Simple optimisation | Year 11

01 / LEARN

The idea, step by step

Use the constraint first

Name the objective and give its units. Use the constraint to replace one variable, leaving a function of one independent variable. State the domain imposed by positive lengths, available resources or other practical conditions.

Differentiate the objective, solve for stationary candidates and determine their type. Substitute back to find every quantity requested, not just the chosen variable.

Check boundaries and assumptions

A stationary candidate outside the domain cannot solve the practical problem. On a closed interval compare endpoints and stationary values. For an open physical domain inspect limiting values when relevant.

An optimum is a result within the model. Integer-only production, wastage, or ignored costs may alter a practical decision. Include units, appropriate rounding and an explanation of the assumptions behind the answer.

A common trap

Differentiating before using the constraint to form the objective.

Instead: Correct modelling comes first; calculus cannot fix the wrong objective or domain.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example 1

A rectangle has perimeter 40m. Find its greatest area.

Let width be x, so length is $20 - x$ and $0 < x < 20$ .
$A = 20 x - x^{2}$ gives $A' = 20 - 2 x = 0$ at 10.
The derivative changes + to −; A(10)=100.

A 10m by 10m square, with maximum area 100m².

Example 2

Worked example 2

A farmer has 60m of fencing for three sides of a rectangle against a straight wall. Maximise area.

Let each perpendicular side be x. The remaining side is $60 - 2 x$ .
$A = x (60 - 2 x)$ on $0 < x < 30$ , so $A' = 60 - 4 x = 0$ at 15.
The maximum area is 15×30=450.

Perpendicular sides 15m, remaining side 30m; area 450m².

Check your understanding

Maximise $P (x) = - x^{2} + 10 x + 3$ on $[0, 8]$ .

Show answer

P=28 at x=5.

$P' = 10 - 2 x$ changes + to − at 5; endpoint values 3 and 19 are smaller.

03 / WORKSHEET

Your turn

6 questions · 26 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

A rectangle has perimeter 24m. Find its maximum area.
4 marks
A rectangle beside a wall has 36m of fencing for the other three sides. Find its maximum area.
4 marks
Maximise $P (x) = - 2 x^{2} + 24 x - 10$ on $[0, 10]$ .
4 marks
Minimise $C (x) = x^{2} - 10 x + 40$ on $[0, 8]$ .
4 marks
Maximise $f (x) = 100 - {(x - 8)}^{2}$ on $[0, 6]$ .
4 marks
An open box is made by removing squares of side x cm from each corner of a 12cm square, then folding. Find the greatest volume.
6 marks

Read the worked solutions

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

A rectangle has perimeter 24m. Find its maximum area.
4 marks
$A = x (12 - x)$ ; $A' = 12 - 2 x = 0$ at 6.
The downward parabola gives the maximum.
36m² with sides 6m and 6m.
A rectangle beside a wall has 36m of fencing for the other three sides. Find its maximum area.
4 marks
$A = x (36 - 2 x)$ ; $A' = 36 - 4 x = 0$ gives 9.
The other side is 18m and area 162m².
162m²; perpendicular sides 9m and remaining side 18m.
Maximise $P (x) = - 2 x^{2} + 24 x - 10$ on $[0, 10]$ .
4 marks
$P' = - 4 x + 24 = 0$ gives 6.
Derivative signs + then − establish the maximum; P(6)=62.
Maximum 62 at x=6.
Minimise $C (x) = x^{2} - 10 x + 40$ on $[0, 8]$ .
4 marks
$C' = 2 x - 10 = 0$ gives 5.
Compare C(0)=40, C(5)=15 and C(8)=24.
Minimum 15 at x=5.
Maximise $f (x) = 100 - {(x - 8)}^{2}$ on $[0, 6]$ .
4 marks
$f' = 16 - 2 x$ is positive throughout the domain.
The stationary input 8 is outside the interval; f(6)=96 is greatest.
96 at x=6.
An open box is made by removing squares of side x cm from each corner of a 12cm square, then folding. Find the greatest volume.
6 marks
$V = x {(12 - 2 x)}^{2} = 144 x - 48 x^{2} + 4 x^{3}$ , with $0 < x < 6$ .
$V' = 12 (x - 2) (x - 6)$ , so the admissible stationary input is 2.
V rises then falls and tends to 0 at both boundaries; V(2)=128.
128cm³ at x=2cm.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics topic 6: rates of change, derivatives and applications. Introductory polynomial calculus with explicit modelling and interpretation.

Official curriculum source ↗

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