, 3)), 5x + 2y = 30 (through (6, 0), (2, 10)) and y = 10. The region lies on the side of x + y = 3 away from the origin and on the side of the other two containing it: x + y ≥ 3, 5x + 2y ≤ 30, y ≤ 10, x ≥ 0, y ≥ 0. (b) The region is bounded, with corners (3, 0), (6, 0), (2, 10), (0, 10), (0, 3); Z there: 12, 24, 18, 10, 3. Maximum 24 at (6, 0); minimum 3 at (0, 3).

• the three constraints with the correct inequality signs (1)
• Z at the five corners (1)
• maximum 24 at (6, 0), minimum 3 at (0, 3) (1)

(a) (a, 2) must satisfy every constraint: a + 2 ≤ 7 gives a ≤ 5; 2a − 2 ≥ 0 gives a ≥ 1; a ≤ 4; a ≥ 0. Together 1 ≤ a ≤ 4. (b) On these points Z = 3a + 2, which increases with a, so it is largest at a = 4: the point (4, 2), Z = 14.

• each constraint turned into a condition on a (1)
• 1 ≤ a ≤ 4 (1)
• (4, 2), Z = 14 (1)

Maximise Z = 300x + 500y subject to x + y ≤ 3, y ≤ x, x ≥ 0, y ≥ 0. The line y = x passes through the origin; testing (1, 0) shows y ≤ x is the side below it. The bounded region has corners (0, 0), (3, 0) and (3/2, 3/2), where y = x meets x + y = 3. Z there: 0, 900, 1200. The maximum is 1200 kcal, by walking 1.5 hours and cycling 1.5 hours.

• formulation: Z = 300x + 500y; x + y ≤ 3, y ≤ x, x, y ≥ 0 (1)
• corners (0, 0), (3, 0), (3/2, 3/2) (1)
• maximum 1200 kcal at 1.5 h walking, 1.5 h cycling (1)