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CBSE Sample Paper 2023-24 · Section A · 1 mark

Q11: Which inequality is not a constraint of the shaded feasible region

The question

The feasible region corresponding to the linear constraints of a Linear Programming Problem is given below.

(0,5) (0,2) (1,0) (2,0) (10,0) O x+2y=10 x−y=1 x+y=2 x x′ y y′
The given graph, redrawn.

Which of the following is not a constraint to the given Linear Programming Problem?

(a) x+y≥2x + y \ge 2

(b) x+2y≤10x + 2y \le 10

(c) x−y≥1x - y \ge 1

(d) x−y≤1x - y \le 1

Exam answer

The shaded region lies on the same side of x−y=1x - y = 1 as the origin O(0,0)O(0, 0)⁠.

Why?

A line ax+by=cax + by = c splits the plane into two half-planes. On one side ax+by<cax + by < c⁠, and on the other ax+by>cax + by > c⁠.

To find which inequality a region satisfies, test one point that is not on the line. The origin is the easiest.

At OO⁠, x−y=0<1x - y = 0 < 1⁠, so the region satisfies x−y≤1x - y \le 1⁠, not x−y≥1x - y \ge 1⁠.

Why?

Check with a corner of the region, (0,2)(0, 2)⁠: x−y=0−2=−2≤1x - y = 0 - 2 = -2 \le 1⁠, so x−y≥1x - y \ge 1 fails there.

Watch out: points on the line itself satisfy both ≤\le and ≥\ge⁠, so always test a point off the line.

For x+y=2x + y = 2⁠: at OO⁠, x+y=0<2x + y = 0 < 2⁠. The region is on the other side, so x+y≥2x + y \ge 2⁠.

For x+2y=10x + 2y = 10⁠: at OO⁠, x+2y=0<10x + 2y = 0 < 10⁠. The region is on the same side, so x+2y≤10x + 2y \le 10⁠.

So (a), (b) and (d) are constraints, and (c) is not.

Why?

Check with the corner (0,5)(0, 5)⁠: 0+5=5≥20 + 5 = 5 \ge 2 and 0+2(5)=10≤100 + 2(5) = 10 \le 10⁠.

The question asks for the inequality that is not a constraint, so the answer is (c).

Answer: (c) x−y≥1x - y \ge 1

Where this comes from

Chapter 12, Linear Programming: Reading constraints from a feasible region.

Question type: Read the rules from a graph

How to spot it: click to see

Type 7. Read the rules from a graph

Spot it: the graph of the region is given, and the question asks for its constraints, or for the rule that is missing.

Here: The region lies on the origin's side of x−y=1x - y = 1⁠, where x−y=0<1x - y = 0 \lt 1⁠, so its rule is x−y≤1x - y \le 1⁠. Checked the same way, (a), (b) and (d) are rules, so (c) is not.

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