Feasible regions | Year 12

01 / LEARN

The idea, step by step

Feasible means every restriction holds

The feasible region is the common overlap of all allowed half-planes, not the union of separate shaded areas. Include non-negativity and any minimum quantities. Some systems are inconsistent; others extend indefinitely. A plausible-looking polygon is not enough without checking the inequalities.

Find the corners from active boundaries

At a vertex, boundary lines meet. Solve pairs of boundary equations, then test the candidates in every constraint. Axis intercepts are only candidates: another resource bound may exclude them. Keep a labelled sketch and an exact coordinate table so a later objective comparison can be checked.

A common trap

Every boundary intersection is a feasible vertex.

Instead: Test each intersection against every restriction, including non-negativity.

02 / WORKED EXAMPLES

See the thinking

Example 1

Find the corners

Find the vertices for $x, y \geq 0$ , $x + y \leq 10$ and $2 x + y \leq 14$ .

The tight x-axis bound is 7 and the tight y-axis bound is 10.
Subtract the two sloping boundary equations to obtain $x = 4$ and then $y = 6$ .
Include the origin and verify each candidate in all constraints.

(0, 0), (7, 0), (4, 6), (0, 10)

Example 2

An empty intersection

Can a point satisfy $x + y \leq 4$ and $x + y \geq 7$ together?

The same sum would have to be at most 4 and at least 7.
These conditions contradict each other.
Changing the objective cannot create a feasible point.

No feasible points.

Check your understanding

Find vertices for $x, y \geq 0$ , $x \leq 5$ , $y \leq 3$ .

Show answer

$(0, 0), (5, 0), (5, 3), (0, 3)$ .

These constraints describe a rectangle.

03 / WORKSHEET

Your turn

6 questions · 12 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Find vertices for $x, y \geq 0$ and $x + y \leq 6$ .
2 marks
Find vertices for $x, y \geq 0$ , $x \leq 4$ , $x + y \leq 7$ .
2 marks
Find where $3 x + y = 15$ meets $x + y = 9$ .
2 marks
Is $(6, 2)$ feasible for $x + y \leq 9$ and $2 x + y \leq 12$ ?
2 marks
Is the region $x, y \geq 0$ , $x + y \geq 5$ bounded?
2 marks
Why do $x \geq 4$ and $x \leq 2$ produce no feasible region?
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find vertices for $x, y \geq 0$ and $x + y \leq 6$ .
2 marks
The boundary meets both axes at 6.
$(0, 0), (6, 0), (0, 6)$ .
Find vertices for $x, y \geq 0$ , $x \leq 4$ , $x + y \leq 7$ .
2 marks
At x=4 the sloping boundary gives y=3; check each corner.
$(0, 0), (4, 0), (4, 3), (0, 7)$ .
Find where $3 x + y = 15$ meets $x + y = 9$ .
2 marks
Subtract to obtain $2 x = 6$ , then substitute.
$(3, 6)$ .
Is $(6, 2)$ feasible for $x + y \leq 9$ and $2 x + y \leq 12$ ?
2 marks
The first sum is 8, but the second is 14.
No.
Is the region $x, y \geq 0$ , $x + y \geq 5$ bounded?
2 marks
The feasible points $(t, 0)$ exist for arbitrarily large t.
No.
Why do $x \geq 4$ and $x \leq 2$ produce no feasible region?
2 marks
A value cannot be at least 4 and at most 2.
No real x meets both bounds.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 General Mathematics: Modelling with Linear Relationships, a compulsory school-assessed topic. The external examination scope differs from the full course.

Official curriculum source ↗

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