Optimisation | Year 12 | Ascent Library

01 / LEARN

The idea, step by step

Model before differentiating

Name the quantity to maximise or minimise. Define variables with units, use the constraint to express the objective in one variable, and state the feasible domain before calculating.

A stationary point is only a candidate. Compare allowed endpoints or use a sign/concavity argument to justify an absolute optimum.

Return to the situation

Report every requested dimension or amount, not only the variable used in the derivative. Check units and whether an endpoint is physically allowed.

Technology can help locate difficult roots, but the model, constraint, domain and interpretation remain mathematical responsibilities. Do not invent a precision unsupported by the original measurements.

A common trap

Finding a stationary number without checking whether it is feasible or whether it maximises the objective.

Instead: Keep the domain visible and justify the classification in the original situation.

02 / WORKED EXAMPLES

See the thinking

Example 1

A rectangle with fixed perimeter

A rectangle has perimeter 20 cm. Find its largest possible area.

Let one side be $x$ cm; the other is $10 - x$ , with $0 < x < 10$ .
$A = x (10 - x)$ , so $A^{'} = 10 - 2 x = 0$ gives $x = 5$ .
$A^{''} = - 2 < 0$ , so this is the maximum.

5 \times 5

cm square, with area

25 {cm}^{2}

Example 2

A pen against a wall

A pen against a wall needs fencing on three sides. With 60 m of fencing, find its maximum area.

Let the two equal perpendicular sides be $x$ m; the third is $60 - 2 x$ .
$A = 60 x - 2 x^{2}$ has derivative $60 - 4 x$ , zero at 15.
The other side is 30 m and the area function is concave down.

450 m^{2}

, using dimensions

15

m by

30

Check your understanding

A rectangle of area 36 cm squared has sides $x$ and $36 / x$ . What expression gives its perimeter?

Show answer

$P (x) = 2 x + 72 / x$ , $x > 0$

Use $P = 2 (length + width)$ after applying the area constraint.

03 / WORKSHEET

Your turn

6 questions · 24 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

A rectangle has perimeter 28 m. Find its maximum area.
3 marks
A three-sided pen beside a wall uses 80 m of fencing. Find optimal dimensions.
3 marks
Squares of side $x$ are cut from each corner of a 12 cm square to form an open box. Find the maximum volume.
4 marks
A product costs $C (x) = x^{2} + 100$ dollars to make $x > 0$ units. Minimise average cost per unit.
4 marks
A point $(x, x^{2})$ lies on a parabola. Find its least distance from $(0, 2)$ .
5 marks
A closed cylinder has total surface area $24 π$ square centimetres. Find its largest volume.
5 marks

Read the worked solutions

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

A rectangle has perimeter 28 m. Find its maximum area.
3 marks
The sides are $x$ and $14 - x$ .
Maximise $14 x - x^{2}$ , whose derivative vanishes at 7.
$49 m^{2}$ for a 7 m square.
A three-sided pen beside a wall uses 80 m of fencing. Find optimal dimensions.
3 marks
Maximise $x (80 - 2 x)$ .
The derivative $80 - 4 x$ is zero at 20.
20 m perpendicular to the wall and 40 m parallel; area $800 m^{2}$ .
Squares of side $x$ are cut from each corner of a 12 cm square to form an open box. Find the maximum volume.
4 marks
The volume is $V = x {(12 - 2 x)}^{2}$ , with $0 < x < 6$ .
Its derivative is $12 (x - 2) (x - 6)$ .
The interior maximum is at 2; both limiting endpoint volumes are zero.
$128 {cm}^{3}$ at $x = 2$ cm.
A product costs $C (x) = x^{2} + 100$ dollars to make $x > 0$ units. Minimise average cost per unit.
4 marks
Average cost is $x + 100 / x$ .
Its derivative is $1 - 100 / x^{2}$ , zero at 10.
The derivative changes from negative to positive.
$20$ per unit at $x = 10$ .
A point $(x, x^{2})$ lies on a parabola. Find its least distance from $(0, 2)$ .
5 marks
Minimise the square of distance: $D^{2} = x^{2} + {(x^{2} - 2)}^{2} = x^{4} - 3 x^{2} + 4$ .
The stationary inputs are 0 and $\pm \sqrt{3 / 2}$ .
The smaller squared distance is $7 / 4$ ; take its square root.
$\sqrt{7} / 2$ , at $x = \pm \sqrt{3 / 2}$ .
A closed cylinder has total surface area $24 π$ square centimetres. Find its largest volume.
5 marks
The constraint gives $h = 12 / r - r$ , where $0 < r < \sqrt{12}$ .
The volume is $π (12 r - r^{3})$ .
Its derivative vanishes at $r = 2$ ; the endpoint volumes tend to zero.
$16 π {cm}^{3}$ at radius 2 cm and height 4 cm.

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Curriculum reference & lesson scope

SACE Stage 2 Mathematical Methods

Further differentiation and applications from the current SACE Methods outline. Logarithmic derivatives have their own chapter; integration follows later. Trigonometric calculus uses radians.

Official curriculum source ↗

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