Topics

Drawing the feasible region and finding its corners

Covers AS91574

What the region is

The feasible region is the set of points satisfying every constraint at once.
Every point in it is a plan the business could actually carry out. Every point outside breaks at least one limit.
It is always a convex polygon (possibly unbounded) — a shape with no dents. That fact is what guarantees the optimum sits at a corner.

Drawing it, efficiently

Step 1 — Draw each boundary line by replacing the inequality with $=$ $=$ and plotting the two intercepts.
- For $2x + y = 100$ : intercepts at $(50, 0)$ and $(0, 100)$ .
- For a line through the origin such as $x = 2y$ , plot $(0,0)$ and one convenient point like $(20, 10)$ .
Step 2 — Choose sensible axis scales first. Look at all the intercepts before drawing; a scale chosen for one constraint often cuts off another.
Step 3 — Test $(0,0)$ in each inequality and mark the allowed side with a small arrow. Where the origin is on the boundary, test another point.
Step 4 — Shade the region satisfied by all of them and label it R.
Number every line to match the constraints. An unlabelled diagram cannot be checked.

Shading convention

Two conventions are in use: shade the region that satisfies each inequality, or shade the region it excludes so that the feasible region is the only unshaded part.
The second is better when there are four or more constraints, because the answer stands out as white space instead of disappearing under overlapping shading.
State which convention you are using on the diagram. Either is acceptable; an unstated one is not.

Finding the corners exactly

Every corner is the intersection of two boundary lines. Find it by solving those two equations simultaneously.
Never read a corner off the graph. The graph tells you which two lines meet at each corner; the algebra tells you where.
Test every candidate in every constraint. Two boundaries can cross outside the region entirely, and that point is not a corner.
Worked through — where $2x + y = 100$ $2 x + y = 100$ meets $x + y = 80$ $x + y = 80$ :
- Subtract: $(2x + y) - (x + y) = 100 - 80$ , so $x = 20$
- Substitute: $20 + y = 80$ , so $y = 60$
- Corner $(20, 60)$ — and it must still be checked against $x \leq 40$ ✓ and $y \geq 0$ ✓

Bounded, unbounded and empty regions

Bounded — enclosed on all sides. A maximum and a minimum both exist.
Unbounded — open in some direction. Typical when all constraints are $\geq$ $\geq$ .
- A minimisation over an unbounded region still has an answer, because the region has corners on the side you are minimising towards.
- A maximisation over an unbounded region may have no answer at all — profit can grow without limit, which in a real context means a constraint has been left out.
Empty — the constraints contradict each other and no plan is possible. Report it, and name the conflicting pair.

Which constraints matter

A constraint is binding (or active) at a point if it holds with equality there — the resource is fully used.
A constraint is slack if it holds strictly — there is some left over.
A redundant constraint never binds anywhere, because another constraint is always tighter. It can be identified on the diagram: its boundary line does not touch the region at all.
Naming the binding constraints at the optimum is the most useful sentence in the whole answer, because it tells the business which resource to buy more of.

Tips and Tricks

Plot all the intercepts before choosing your scale. Redrawing the axes half-way through because one line runs off the page wastes more time than the check costs.

Mark the allowed side of every line with an arrow as you draw it. Doing all the testing at the end, on a diagram of five lines, is how the wrong overlap gets shaded.

Solve for corners, never read them. A corner read as "about $(20, 60)$ " gives an objective value that cannot be trusted to the nearest hundred dollars.

Check every candidate corner against every constraint. Rejecting a crossing point that lies outside the region is part of the working and should be visible.

An empty region is a real answer. Say which two constraints conflict, and by how much — that is what the business needs to know.

Identify the binding constraints at your answer. "Both vat time and packing time are fully used" is worth more than the profit figure alone.

Worked ExampleDrawing the region for the dairy model

Draw the feasible region for the dairy model from the previous page, and find every corner exactly.

$\text{Maximise } P = 40x + 30y \ \text{ subject to } \begin{cases} 2x + y \leq 100 & (1) \\ x + y \leq 80 & (2) \\ x \leq 40 & (3) \\ y \geq 10 & (4) \\ x \geq 0,\ y \geq 0 \end{cases}$

Step 1 — Find the intercepts of each boundary

Line (1): $2x + y = 100$

$x = 0$ : $y = 100$ → $(0, 100)$
$y = 0$ : $x = 50$ → $(50, 0)$

Line (2): $x + y = 80$

$x = 0$ : $y = 80$ → $(0, 80)$
$y = 0$ : $x = 80$ → $(80, 0)$

Line (3): $x = 40$ — vertical. Line (4): $y = 10$ — horizontal.

Step 2 — Test the origin in each constraint

Constraint	Test at $(0,0)$	Allowed side
(1) $2x + y \leq 100$	$0 \leq 100$ TRUE	towards origin
(2) $x + y \leq 80$	$0 \leq 80$ TRUE	towards origin
(3) $x \leq 40$	$0 \leq 40$ TRUE	towards origin
(4) $y \geq 10$	$0 \geq 10$ FALSE	away from origin

Step 3 — Identify which lines form each corner, from the diagram

Reading the shape of the region, its boundary runs: up the line $x = 0$ , along $x + y = 80$ , along $2x + y = 100$ , down $x = 40$ , and back along $y = 10$ .

Step 4 — Corner A: $x = 0$ meets $y = 10$

$A = (0,\ 10)$

Check all constraints: (1) $0 + 10 = 10 \leq 100$ ✓ (2) $10 \leq 80$ ✓ (3) $0 \leq 40$ ✓ (4) $10 \geq 10$ ✓ Valid.

Step 5 — Corner B: $x = 0$ meets $x + y = 80$

$B = (0,\ 80)$

Check: (1) $0 + 80 = 80 \leq 100$ ✓ (3) ✓ (4) $80 \geq 10$ ✓ Valid.

Step 6 — Corner C: $x + y = 80$ meets $2x + y = 100$

Subtract the first from the second:

$(2x + y) - (x + y) = 100 - 80$

$x = 20$

Substitute into $x + y = 80$ :

$y = 60$

$C = (20,\ 60)$

Check: (3) $20 \leq 40$ ✓ (4) $60 \geq 10$ ✓ Valid.

Step 7 — Corner D: $2x + y = 100$ meets $x = 40$

$2(40) + y = 100 \quad \Rightarrow \quad y = 20$

$D = (40,\ 20)$

Check: (2) $40 + 20 = 60 \leq 80$ ✓ (4) $20 \geq 10$ ✓ Valid.

Step 8 — Corner E: $x = 40$ meets $y = 10$

$E = (40,\ 10)$

Check: (1) $80 + 10 = 90 \leq 100$ ✓ (2) $50 \leq 80$ ✓ Valid.

Step 9 — Test a crossing point that is NOT a corner

Where does $x + y = 80$ meet $x = 40$ ? At $(40, 40)$ .

Check constraint (1): $2(40) + 40 = 120 > 100$ ✗ Fails the vat limit.

Step 10 — State the region

It is bounded, so both a maximum and a minimum of the profit function exist and will be found at corners.

Step 11 — Note which constraints bind where

Corner	Binding constraints	Meaning
$(0, 10)$	$x = 0$ , $y = 10$	no Halloumi, contract minimum only
$(0, 80)$	$x = 0$ , packing	no Halloumi, all packing used
$(20, 60)$	packing and vat	both main resources fully used
$(40, 20)$	storage and vat	storage full, vat full
$(40, 10)$	storage, contract	storage full, contract at minimum

Drawing the feasible region and finding its corners

Covers AS91574

What the region is

The feasible region is the set of points satisfying every constraint at once.
Every point in it is a plan the business could actually carry out. Every point outside breaks at least one limit.
It is always a convex polygon (possibly unbounded) — a shape with no dents. That fact is what guarantees the optimum sits at a corner.

Drawing it, efficiently

Step 1 — Draw each boundary line by replacing the inequality with $=$ $=$ and plotting the two intercepts.
- For $2x + y = 100$ : intercepts at $(50, 0)$ and $(0, 100)$ .
- For a line through the origin such as $x = 2y$ , plot $(0,0)$ and one convenient point like $(20, 10)$ .
Step 2 — Choose sensible axis scales first. Look at all the intercepts before drawing; a scale chosen for one constraint often cuts off another.
Step 3 — Test $(0,0)$ in each inequality and mark the allowed side with a small arrow. Where the origin is on the boundary, test another point.
Step 4 — Shade the region satisfied by all of them and label it R.
Number every line to match the constraints. An unlabelled diagram cannot be checked.

Shading convention

Two conventions are in use: shade the region that satisfies each inequality, or shade the region it excludes so that the feasible region is the only unshaded part.
The second is better when there are four or more constraints, because the answer stands out as white space instead of disappearing under overlapping shading.
State which convention you are using on the diagram. Either is acceptable; an unstated one is not.

Finding the corners exactly

Every corner is the intersection of two boundary lines. Find it by solving those two equations simultaneously.
Never read a corner off the graph. The graph tells you which two lines meet at each corner; the algebra tells you where.
Test every candidate in every constraint. Two boundaries can cross outside the region entirely, and that point is not a corner.
Worked through — where $2x + y = 100$ $2 x + y = 100$ meets $x + y = 80$ $x + y = 80$ :
- Subtract: $(2x + y) - (x + y) = 100 - 80$ , so $x = 20$
- Substitute: $20 + y = 80$ , so $y = 60$
- Corner $(20, 60)$ — and it must still be checked against $x \leq 40$ ✓ and $y \geq 0$ ✓

Bounded, unbounded and empty regions

Bounded — enclosed on all sides. A maximum and a minimum both exist.
Unbounded — open in some direction. Typical when all constraints are $\geq$ $\geq$ .
- A minimisation over an unbounded region still has an answer, because the region has corners on the side you are minimising towards.
- A maximisation over an unbounded region may have no answer at all — profit can grow without limit, which in a real context means a constraint has been left out.
Empty — the constraints contradict each other and no plan is possible. Report it, and name the conflicting pair.

Which constraints matter

A constraint is binding (or active) at a point if it holds with equality there — the resource is fully used.
A constraint is slack if it holds strictly — there is some left over.
A redundant constraint never binds anywhere, because another constraint is always tighter. It can be identified on the diagram: its boundary line does not touch the region at all.
Naming the binding constraints at the optimum is the most useful sentence in the whole answer, because it tells the business which resource to buy more of.

Tips and Tricks

Plot all the intercepts before choosing your scale. Redrawing the axes half-way through because one line runs off the page wastes more time than the check costs.

Mark the allowed side of every line with an arrow as you draw it. Doing all the testing at the end, on a diagram of five lines, is how the wrong overlap gets shaded.

Solve for corners, never read them. A corner read as "about $(20, 60)$ " gives an objective value that cannot be trusted to the nearest hundred dollars.

Check every candidate corner against every constraint. Rejecting a crossing point that lies outside the region is part of the working and should be visible.

An empty region is a real answer. Say which two constraints conflict, and by how much — that is what the business needs to know.

Identify the binding constraints at your answer. "Both vat time and packing time are fully used" is worth more than the profit figure alone.

Worked ExampleDrawing the region for the dairy model

Draw the feasible region for the dairy model from the previous page, and find every corner exactly.

$\text{Maximise } P = 40x + 30y \ \text{ subject to } \begin{cases} 2x + y \leq 100 & (1) \\ x + y \leq 80 & (2) \\ x \leq 40 & (3) \\ y \geq 10 & (4) \\ x \geq 0,\ y \geq 0 \end{cases}$

Step 1 — Find the intercepts of each boundary

Line (1): $2x + y = 100$

$x = 0$ : $y = 100$ → $(0, 100)$
$y = 0$ : $x = 50$ → $(50, 0)$

Line (2): $x + y = 80$

$x = 0$ : $y = 80$ → $(0, 80)$
$y = 0$ : $x = 80$ → $(80, 0)$

Line (3): $x = 40$ — vertical. Line (4): $y = 10$ — horizontal.

Step 2 — Test the origin in each constraint

Constraint	Test at $(0,0)$	Allowed side
(1) $2x + y \leq 100$	$0 \leq 100$ TRUE	towards origin
(2) $x + y \leq 80$	$0 \leq 80$ TRUE	towards origin
(3) $x \leq 40$	$0 \leq 40$ TRUE	towards origin
(4) $y \geq 10$	$0 \geq 10$ FALSE	away from origin

Step 3 — Identify which lines form each corner, from the diagram

Reading the shape of the region, its boundary runs: up the line $x = 0$ , along $x + y = 80$ , along $2x + y = 100$ , down $x = 40$ , and back along $y = 10$ .

Step 4 — Corner A: $x = 0$ meets $y = 10$

$A = (0,\ 10)$

Check all constraints: (1) $0 + 10 = 10 \leq 100$ ✓ (2) $10 \leq 80$ ✓ (3) $0 \leq 40$ ✓ (4) $10 \geq 10$ ✓ Valid.

Step 5 — Corner B: $x = 0$ meets $x + y = 80$

$B = (0,\ 80)$

Check: (1) $0 + 80 = 80 \leq 100$ ✓ (3) ✓ (4) $80 \geq 10$ ✓ Valid.

Step 6 — Corner C: $x + y = 80$ meets $2x + y = 100$

Subtract the first from the second:

$(2x + y) - (x + y) = 100 - 80$

$x = 20$

Substitute into $x + y = 80$ :

$y = 60$

$C = (20,\ 60)$

Check: (3) $20 \leq 40$ ✓ (4) $60 \geq 10$ ✓ Valid.

Step 7 — Corner D: $2x + y = 100$ meets $x = 40$

$2(40) + y = 100 \quad \Rightarrow \quad y = 20$

$D = (40,\ 20)$

Check: (2) $40 + 20 = 60 \leq 80$ ✓ (4) $20 \geq 10$ ✓ Valid.

Step 8 — Corner E: $x = 40$ meets $y = 10$

$E = (40,\ 10)$

Check: (1) $80 + 10 = 90 \leq 100$ ✓ (2) $50 \leq 80$ ✓ Valid.

Step 9 — Test a crossing point that is NOT a corner

Where does $x + y = 80$ meet $x = 40$ ? At $(40, 40)$ .

Check constraint (1): $2(40) + 40 = 120 > 100$ ✗ Fails the vat limit.

Step 10 — State the region

It is bounded, so both a maximum and a minimum of the profit function exist and will be found at corners.

Step 11 — Note which constraints bind where

Corner	Binding constraints	Meaning
$(0, 10)$	$x = 0$ , $y = 10$	no Halloumi, contract minimum only
$(0, 80)$	$x = 0$ , packing	no Halloumi, all packing used
$(20, 60)$	packing and vat	both main resources fully used
$(40, 20)$	storage and vat	storage full, vat full
$(40, 10)$	storage, contract	storage full, contract at minimum

Drawing the feasible region and finding its corners

What the region is

Drawing it, efficiently

Shading convention

Finding the corners exactly

Bounded, unbounded and empty regions

Which constraints matter

Step 1 — Find the intercepts of each boundary

Step 2 — Test the origin in each constraint

Step 3 — Identify which lines form each corner, from the diagram

Step 4 — Corner A: x=0x = 0x=0 meets y=10y = 10y=10

Step 5 — Corner B: x=0x = 0x=0 meets x+y=80x + y = 80x+y=80

Step 6 — Corner C: x+y=80x + y = 80x+y=80 meets 2x+y=1002x + y = 1002x+y=100

Step 7 — Corner D: 2x+y=1002x + y = 1002x+y=100 meets x=40x = 40x=40

Step 8 — Corner E: x=40x = 40x=40 meets y=10y = 10y=10

Step 9 — Test a crossing point that is NOT a corner

Step 10 — State the region

Step 11 — Note which constraints bind where

Drawing the feasible region and finding its corners

What the region is

Drawing it, efficiently

Shading convention

Finding the corners exactly

Bounded, unbounded and empty regions

Which constraints matter

Step 1 — Find the intercepts of each boundary

Step 2 — Test the origin in each constraint

Step 3 — Identify which lines form each corner, from the diagram

Step 4 — Corner A: x=0x = 0x=0 meets y=10y = 10y=10

Step 5 — Corner B: x=0x = 0x=0 meets x+y=80x + y = 80x+y=80

Step 6 — Corner C: x+y=80x + y = 80x+y=80 meets 2x+y=1002x + y = 1002x+y=100

Step 7 — Corner D: 2x+y=1002x + y = 1002x+y=100 meets x=40x = 40x=40

Step 8 — Corner E: x=40x = 40x=40 meets y=10y = 10y=10

Step 9 — Test a crossing point that is NOT a corner

Step 10 — State the region

Step 11 — Note which constraints bind where

Step 4 — Corner A: $x = 0$ meets $y = 10$

Step 5 — Corner B: $x = 0$ meets $x + y = 80$

Step 6 — Corner C: $x + y = 80$ meets $2x + y = 100$

Step 7 — Corner D: $2x + y = 100$ meets $x = 40$

Step 8 — Corner E: $x = 40$ meets $y = 10$

Step 4 — Corner A: $x = 0$ meets $y = 10$

Step 5 — Corner B: $x = 0$ meets $x + y = 80$

Step 6 — Corner C: $x + y = 80$ meets $2x + y = 100$

Step 7 — Corner D: $2x + y = 100$ meets $x = 40$

Step 8 — Corner E: $x = 40$ meets $y = 10$