A feasible region is where every constraint is true at the same time. Deal with each inequality on its own, then keep only the overlap.
This builds on drawing a boundary for a linear inequality. After it, checking a proposed point uses the region to decide whether a given point is acceptable.
How do you find the region?
- Draw each boundary line, solid or dashed.
- Test a point for each inequality and mark which side is wanted.
- Keep only the part where all the wanted sides overlap.
- Find the corners by reading the graph or solving pairs of boundary equations.
- Check one point inside the region against every inequality.
Some questions ask you to shade the required region. Others ask you to shade the unwanted regions, so that the feasible region is left blank. Read the instruction before you begin.
Worked example
Find the feasible region for x ≥ 1, y ≥ 2 and x + y ≤ 6.
Step 1, boundaries: x = 1 is a vertical solid line. y = 2 is a horizontal solid line. x + y = 6 is a solid line through (6, 0) and (0, 6).
Step 2, sides: x ≥ 1 wants the right of x = 1. y ≥ 2 wants above y = 2. x + y ≤ 6 wants the origin side, because 0 ≤ 6 is true.
Step 3, overlap: the points to the right of x = 1, above y = 2 and below x + y = 6 form a triangle.
Step 4, corners:
- x = 1 and y = 2 meet at (1, 2).
- x = 1 and x + y = 6: y = 5, so (1, 5).
- y = 2 and x + y = 6: x = 4, so (4, 2).
Step 5, integer points: for x = 1, y can be 2, 3, 4, 5, which is 4 points. For x = 2, y is 2, 3, 4, which is 3 points. For x = 3, y is 2, 3, which is 2 points. For x = 4, y is 2, which is 1 point. The total is 4 + 3 + 2 + 1 = 10 integer points, counting those on the boundary because the lines are solid.
Check: (2, 3) gives 2 ≥ 1, 3 ≥ 2 and 5 ≤ 6. All true.
The mistake to watch for
Mistaken working: shading the wanted side of every inequality and then keeping the whole shaded area, including parts covered by only one or two of them.
The student treated the constraints as “or” instead of “and”.
The feasible region must satisfy all the constraints, so it is the overlap only. A test point such as (5, 5) may satisfy x ≥ 1 and y ≥ 2, but x + y = 10 is not ≤ 6, so (5, 5) is not feasible. The correction is to test every point you claim is inside against every inequality, not just the first one.
Check yourself
1. For x ≥ 1, y ≥ 2 and x + y ≤ 6 above, is (4, 3) in the feasible region?
Show answer
4 ≥ 1 is true and 3 ≥ 2 is true, but 4 + 3 = 7 and 7 ≤ 6 is false.
No, because it breaks one constraint.
2. Find the corners of the region x ≥ 0, y ≥ 0 and x + 2y ≤ 8.
Show answer
x = 0 and y = 0 meet at (0, 0). With y = 0: x = 8, so (8, 0). With x = 0: 2y = 8, so y = 4, giving (0, 4).
(0, 0), (8, 0) and (0, 4)
3. How many integer points satisfy x ≥ 1, y ≥ 1 and x + y ≤ 4?
Show answer
x = 1: y = 1, 2, 3 gives 3 points. x = 2: y = 1, 2 gives 2 points. x = 3: y = 1 gives 1 point.
Total: 3 + 2 + 1 = 6 points.
Where this leads next
Next, check whether a proposed point satisfies every constraint, which uses the same tests without drawing. The non-calculator working trainer helps with corner arithmetic. Earlier work on simultaneous relationships explains how to find where two lines meet.
When several constraints overlap, a teacher in online one-to-one Mathematics tuition can watch how you organise the graph and suggest an order that keeps it readable.