Topics

Systems of inequations and the feasible region

Covers AS91269

What a system of inequations describes

Real situations impose several restrictions at once: a budget, a time limit, a minimum order, a storage capacity.
Each restriction is one inequation, and each inequation cuts the plane down to a half-plane.
The feasible region is the overlap — the set of points satisfying every inequation simultaneously.
Every point in the feasible region is a genuinely possible plan. Every point outside breaks at least one restriction.

Building the region

Step 1 — Define the variables, in words, with units.
Step 2 — Write one inequation per restriction, and add $x \geq 0$ , $y \geq 0$ .
Step 3 — Draw each boundary line from its intercepts, solid or dashed as required.
Step 4 — Test the origin in each inequation and mark which side is allowed.
Step 5 — Identify the overlap and shade it clearly.
Number your lines on the diagram to match the inequations. A region bounded by five unlabelled lines cannot be checked by anyone, including you.

Finding the corner points

The corners (vertices) of the region are where two boundary lines cross.
Find them algebraically, by solving the two boundary equations simultaneously — not by reading the graph.
Which two lines? Only those that actually meet on the edge of the region. Two boundaries can cross at a point that is outside the region entirely, and that crossing is not a corner.
Check every candidate corner by substituting it into all the inequations. If it fails one, it is not a corner of the feasible region.
The corners matter because they are the extreme combinations — the plans that use one or more resources completely.

Bounded and unbounded regions

A bounded region is enclosed on all sides — it is a polygon with finite area.
An unbounded region extends forever in some direction. That is normal when all the restrictions are "at least" conditions.
An unbounded region is not an error, but it is worth saying so, because it means no restriction stops the quantities growing in that direction.

Whole-number solutions

When the variables count discrete items — trucks, workers, tickets, sacks — only points with whole-number coordinates are usable.
The corner points are often not whole numbers. A corner at $(6.4, 3.2)$ is a genuine corner of the region and an impossible plan.
List the whole-number points near a corner and test them in every inequation to find the usable options.
Say this in the write-up. Recognising that the model's continuous region overstates the real set of choices is exactly the interpretation EN4 asks for.

Interpreting the region in context

A point inside — a plan that satisfies every restriction, with something to spare.
A point on a boundary — a plan that uses one resource exactly.
A corner — a plan that uses two resources exactly, at the same time.
An empty region — the restrictions contradict each other, so no plan is possible at all. Report that as the answer, and say which pair of conditions conflicts.
Always translate back. "The region has corners at $(0,8)$ , $(4,6)$ and $(9,0)$ " means "the factory can make anything from 0 to 9 of the first product, and the best combinations use both machines fully".

Tips and Tricks

Number each line on your diagram to match its inequation. Unlabelled boundaries make a region impossible to check and impossible to mark.

Find corners by solving equations, not by reading the graph. A corner read off as "about $(3.5, 4)$ " is a guess; solving the pair of boundary equations gives the exact point.

Test every corner in every inequation. Two boundary lines can cross well outside the region — that intersection is not a corner, and including it is a common error.

Watch for whole numbers. If the variables count things, say explicitly that only lattice points in the region are real options, and give one.

An empty region is a valid answer. If the restrictions cannot all be met, say so and name the two that conflict — do not force a region that is not there.

Shade lightly, or shade what is excluded. With four or five inequations the satisfying region can disappear under the pencil. Whichever convention you use, state it on the diagram.

Worked ExamplePlanning a production run

A small workshop makes wooden chairs and tables. Each chair needs 2 hours of machining and 3 hours of finishing. Each table needs 5 hours of machining and 2 hours of finishing. There are at most 40 machining hours and at most 36 finishing hours available this week. The workshop has already promised to deliver at least 2 tables.

Write the system of inequations, draw the feasible region, find its corner points, and interpret the result for the workshop.

Step 1 — Define the variables

Step 2 — Form one inequation per restriction

Machining hours — 2 per chair, 5 per table, at most 40:

$2c + 5t \leq 40 \qquad (1)$

Finishing hours — 3 per chair, 2 per table, at most 36:

$3c + 2t \leq 36 \qquad (2)$

The promised tables — at least 2:

$t \geq 2 \qquad (3)$

Non-negativity — you cannot make a negative number of chairs:

$c \geq 0 \qquad (4)$

Step 3 — Find the intercepts of each boundary

Line (1): $2c + 5t = 40$

$c = 0$ : $t = 8$ → $(0, 8)$
$t = 0$ : $c = 20$ → $(20, 0)$

Line (2): $3c + 2t = 36$

$c = 0$ : $t = 18$ → $(0, 18)$
$t = 0$ : $c = 12$ → $(12, 0)$

Line (3): $t = 2$ — a horizontal line.

Step 4 — Test the origin in each inequation

(1): $2(0) + 5(0) = 0 \leq 40$ TRUE → shade towards the origin
(2): $3(0) + 2(0) = 0 \leq 36$ TRUE → shade towards the origin
(3): $0 \geq 2$ FALSE → shade away from the origin, above the line $t = 2$

Step 5 — Identify the corner points

The region is bounded below by $t = 2$ , on the left by $c = 0$ , and above/right by lines (1) and (2).

Corner A — where $c = 0$ meets $t = 2$ :

$A = (0, 2)$

Corner B — where $t = 2$ meets line (1)? Check both:

Line (1) at $t = 2$ : $2c + 10 = 40$ , so $c = 15$ → $(15, 2)$
Line (2) at $t = 2$ : $3c + 4 = 36$ , so $c = \frac{32}{3} = 10.67$ → $(10.67, 2)$

Test $(15, 2)$ in inequation (2): $3(15) + 2(2) = 49 > 36$ FAILS — so $(15,2)$ is not in the region.

So the corner on $t = 2$ is $B = \left(\frac{32}{3}, 2\right) \approx (10.67, 2)$ , from line (2).

Corner C — where lines (1) and (2) cross. Solve simultaneously:

$2c + 5t = 40 \qquad (1)$ $3c + 2t = 36 \qquad (2)$

Eliminate $c$ : $(1) \times 3$ gives $6c + 15t = 120$ ; $(2) \times 2$ gives $6c + 4t = 72$ . Subtract:

$11t = 48 \quad \Rightarrow \quad t = \frac{48}{11} \approx 4.36$

Substitute back into (1):

$2c + 5\left(\frac{48}{11}\right) = 40$

$2c = 40 - \frac{240}{11} = \frac{440 - 240}{11} = \frac{200}{11}$

$c = \frac{100}{11} \approx 9.09$

$C = \left(\frac{100}{11}, \frac{48}{11}\right) \approx (9.09, 4.36)$

Check $C$ in all inequations: $t \approx 4.36 \geq 2$ ✓, $c \geq 0$ ✓, and it lies on both (1) and (2) by construction ✓

Corner D — where line (1) meets $c = 0$ :

$5t = 40 \quad \Rightarrow \quad t = 8, \qquad D = (0, 8)$

Check $D$ in (2): $3(0) + 2(8) = 16 \leq 36$ ✓

Step 6 — State the region

Step 7 — Interpret in context, including whole numbers

Two of the corners are not whole numbers, so they are not achievable plans — the workshop cannot make $9.09$ chairs.

Useful whole-number plans near the corners:

$(10, 2)$ — machining $2(10) + 5(2) = 30 \leq 40$ ✓, finishing $3(10) + 2(2) = 34 \leq 36$ ✓ Feasible.
$(9, 4)$ — machining $18 + 20 = 38 \leq 40$ ✓, finishing $27 + 8 = 35 \leq 36$ ✓ Feasible.
$(10, 4)$ — machining $20 + 20 = 40 \leq 40$ ✓, finishing $30 + 8 = 38 > 36$ ✗ Not feasible — 2 hours short on finishing.

The binding restriction depends where you are. Along the edge from $(10.67, 2)$ to $(9.09, 4.36)$ the workshop uses all 36 finishing hours; along the edge from $(9.09, 4.36)$ to $(0,8)$ it uses all 40 machining hours. Only at the corner $(9.09, 4.36)$ would both be used completely — and since that point is not a whole number of items, no achievable plan uses both resources exactly.

Systems of inequations and the feasible region

Covers AS91269

What a system of inequations describes

Real situations impose several restrictions at once: a budget, a time limit, a minimum order, a storage capacity.
Each restriction is one inequation, and each inequation cuts the plane down to a half-plane.
The feasible region is the overlap — the set of points satisfying every inequation simultaneously.
Every point in the feasible region is a genuinely possible plan. Every point outside breaks at least one restriction.

Building the region

Step 1 — Define the variables, in words, with units.
Step 2 — Write one inequation per restriction, and add $x \geq 0$ , $y \geq 0$ .
Step 3 — Draw each boundary line from its intercepts, solid or dashed as required.
Step 4 — Test the origin in each inequation and mark which side is allowed.
Step 5 — Identify the overlap and shade it clearly.
Number your lines on the diagram to match the inequations. A region bounded by five unlabelled lines cannot be checked by anyone, including you.

Finding the corner points

The corners (vertices) of the region are where two boundary lines cross.
Find them algebraically, by solving the two boundary equations simultaneously — not by reading the graph.
Which two lines? Only those that actually meet on the edge of the region. Two boundaries can cross at a point that is outside the region entirely, and that crossing is not a corner.
Check every candidate corner by substituting it into all the inequations. If it fails one, it is not a corner of the feasible region.
The corners matter because they are the extreme combinations — the plans that use one or more resources completely.

Bounded and unbounded regions

A bounded region is enclosed on all sides — it is a polygon with finite area.
An unbounded region extends forever in some direction. That is normal when all the restrictions are "at least" conditions.
An unbounded region is not an error, but it is worth saying so, because it means no restriction stops the quantities growing in that direction.

Whole-number solutions

When the variables count discrete items — trucks, workers, tickets, sacks — only points with whole-number coordinates are usable.
The corner points are often not whole numbers. A corner at $(6.4, 3.2)$ is a genuine corner of the region and an impossible plan.
List the whole-number points near a corner and test them in every inequation to find the usable options.
Say this in the write-up. Recognising that the model's continuous region overstates the real set of choices is exactly the interpretation EN4 asks for.

Interpreting the region in context

A point inside — a plan that satisfies every restriction, with something to spare.
A point on a boundary — a plan that uses one resource exactly.
A corner — a plan that uses two resources exactly, at the same time.
An empty region — the restrictions contradict each other, so no plan is possible at all. Report that as the answer, and say which pair of conditions conflicts.
Always translate back. "The region has corners at $(0,8)$ , $(4,6)$ and $(9,0)$ " means "the factory can make anything from 0 to 9 of the first product, and the best combinations use both machines fully".

Tips and Tricks

Number each line on your diagram to match its inequation. Unlabelled boundaries make a region impossible to check and impossible to mark.

Find corners by solving equations, not by reading the graph. A corner read off as "about $(3.5, 4)$ " is a guess; solving the pair of boundary equations gives the exact point.

Test every corner in every inequation. Two boundary lines can cross well outside the region — that intersection is not a corner, and including it is a common error.

Watch for whole numbers. If the variables count things, say explicitly that only lattice points in the region are real options, and give one.

An empty region is a valid answer. If the restrictions cannot all be met, say so and name the two that conflict — do not force a region that is not there.

Shade lightly, or shade what is excluded. With four or five inequations the satisfying region can disappear under the pencil. Whichever convention you use, state it on the diagram.

Worked ExamplePlanning a production run

Write the system of inequations, draw the feasible region, find its corner points, and interpret the result for the workshop.

Step 1 — Define the variables

Step 2 — Form one inequation per restriction

Machining hours — 2 per chair, 5 per table, at most 40:

$2c + 5t \leq 40 \qquad (1)$

Finishing hours — 3 per chair, 2 per table, at most 36:

$3c + 2t \leq 36 \qquad (2)$

The promised tables — at least 2:

$t \geq 2 \qquad (3)$

Non-negativity — you cannot make a negative number of chairs:

$c \geq 0 \qquad (4)$

Step 3 — Find the intercepts of each boundary

Line (1): $2c + 5t = 40$

$c = 0$ : $t = 8$ → $(0, 8)$
$t = 0$ : $c = 20$ → $(20, 0)$

Line (2): $3c + 2t = 36$

$c = 0$ : $t = 18$ → $(0, 18)$
$t = 0$ : $c = 12$ → $(12, 0)$

Line (3): $t = 2$ — a horizontal line.

Step 4 — Test the origin in each inequation

(1): $2(0) + 5(0) = 0 \leq 40$ TRUE → shade towards the origin
(2): $3(0) + 2(0) = 0 \leq 36$ TRUE → shade towards the origin
(3): $0 \geq 2$ FALSE → shade away from the origin, above the line $t = 2$

Step 5 — Identify the corner points

The region is bounded below by $t = 2$ , on the left by $c = 0$ , and above/right by lines (1) and (2).

Corner A — where $c = 0$ meets $t = 2$ :

$A = (0, 2)$

Corner B — where $t = 2$ meets line (1)? Check both:

Line (1) at $t = 2$ : $2c + 10 = 40$ , so $c = 15$ → $(15, 2)$
Line (2) at $t = 2$ : $3c + 4 = 36$ , so $c = \frac{32}{3} = 10.67$ → $(10.67, 2)$

Test $(15, 2)$ in inequation (2): $3(15) + 2(2) = 49 > 36$ FAILS — so $(15,2)$ is not in the region.

So the corner on $t = 2$ is $B = \left(\frac{32}{3}, 2\right) \approx (10.67, 2)$ , from line (2).

Corner C — where lines (1) and (2) cross. Solve simultaneously:

$2c + 5t = 40 \qquad (1)$ $3c + 2t = 36 \qquad (2)$

Eliminate $c$ : $(1) \times 3$ gives $6c + 15t = 120$ ; $(2) \times 2$ gives $6c + 4t = 72$ . Subtract:

$11t = 48 \quad \Rightarrow \quad t = \frac{48}{11} \approx 4.36$

Substitute back into (1):

$2c + 5\left(\frac{48}{11}\right) = 40$

$2c = 40 - \frac{240}{11} = \frac{440 - 240}{11} = \frac{200}{11}$

$c = \frac{100}{11} \approx 9.09$

$C = \left(\frac{100}{11}, \frac{48}{11}\right) \approx (9.09, 4.36)$

Check $C$ in all inequations: $t \approx 4.36 \geq 2$ ✓, $c \geq 0$ ✓, and it lies on both (1) and (2) by construction ✓

Corner D — where line (1) meets $c = 0$ :

$5t = 40 \quad \Rightarrow \quad t = 8, \qquad D = (0, 8)$

Check $D$ in (2): $3(0) + 2(8) = 16 \leq 36$ ✓

Step 6 — State the region

Step 7 — Interpret in context, including whole numbers

Two of the corners are not whole numbers, so they are not achievable plans — the workshop cannot make $9.09$ chairs.

Useful whole-number plans near the corners:

$(10, 2)$ — machining $2(10) + 5(2) = 30 \leq 40$ ✓, finishing $3(10) + 2(2) = 34 \leq 36$ ✓ Feasible.
$(9, 4)$ — machining $18 + 20 = 38 \leq 40$ ✓, finishing $27 + 8 = 35 \leq 36$ ✓ Feasible.
$(10, 4)$ — machining $20 + 20 = 40 \leq 40$ ✓, finishing $30 + 8 = 38 > 36$ ✗ Not feasible — 2 hours short on finishing.