Linear programming | Year 12

01 / LEARN

The idea, step by step

Organise the model before solving

Define variables and units, identify each resource bound and state whether decisions are continuous or whole numbers. A per-unit resource table helps prevent mixing coefficients from different resources. Form the objective separately and include non-negativity, minimum quantities and capacity limits supported by the context.

Explain the recommendation

Find feasible vertices and evaluate the objective at each. State the decision and objective value in context, then check original resource use. Identify unused resources when relevant. An unbounded region can still have a minimum: what matters is whether the objective can improve without limit in a feasible direction.

A common trap

An unbounded region never has an optimum.

Instead: It may have a finite minimum or maximum; examine how the objective changes in permitted directions.

02 / WORKED EXAMPLES

See the thinking

Example 1

A complete production model

Maximise $P = 5 x + 3 y$ subject to $x, y \geq 0$ , $x + y \leq 10$ and $2 x + y \leq 14$ .

Feasible vertices are $(0, 0), (7, 0), (4, 6), (0, 10)$ .
Profit values are 0,35,38,30.
At $(4, 6)$ both resource bounds are used exactly: 10 and 14.

Maximum 38 at

(4, 6)

Example 2

A minimum-cost mix

Minimise $C = 3 x + 4 y$ subject to $x, y \geq 0$ , $x + y \geq 8$ and $x + 2 y \geq 12$ .

The lower-bound vertices are $(12, 0), (4, 4), (0, 8)$ .
Costs are 36,28,32.
On the horizontal ray to the right of (12,0), cost increases; on the vertical ray above (0,8), cost increases. The connecting boundary segments are linear, so compare their endpoints.

Minimum 28 at

(4, 4)

Check your understanding

Maximise $P = 2 x + 3 y$ for $x, y \geq 0$ , $x + y \leq 5$ .

Show answer

Maximum 15 at $(0, 5)$ .

Vertex values are 0,10,15.

03 / WORKSHEET

Your turn

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

Worksheet PDF Worked solutions

Read the questions online

Maximise $P = 4 x + 3 y$ for $x, y \geq 0$ , $x + y \leq 8$ , $x \leq 5$ .
2 marks
Maximise $P = 3 x + 5 y$ for $x, y \geq 0$ , $x + y \leq 9$ , $x + 2 y \leq 14$ .
2 marks
Minimise $C = 2 x + 5 y$ for $x, y \geq 0$ , $x + y \geq 6$ .
2 marks
Why has $P = x + y$ no finite maximum when only $x, y \geq 0$ is required?
2 marks
A high-profit proposal uses 18 hours when at most 16 are available. Is it acceptable?
2 marks
The optimum for $P = 5 x + 3 y$ is $(4, 6)$ . If P is in hundreds of dollars, interpret the result.
2 marks

Read the worked solutions

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

Maximise $P = 4 x + 3 y$ for $x, y \geq 0$ , $x + y \leq 8$ , $x \leq 5$ .
2 marks
The vertices are $(0, 0), (5, 0), (5, 3), (0, 8)$ .
Their values are 0,20,29,24.
Maximum 29 at $(5, 3)$ .
Maximise $P = 3 x + 5 y$ for $x, y \geq 0$ , $x + y \leq 9$ , $x + 2 y \leq 14$ .
2 marks
Vertices are $(0, 0), (9, 0), (4, 5), (0, 7)$ .
Values are 0,27,37,35.
Maximum 37 at $(4, 5)$ .
Minimise $C = 2 x + 5 y$ for $x, y \geq 0$ , $x + y \geq 6$ .
2 marks
Rewrite the objective as $C = 2 (x + y) + 3 y$ .
Since $x + y \geq 6$ and $y \geq 0$ , $C \geq 12$ ; (6,0) attains that bound.
Minimum 12 at $(6, 0)$ .
Why has $P = x + y$ no finite maximum when only $x, y \geq 0$ is required?
2 marks
Feasible points $(t, 0)$ have arbitrarily large objective t.
It is unbounded above.
A high-profit proposal uses 18 hours when at most 16 are available. Is it acceptable?
2 marks
An infeasible decision is excluded regardless of its profit.
No.
The optimum for $P = 5 x + 3 y$ is $(4, 6)$ . If P is in hundreds of dollars, interpret the result.
2 marks
$P = 38$ hundreds, so multiply by 100.
AUD 3,800.

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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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