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The Vault
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The Vault
CBSE Class 12 Maths · 11 questions · 26 marks
Linear programming turns a real business or resource-allocation problem into a search for the best value of a linear objective function, subject to constraints that also happen to be linear. This chapter's central idea, the corner point theorem, means you never need to check the entire feasible region — only its vertices, found by graphing the boundary lines — to find the maximum or minimum, whether the goal is maximising profit or minimising cost.
For the LPP: maximize Z = 3x + 4y subject to x+y≤4, x≥0, y≥0, the maximum value of Z is:
Answer
The maximum value is 16. The corner points of the feasible region are (0,0), (4,0) and (0,4). Z(0,0)=0, Z(4,0)=12, Z(0,4)=16. The largest of these is 16, at (0,4).
The feasible region of a linear programming problem is always a:
Answer
The feasible region is always a convex polygon (when bounded). Each linear constraint defines a half-plane, and the intersection of half-planes is always a convex set, since a straight line joining any two points of the region stays within it.
In a linear programming problem, the optimal value of the objective function is attained at:
Answer
The optimal value is attained at a corner point of the feasible region. This is the Corner Point Theorem: for a bounded feasible region, the maximum or minimum of a linear objective function always occurs at one of the vertices, not at an interior point.
If the feasible region of an LPP is unbounded, then the maximum value of Z=ax+by (with a,b>0):
Answer
The maximum value may or may not exist. On an unbounded feasible region, Z can sometimes increase without limit as x or y grows, in which case no maximum exists; in other cases the region's shape does cap Z, in which case a genuine maximum can be found. Each unbounded case must be checked individually.
Assertion (A): For the LPP maximize Z = x + y subject to x+y≤4, x≥0, y≥0, the maximum value Z=4 is attained at every point of the line segment joining (4,0) and (0,4), not just a single corner. Reason (R): This happens because the objective function's line x+y=4 coincides with one of the boundary edges of the feasible region, so the entire edge is optimal.
Answer
Both A and R are true and R is the correct explanation of A. The constraint boundary is exactly x+y=4, which is the same line as the objective function Z=x+y=4. So every point on the segment from (4,0) to (0,4) satisfies Z=4, and since this is the largest value Z takes over the whole feasible region, all these points are simultaneously optimal. Reason R correctly explains why this coincidence of lines produces multiple optimal solutions, matching Assertion A.
Find the corner points of the feasible region determined by x≥0, y≥0, x+y≤6, and evaluate Z=5x+3y at each. State the maximum value.
Answer
The corner points are (0,0), (6,0) and (0,6). Z(0,0) = 0, Z(6,0) = 30, Z(0,6) = 18. The maximum value of Z is 30, attained at (6,0).
Find the corner points of the feasible region determined by x≥0, y≥0, x+2y≤8, and evaluate Z=3x+2y at each. State the maximum value.
Answer
The corner points are (0,0), (8,0) and (0,4). Z(0,0) = 0, Z(8,0) = 24, Z(0,4) = 8. The maximum value of Z is 24, attained at (8,0).
A feasible region has corner points (0,0), (5,0), (3,4) and (0,6). If Z = 4x + 6y, find the maximum and minimum values of Z over this region.
Answer
Evaluate Z at each corner point. Z(0,0) = 0. Z(5,0) = 20. Z(3,4) = 4(3)+6(4) = 12+24 = 36. Z(0,6) = 6(6) = 36. The minimum value is 0, at (0,0). The maximum value is 36, attained at both (3,4) and (0,6).
A furniture dealer has ₹50,000 to invest and storage space for at most 60 pieces of furniture. A table costs ₹2500 and a chair costs ₹500. He estimates a profit of ₹250 on a table and ₹75 on a chair. Formulate this as an LPP to maximize profit and solve it graphically.
Answer
Let x = number of tables and y = number of chairs. Maximize Z = 250x + 75y, subject to: 2500x + 500y ≤ 50000, i.e. 5x + y ≤ 100; x + y ≤ 60; and x≥0, y≥0. Find the corner points. Solving 5x+y=100 and x+y=60 together: subtracting gives 4x=40, so x=10, y=50. Other corners: (0,0), (20,0) from 5x+y=100 with y=0, and (0,60) from x+y=60 with x=0. Evaluate Z: Z(0,0)=0, Z(20,0)=5000, Z(10,50)=250(10)+75(50)=2500+3750=6250, Z(0,60)=75(60)=4500. The maximum profit is ₹6250, achieved by buying 10 tables and 50 chairs.
Minimize Z = 200x + 500y subject to the constraints x+2y≥10, 3x+4y≤24, x≥0, y≥0. Find the minimum value graphically.
Answer
Find the corner points of the feasible region by solving pairs of boundary equations. Solving x+2y=10 and 3x+4y=24 together: doubling the first gives 2x+4y=20; subtracting from the second gives x=4, so 2y=6, y=3. Corner: (4,3). x+2y=10 meets the y-axis (x=0) at y=5; checking 3(0)+4(5)=20≤24, this point (0,5) is feasible. 3x+4y=24 meets the y-axis (x=0) at y=6; checking 0+2(6)=12≥10, this point (0,6) is feasible. So the feasible region is the triangle with corners (0,5), (4,3), (0,6). Evaluate Z: Z(0,5)=2500, Z(4,3)=200(4)+500(3)=800+1500=2300, Z(0,6)=3000. The minimum value of Z is 2300, attained at (4,3).
Read the following and answer the questions that follow: A dietician wants to plan a diet containing at least 8 units of vitamin A and at least 10 units of vitamin B, using two food items. Each unit of Food 1 supplies 2 units of vitamin A and 1 unit of vitamin B, and costs ₹5. Each unit of Food 2 supplies 1 unit of vitamin A and 2 units of vitamin B, and costs ₹7. (a) Formulate the LPP to minimize cost, writing the objective function and constraints. (b) Find the corner points of the feasible region. (c) Evaluate the objective function at each corner point. (d) State the minimum cost and the corresponding quantities of Food 1 and Food 2.
Answer
(a) Let x = units of Food 1, y = units of Food 2. Minimize Z = 5x + 7y, subject to 2x+y≥8, x+2y≥10, x≥0, y≥0. (b) Solving 2x+y=8 and x+2y=10 together: doubling the first gives 4x+2y=16; subtracting the second gives 3x=6, so x=2, y=4 — corner (2,4). The line 2x+y=8 meets the y-axis at (0,8), which satisfies x+2y=16≥10. The line x+2y=10 meets the x-axis at (10,0), which satisfies 2x+y=20≥8. So the corners are (0,8), (2,4), (10,0). (c) Z(0,8) = 7(8) = 56. Z(2,4) = 5(2)+7(4) = 10+28 = 38. Z(10,0) = 5(10) = 50. (d) The minimum cost is ₹38, using 2 units of Food 1 and 4 units of Food 2.
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