Feasible regions | Year 12

01 / LEARN

Understand the idea.

Feasible means every restriction holds

The feasible region contains only points satisfying every restriction together. Shade the common overlap of the allowed half-planes. Non-negativity and minimum quantities matter just as much as the resource bounds.

The overlap might be a bounded polygon, an unbounded region or empty. Test a point against every inequality before calling it feasible; satisfying one resource limit cannot compensate for breaking another.

Find the corners from active boundaries

Find possible vertices by intersecting boundary lines, including the coordinate axes. Then reject candidates violating any restriction. An axis intercept may be cut off by a tighter constraint before it becomes part of the feasible boundary.

Keep exact coordinates while solving and label them on the sketch. These verified vertices will later be used to compare objective values; inaccurate graph readings can change which decision appears best.

A first-quadrant grid runs from zero to twelve on both axes. The region with vertices (0,0), (7,0), (4,6), (0,10) is shaded under the supplied lines x+y=10 and 2x+y=14. No objective values are marked.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Find the corners

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

Worked solution

1
Find the feasible axis endpoints.
$\begin{aligned} y = 0 & : x \leq 10, x \leq 7 \\ x = 0 & : y \leq 10, y \leq 14 \end{aligned}$
Both sloping constraints must hold, so the tighter bounds give $(7, 0)$ and $(0, 10)$ .
2
Intersect the sloping boundaries.
$\begin{aligned} x + y & = 10 \\ 2 x + y & = 14 \\ x & = 4, y = 6 \end{aligned}$
Subtracting the first equality from the second finds the shared corner.
3
Check the non-axis corner.
$4 + 6 = 10, 2 (4) + 6 = 14$
Both resources are exactly used and both coordinates are non-negative.
4
Collect the boundary corners.
$(0, 0), (7, 0), (4, 6), (0, 10)$
The origin and axis endpoints also satisfy both upper bounds; these are the four corners of their common overlap.

Answer

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

EXAMPLE 02

An empty intersection

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

Worked solution

1
Combine the demands on the same sum.
$7 \leq x + y \leq 4$
Any feasible point would have to satisfy both inequalities simultaneously.
2
Identify the contradiction.
$7 > 4$
No real number can be at least 7 and at most 4.
3
State the feasible-set conclusion.
There are no feasible points. Choosing a different objective cannot repair inconsistent constraints.

Answer

No feasible points.

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

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

The bounds are $0 \leq x \leq 5$ and $0 \leq y \leq 3$ . Combining their endpoint values gives the rectangle corners $(0, 0), (5, 0), (5, 3), (0, 3)$ .

Keep practising

6 worksheet questions · 12 marks

Open worksheet 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 ↗

In this lesson

Understand the idea.

Feasible means every restriction holds

Find the corners from active boundaries

See the method in action.

Find the corners

Find the feasible axis endpoints.

Intersect the sloping boundaries.

Check the non-axis corner.

Collect the boundary corners.

An empty intersection

Combine the demands on the same sum.

Identify the contradiction.

State the feasible-set conclusion.

Try it yourself.

Check your understanding

Keep practising