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Module 3 · L09 of 15 ~35 min +95 XP available

Optimisation

Calculus lets us find the best outcome: maximum profit, minimum cost, or greatest efficiency. Like a business deciding the exact price that maximises revenue, optimisation finds the sweet spot where a function reaches its peak or trough.

Today's hook — A farmer has 200 m of fencing and wants to enclose the largest possible rectangular paddock using a river as one side. What dimensions give the maximum area? Is it always best to make a square?

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Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Build custom Build your own from any module question

Recall — your gut answer first

+5 XP warm-up

A farmer has 200 m of fencing and wants to enclose the largest possible rectangular paddock using a river as one side. What dimensions do you think give the maximum area? Is it always best to make a square?

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The two moves

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There are only two moves in every optimisation problem. Lock them in and the rest is just algebra.

Move 1 — Model then reduce: define variables, write the objective function (what you want to maximise or minimise), then use a constraint to express it in one variable.

Move 2 — Differentiate and verify: set the derivative to zero, solve for the critical value, and confirm it is the correct type (maximum or minimum) using the second derivative or a sign table.

$$\frac{dA}{dx} = 0 \implies \text{stationary point}$$

Always check endpoints

The maximum or minimum may occur at the boundary of the valid domain, not at an interior stationary point. Check all candidates.

Draw a diagram

Visualising the problem helps you identify variables and constraints. Always sketch before writing equations.

Verify the nature

Use the second derivative test: if $f''(x) < 0$ at the stationary point it is a maximum; if $f''(x) > 0$ it is a minimum.

What you will master

Know

Key facts

The five-step optimisation process
The second derivative test for maximum/minimum
That a constraint reduces the number of variables

Understand

Concepts

Why setting the derivative to zero finds stationary points
How a physical constraint links two variables
Why verification is essential, not optional

Can do

Skills

Translate a word problem into an objective function
Use calculus to find maximum and minimum values in practical contexts
Verify that a solution is a maximum or minimum

Key terms

OptimisationThe process of finding the maximum or minimum value of a function subject to given conditions.

ConstraintA condition that limits the possible values of variables, used to express one variable in terms of another.

Objective functionThe function that needs to be maximised or minimised.

Stationary pointA point where the derivative equals zero; may be a maximum, minimum, or inflection.

Second derivative testIf $f''(x) < 0$ at a stationary point, it is a local maximum; if $f''(x) > 0$, a local minimum.

DomainThe set of valid input values; physical contexts restrict variables to positive values.

The optimisation process

core concept

Optimisation problems involve finding the best value of some quantity. The systematic process is:

Define variables — assign letters to the unknown quantities
Write the objective function — the expression to be maximised or minimised
Apply the constraint — use any given condition to express the objective function in one variable
Differentiate and set to zero — find stationary points
Verify and conclude — confirm the nature of the stationary point and state the answer in context

Common contexts. Optimisation appears in area maximisation (fencing problems), cost minimisation (material problems), profit maximisation (economics), and volume problems (packaging design). The method is identical in each case — only the objective function changes.

Five steps: define → write objective function → apply constraint → differentiate → verify; Constraint: use it to eliminate one variable so the objective function has only one variable

Pause — copy the five-step optimisation process (define → objective function → apply constraint to reduce to one variable → differentiate → verify) into your book.

Quick check: True or false — to find the maximum area of a rectangle with fixed perimeter, you only need to differentiate; no verification step is required.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · MAXIMUM AREA WITH FIXED PERIMETER

Find the maximum area of a rectangle with perimeter 40 m.

Let width $= w$ and length $= l$. Perimeter: $2w + 2l = 40$, so $l = 20 - w$.

Define variables and use the perimeter constraint to express $l$ in terms of $w$.

PROBLEM 2 · FENCING WITH A RIVER BOUNDARY

A farmer has 200 m of fencing for three sides of a rectangular paddock (one side is a river). Find the maximum area.

Let the sides perpendicular to the river be $x$ each, and the side parallel be $y$. Fencing: $2x + y = 200$, so $y = 200 - 2x$.

Set up variables and constraint from the available fencing. Only three sides need fencing because the river forms the fourth side.

PROBLEM 3 · MINIMUM SURFACE AREA OF A CYLINDER

Find the minimum surface area of a cylinder with volume $1000\pi$ cm$^3$.

Volume $V = \pi r^2 h = 1000\pi$, so $h = \dfrac{1000}{r^2}$

Use the volume constraint to express height in terms of radius.

Quick check: A rectangle has perimeter 60 m. What dimensions give the maximum area?

Common errors · the 3 traps that cost marks

Trap 01

Forgetting to verify the nature of the stationary point

In optimisation, you must confirm that your stationary point is actually a maximum or minimum using the second derivative or a sign table. The problem asks for a specific type and a bare critical value is not a complete answer.

Trap 02

Ignoring domain restrictions

Physical quantities like length and area must be positive. Always state the valid domain for your variable and check that your answer lies within it. A negative width is not a valid answer even if the algebra produces it.

Trap 03

Using the wrong constraint

Students sometimes use the wrong equation to eliminate a variable. Make sure your constraint exactly matches the physical situation described in the problem — read it carefully before writing equations.

Odd one out: Three of the following are steps in the optimisation process. Which one is NOT?

Work mode · how are you completing this lesson?

Quick-fire practice · 5 problems

Find two positive numbers whose sum is 20 and whose product is maximum.

A box with a square base and open top has volume 32 cm$^3$. Find the dimensions that minimise surface area.

Find the maximum value of $f(x) = x(10 - x)$ for $0 \le x \le 10$.

A rectangular page must contain 24 cm$^2$ of print with 2 cm margins on all sides. Find the minimum page area.

The profit function is $P(x) = -2x^2 + 40x - 100$. Find the number of units that maximises profit.

Fill the blanks: drag each token into the matching blank.

constraint objective zero second

The ___ function is what you want to maximise or minimise. Use the ___ to reduce it to one variable. Set the derivative to ___ and find the stationary point. Use the ___ derivative to verify the nature.

Think aloud: A profit function is $P(x) = -x^2 + 12x - 20$. What value of $x$ maximises profit? Explain your steps.

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Revisit your thinking

Earlier you were asked: What dimensions of a three-sided paddock give the maximum area? With $2x + y = 200$, the optimal value is $x = 50$ m and $y = 100$ m, giving area $= 5000$ m$^2$. This is not a square — when one side is free the optimal ratio changes. Optimisation with calculus gives the exact answer without guessing.

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Multiple choice

+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 4

Maximum area with fixed perimeter

A rectangular garden is to be fenced on all four sides with 60 m of fencing. Find the dimensions that give the maximum area, and state this maximum area. 4 MARKS

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View comprehensive answer

Let width $= w$, length $= l$. $2w + 2l = 60$, so $l = 30 - w$ [0.5]

$A = w(30 - w) = 30w - w^2$ [0.5]

$\frac{dA}{dw} = 30 - 2w = 0$, so $w = 15$ [1]

$l = 15$, so maximum area $= 225$ m$^2$ [1]

Verify: $\frac{d^2A}{dw^2} = -2 < 0$, confirming maximum [1]

ApplyBand 4

Maximise profit

The cost of producing $x$ items is $C(x) = x^2 + 10x + 50$. The selling price per item is $\$30$. Find the number of items that maximises profit. 3 MARKS

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View comprehensive answer

Revenue $R(x) = 30x$ [0.5]

Profit $P(x) = R(x) - C(x) = 30x - x^2 - 10x - 50 = -x^2 + 20x - 50$ [0.5]

$P'(x) = -2x + 20 = 0$, so $x = 10$ [1]

$P''(x) = -2 < 0$, confirming maximum profit at 10 items [1]

AnalyseBand 5

Optimisation with a constraint

A closed cylindrical can must have volume $500\pi$ cm$^3$. Show that the surface area is $S = 2\pi r^2 + \frac{1000\pi}{r}$ and find the value of $r$ that minimises $S$. 5 MARKS

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View comprehensive answer

$V = \pi r^2 h = 500\pi$, so $h = \frac{500}{r^2}$ [1]

$S = 2\pi r^2 + 2\pi rh = 2\pi r^2 + 2\pi r \cdot \frac{500}{r^2} = 2\pi r^2 + \frac{1000\pi}{r}$ [1.5]

$\frac{dS}{dr} = 4\pi r - \frac{1000\pi}{r^2} = 0$ [1]

$4\pi r^3 = 1000\pi$, so $r^3 = 250$, $r = \sqrt[3]{250} = 5\sqrt[3]{2}$ cm [1.5]

📖 Comprehensive answers (click to reveal)

Drill 1: Let $x$ and $20-x$ be the numbers. Product $P = x(20-x)$, $P' = 20-2x = 0 \implies x = 10$. Maximum product $= 100$.

Drill 2: Let side of square base $= s$, height $= h$. Volume: $s^2 h = 32 \implies h = 32/s^2$. Surface area $= s^2 + 4sh = s^2 + 128/s$. $SA' = 2s - 128/s^2 = 0 \implies s^3 = 64 \implies s = 4$ cm, $h = 2$ cm.

Drill 3: $f'(x) = 10 - 2x = 0 \implies x = 5$. Maximum $= 5(5) = 25$.

Drill 4: Let print width $= x$, height $= y$. $xy = 24 \implies y = 24/x$. Page dimensions: $(x+4)(y+4)$. Expand and minimise: $A = (x+4)(24/x+4) = 24 + 4x + 96/x + 16$. $A' = 4 - 96/x^2 = 0 \implies x^2 = 24 \implies x = 2\sqrt{6}$. Minimum page area $= (2\sqrt{6}+4)^2 \approx 56.98$ cm$^2$.

Drill 5: $P'(x) = -4x + 40 = 0 \implies x = 10$ units.

Boss battle · The Optimiser

earn bronze · silver · gold

Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena

Science Jump · platform challenge

Climb platforms by answering optimisation questions. Lighter alternative to the boss.

Mark lesson as complete

Tick when you have finished the practice and review.

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Module overview · Maths Advanced · Checkpoint 1