Simple optimisation | Year 11

01 / LEARN

Understand the idea.

Use the constraint first

Optimisation begins by identifying the quantity to maximise or minimise. Name each variable and use the physical constraint to write the objective using one variable. A fixed perimeter constrains lengths, while the area is the separate quantity being optimised.

State the feasible domain before differentiating. Positive lengths, budgets and available materials restrict which algebraic solutions are meaningful. Solve the derivative equation only after forming the correct objective, then substitute back to recover all requested dimensions and the objective value.

Check boundaries and assumptions

A derivative zero supplies a candidate, not a complete maximum or minimum argument. Use derivative signs to justify its type. On a closed interval, compare it with every endpoint; on an open physical interval, consider what happens as dimensions approach the excluded boundaries.

Report the optimum with units and explain what it means in the model. If quantities must be whole numbers, nearby feasible integers may need comparison. Calculus answers depend on the stated assumptions, such as negligible fence thickness and no additional material losses.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example 1

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

Worked solution

1
Use the perimeter constraint.
$\begin{aligned} 2 x + 2 y & = 40 \\ y & = 20 - x, 0 < x < 20 \end{aligned}$
Two widths and two lengths use 40 metres; both dimensions must be positive.
2
Write area as a function of one side.
$A (x) = x y = x (20 - x) = 20 x - x^{2}$
Replacing the other side using the constraint makes area a function of one independent variable.
3
Find the stationary candidate.
$\begin{aligned} A^{'} (x) & = 20 - 2 x \\ A^{'} (x) = 0 & ⟹ x = 10 \end{aligned}$
A smooth interior optimum must have zero derivative, and 10 lies in the feasible domain.
4
Justify the global maximum.
$\begin{aligned} 0 < x < 10 : & A^{'} (x) > 0 \\ 10 < x < 20 : & A^{'} (x) < 0 \end{aligned}$
Area increases up to 10 and decreases afterwards throughout the domain; it tends to zero at the excluded boundaries.
5
Find the dimensions and greatest area.
$y = 20 - 10 = 10 m, A (10) = 10 (10) = 100 m^{2}$
The optimum is a 10 metre by 10 metre square with maximum area 100 square metres.

Answer

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

EXAMPLE 02

Worked example 2

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

Worked solution

1
Count the three fenced sides.
$2 x + y = 60, y = 60 - 2 x, 0 < x < 30$
The wall supplies the fourth side, leaving two perpendicular sides of length $x$ and one fenced side of length $y$ .
2
Express the area using the constraint.
$A (x) = x (60 - 2 x) = 60 x - 2 x^{2}$
Area is length times width; substituting the fence constraint leaves one variable.
3
Solve for the stationary width.
$\begin{aligned} A^{'} (x) & = 60 - 4 x \\ 60 - 4 x = 0 & ⟹ x = 15 \end{aligned}$
The feasible interior candidate occurs where area stops increasing and begins decreasing.
4
Verify it gives the greatest area.
$\begin{aligned} 0 < x < 15 : & A^{'} (x) > 0 \\ 15 < x < 30 : & A^{'} (x) < 0 \end{aligned}$
The derivative signs show an increase followed by a decrease across the full feasible domain; area tends to zero at either boundary.
5
Calculate the remaining length and area.
$y = 60 - 2 (15) = 30 m, A (15) = 15 (30) = 450 m^{2}$
The two perpendicular sides are 15 metres each and the remaining fenced side is 30 metres.

Answer

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

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

P=28 at x=5.

$P^{'} (x) = 10 - 2 x = 0$ gives $x = 5$ , which lies in $[0, 8]$ . Compare $P (0) = 3$ , $P (5) = 28$ and $P (8) = 19$ . The largest value is 28 at $x = 5$ .

Keep practising

6 worksheet questions · 26 marks

Open worksheet 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 $x = 2$ cm.

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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 ↗

In this lesson

Understand the idea.

Use the constraint first

Check boundaries and assumptions

See the method in action.

Worked example 1

Use the perimeter constraint.

Write area as a function of one side.

Find the stationary candidate.

Justify the global maximum.

Find the dimensions and greatest area.

Worked example 2

Count the three fenced sides.

Express the area using the constraint.

Solve for the stationary width.

Verify it gives the greatest area.

Calculate the remaining length and area.

Try it yourself.

Check your understanding

Keep practising