Objective functions | Year 12

01 / LEARN

The idea, step by step

Define what best means

Constraints say what is allowed; the objective says what to improve. State whether to maximise or minimise. Profit uses selling price minus unit cost, not selling price alone. A fixed cost shifts every objective value by the same amount and therefore does not change which feasible decision is best when all other conditions stay unchanged.

A family of parallel lines

For $P = a x + b y$ , each fixed P-value produces a straight line. Changing P shifts it without changing its gradient. On a non-empty bounded polygon a linear objective has an optimum at a vertex. If it is constant along a feasible edge, all points on that edge are optimal; the result is not necessarily unique.

P = a x + b y

A common trap

A resource expression is automatically the objective.

Instead: A resource expression measures something limited; an objective measures the quantity being optimised.

02 / WORKED EXAMPLES

See the thinking

Example 1

Profit rather than revenue

Basic kits sell for AUD 18 and cost AUD 11. Deluxe kits sell for AUD 26 and cost AUD 15. Write profit for x basic and y deluxe kits; evaluate at $(6, 4)$ .

Unit profits are 7 and 11 dollars.
Thus $P = 7 x + 11 y$ .
At $(6, 4)$ , $P = 42 + 44 = 86$ .

P = 7 x + 11 y

; AUD 86.

Example 2

Compare corners

A region has vertices $(0, 0), (5, 0), (3, 4), (0, 6)$ . Maximise $P = 4 x + 5 y$ .

Vertex	P
(0,0)	0
(5,0)	20
(3,4)	32
(0,6)	30

Objective values are 0,20,32,30 in the given order.
The largest value is 32.
A linear objective on a bounded polygon attains an optimum at a vertex.

Maximum 32 at

(3, 4)

Check your understanding

Write profit when each x earns AUD 9 and each y earns AUD 4. Evaluate at $(2, 7)$ .

Show answer

$P = 9 x + 4 y$ ; AUD 46.

$18 + 28 = 46$ .

03 / WORKSHEET

Your turn

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

Worksheet PDF Worked solutions

Read the questions online

A has unit revenue 35 and cost 23; B has revenue 28 and cost 19. Write profit.
2 marks
Evaluate $C = 8 x + 13 y$ at $(4, 3)$ .
2 marks
At vertices $(0, 0), (8, 0), (4, 5), (0, 7)$ , maximise $P = 3 x + 4 y$ .
2 marks
At vertices $(2, 3), (7, 3), (4, 6)$ , minimise $C = 5 x + 2 y$ .
2 marks
Does subtracting a fixed AUD 100 change the maximiser of $P = 7 x + 5 y$ ?
2 marks
Why does every point from $(0, 6)$ to $(6, 0)$ have the same $P = 2 x + 2 y$ ?
2 marks

Read the worked solutions

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

A has unit revenue 35 and cost 23; B has revenue 28 and cost 19. Write profit.
2 marks
Subtract each unit cost from revenue first.
$P = 12 x + 9 y$ .
Evaluate $C = 8 x + 13 y$ at $(4, 3)$ .
2 marks
$32 + 39 = 71$ .
71.
At vertices $(0, 0), (8, 0), (4, 5), (0, 7)$ , maximise $P = 3 x + 4 y$ .
2 marks
The values are 0,24,32,28.
Maximum 32 at $(4, 5)$ .
At vertices $(2, 3), (7, 3), (4, 6)$ , minimise $C = 5 x + 2 y$ .
2 marks
The values are 16,41,32.
Minimum 16 at $(2, 3)$ .
Does subtracting a fixed AUD 100 change the maximiser of $P = 7 x + 5 y$ ?
2 marks
All objective values decrease equally, preserving their order.
No, if the feasible set stays unchanged.
Why does every point from $(0, 6)$ to $(6, 0)$ have the same $P = 2 x + 2 y$ ?
2 marks
On that edge $x + y = 6$ , so $P = 2 (x + y) = 12$ .
Each has P=12.

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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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