Optimisation | Year 12 | Ascent Library

01 / LEARN

Understand the idea.

Model before differentiating

Start an optimisation problem by defining the quantity to maximise or minimise. Use the constraint to express it in one variable, with units and a feasible domain, before finding a derivative.

A stationary input is a candidate, not a complete answer. Check that it is feasible and justify the absolute optimum using derivative signs, concavity or endpoint comparisons, including any allowed boundary cases.

Return to the situation

Return from the calculus variable to the original situation. Report all requested dimensions and the resulting area, volume or cost. An optimal value needs the appropriate units, not just the number found from $f^{'} = 0$ .

Positive lengths usually make degenerate endpoints inadmissible. Keep that restriction visible and check behaviour near the boundaries. Technology can locate difficult candidates, but it cannot decide whether a model or its domain matches the situation.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A rectangle with fixed perimeter

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

Worked solution

1
Use the perimeter constraint.
$2 (x + y) = 20 ⟹ y = 10 - x, 0 < x < 10$
Both rectangle side lengths must be positive.
2
Express and differentiate the area.
$A (x) = x (10 - x) = 10 x - x^{2}, A^{'} (x) = 10 - 2 x$
The constraint reduces the area to a function of one variable.
3
Find the feasible stationary input.
$10 - 2 x = 0 ⟹ x = 5, y = 5$
Both dimensions satisfy the perimeter constraint and lie in the feasible range.
4
Justify the maximum and evaluate it.
$A^{''} (x) = - 2 < 0, A (5) = 25 {cm}^{2}$
The quadratic is concave down throughout the feasible interval, so its stationary point is the absolute maximum.

Answer

A $5 \times 5$ cm square, with area $25 {cm}^{2}$ .

EXAMPLE 02

A pen against a wall

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

Worked solution

1
Use only the three fenced sides.
$2 x + y = 60 ⟹ y = 60 - 2 x, 0 < x < 30$
The wall supplies the fourth side, while the two perpendicular sides each use x metres of fencing.
2
Form and differentiate the area.
$A (x) = x (60 - 2 x) = 60 x - 2 x^{2}, A^{'} (x) = 60 - 4 x$
The available fencing fixes the third side once x is chosen.
3
Find the dimensions at the stationary input.
$60 - 4 x = 0 ⟹ x = 15, y = 60 - 2 (15) = 30$
The candidate gives positive side lengths within the feasible domain.
4
Verify and report the maximum area.
$A^{''} (x) = - 4 < 0, A (15) = 15 (30) = 450 m^{2}$
The area is concave down everywhere on the domain, so dimensions 15 m by 30 m give its absolute maximum.

Answer

$450 m^{2}$ , using dimensions $15$ m by $30$ m.

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

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

Area $36$ gives the second side $\frac{36}{x}$ for $x > 0$ . Perimeter is twice the sum of the two sides: $P (x) = 2 (x + \frac{36}{x}) = 2 x + \frac{72}{x}$ .

Keep practising

6 worksheet questions · 24 marks

Open worksheet 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 $x + \frac{100}{x}$ .
Its derivative is $1 - \frac{100}{x^{2}}$ , 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 $\pm \sqrt{\frac{3}{2}}$ .
The smaller squared distance is $\frac{7}{4}$ ; take its square root.
$\frac{\sqrt{7}}{2}$ , at $x = \pm \sqrt{\frac{3}{2}}$ .
A closed cylinder has total surface area $24 π$ square centimetres. Find its largest volume.
5 marks
The constraint gives $h = \frac{12}{r} - r$ , 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.

Checking saved progress…

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 ↗

In this lesson

Understand the idea.

Model before differentiating

Return to the situation

See the method in action.

A rectangle with fixed perimeter

Use the perimeter constraint.

Express and differentiate the area.

Find the feasible stationary input.

Justify the maximum and evaluate it.

A pen against a wall

Use only the three fenced sides.

Form and differentiate the area.

Find the dimensions at the stationary input.

Verify and report the maximum area.

Try it yourself.

Check your understanding

Keep practising