Optimisation problems (SACE Stage 2 Mathematical Methods)

Syllabus dot point

“Optimisation problems use the first derivative to locate the maximum or minimum value of a quantity subject to a constraint.”

8 min readUpdated 2 Jun 2026

Quick answer

To optimise a quantity, write it as a function of one variable (using a constraint to eliminate any extra variables), differentiate, set the derivative to zero to find stationary points, and then justify whether each is a maximum or minimum using the second derivative or a sign test. Always check the endpoints of the allowed domain.

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What this dot point is asking
The standard method
Reducing to one variable
Justifying the nature of the stationary point
A volume problem with a chain of variables
Common errors
Why it matters
Exam-style questions

What this dot point is asking

An optimisation problem asks for the largest or smallest possible value of some quantity - maximum volume, minimum cost, shortest distance, greatest area. Calculus solves these because a smooth maximum or minimum occurs where the gradient is zero, i.e. at a stationary point where $f'(x)=0$ .

The standard method

Almost every optimisation problem follows the same five steps:

Define variables and write the quantity to optimise as a formula. This is the objective function.
Use the constraint to express the objective in terms of a single variable.
Differentiate and set the derivative equal to zero.
Solve for the stationary point(s).
Justify that you have the required maximum or minimum (second-derivative test or sign of $f'$ ), and state the domain so you can also check endpoints.

Where extrema occur

On a closed interval, the global maximum and minimum of a continuous function occur either at a stationary point ( $f'(x)=0$ ) or at an endpoint of the interval. Both must be checked.

Reducing to one variable

The step students find hardest is using the constraint. Suppose a rectangle has perimeter $40$ m, so $2x+2y=40$ , giving $y=20-x$ . The area $A=xy$ then becomes a function of $x$ alone:

A(x)=x(20-x)=20x-x^2.

Now there is only one variable to differentiate with respect to.

Maximum area of a rectangular paddock

Worked example

A farmer has $40$ m of fencing to enclose a rectangular paddock. What dimensions give the maximum area?

Let the sides be $x$ and $y$ . The constraint is the perimeter:

2x+2y=40 \quad\Rightarrow\quad y=20-x.

The objective is the area:

A(x)=x(20-x)=20x-x^2, \qquad 0<x<20.

Differentiate and set to zero:

A'(x)=20-2x=0 \quad\Rightarrow\quad x=10.

Confirm it is a maximum with the second derivative:

A''(x)=-2<0,

x=10

gives a maximum. Then

y=20-10=10

The maximum area is a $10\text{ m}\times 10\text{ m}$ square with area $A=100\ \text{m}^2$ .

Justifying the nature of the stationary point

You must show why your answer is a maximum (or minimum) - marks are lost for finding $x$ but not justifying it.

Second-derivative test: if $f''(x)<0$ at the stationary point it is a local maximum; if $f''(x)>0$ it is a local minimum.
Sign test: check the sign of $f'(x)$ just before and just after the stationary point. A change from $+$ to $-$ is a maximum; $-$ to $+$ is a minimum.

Second-derivative test

At a stationary point where $f'(a)=0$ :

f''(a)<0 \Rightarrow \text{local maximum}, \qquad f''(a)>0 \Rightarrow \text{local minimum}.

A volume problem with a chain of variables

Largest open box from a square sheet

A square sheet of card of side $12$ cm has equal squares of side $x$ cut from each corner. The sides are folded up to form an open box. Find the value of $x$ that maximises the volume of the box.

Step 1: Write the volume as a function of $x$ and state the domain

When a square of side $x$ is cut from each corner, the base of the box has side length $12-2x$ in both directions, and the height of the box is $x$ . The volume is therefore:

V(x)=x(12-2x)^2.

Since

x

must be positive and the base length

12-2x

must also be positive, the valid domain is

0<x<6

Step 2: Expand and differentiate

Expand the volume function so it is in a form that is straightforward to differentiate:

V(x)=x(144-48x+4x^2)=144x-48x^2+4x^3.

Differentiating term by term gives:

V'(x)=144-96x+12x^2.

Step 3: Solve $V'(x)=0$ to find the stationary points

Factor out $12$ to simplify the quadratic:

V'(x)=12(x^2-8x+12)=12(x-2)(x-6).

Setting $V'(x)=0$ gives $x=2$ or $x=6$ . Because the domain is $0<x<6$ , the value $x=6$ lies on the boundary (not inside the open interval) and produces a box with zero volume, so only $x=2$ is a valid candidate for the maximum.

Step 4: Confirm $x=2$ gives a maximum

Compute the second derivative to test the nature of the stationary point:

V''(x)=-96+24x.

Substituting $x=2$ : $V''(2)=-96+48=-48<0$ . Because the second derivative is negative, the curve is concave down at this point, confirming that $x=2$ gives a local maximum.

Step 5: Calculate the maximum volume

Substitute $x=2$ back into the original volume formula:

V(2)=2(12-4)^2=2\times 64=128\ \text{cm}^3.

Final answer: cutting squares of side $x=2$ cm from each corner gives the maximum volume of $128\ \text{cm}^3$ .

Common errors

Forgetting to use the constraint or to check the domain

Two slips dominate. First, students try to differentiate an objective that still has two variables in it - you must substitute the constraint first to get a single-variable function. Second, they find a stationary point but ignore the physical domain. A length cannot be negative and cannot exceed the size of the material, so always state the valid interval (e.g. $0<x<6$ ) and reject solutions outside it.

State the domain before differentiating

Write the allowed values of your variable (e.g. $0<x<20$ ) as soon as you form the objective function. This makes it automatic to discard impossible stationary points and reminds you to test the endpoints when the question asks for a global extremum on a closed interval.

Why it matters

Optimisation is the headline application of differentiation in Stage 2 and a reliable source of extended-response marks in both the Skills and Applications Tasks and the external exam. The same five-step structure - model, constrain, differentiate, solve, justify - recurs in your Mathematical Investigation, where you may optimise a quantity in a modelling context of your own choosing.

Exam-style questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

SACE 2022-style4 marks

Calculator-assumed. For

f(x) = x\,e^{-x}

with

x \ge 0

, the tangent to the graph at the point where

x = a

has

y

-intercept

Y(a) = a^2 e^{-a}

. Determine the value of

a

that maximises the

y

-intercept of the tangent, and justify that it is a maximum.

Show worked answer →

Optimise $Y(a) = a^2 e^{-a}$ over $a \ge 0$ using calculus.

Differentiate with the product rule. Let $u = a^2$ and $v = e^{-a}$ , so $u' = 2a$ and $v' = -e^{-a}$ :

Y'(a) = 2a\,e^{-a} + a^2(-e^{-a}) = e^{-a}(2a - a^2) = a\,e^{-a}(2 - a).

Set $Y'(a) = 0$ . Since $e^{-a} > 0$ always, the stationary points are $a = 0$ and $a = 2$ .

Justify: for $0 < a < 2$ , $(2 - a) > 0$ so $Y'(a) > 0$ (increasing); for $a > 2$ , $(2 - a) < 0$ so $Y'(a) < 0$ (decreasing). The gradient changes from positive to negative at $a = 2$ , confirming a local maximum.

So the $y$ -intercept is maximised at $a = 2$ . Marks: one for differentiating, one for $Y'(a) = 0$ , one for the stationary points, and one for the sign-test justification.

SACE 2021-style5 marks

Calculator-assumed. A closed cylindrical can has volume

500\text{ cm}^3

. Its surface area is

S = 2\pi r^2 + \dfrac{1000}{r}

, where

r

is the base radius in cm. Find the value of

r

that minimises the surface area, and the minimum surface area, giving answers to one decimal place.

Show worked answer →

Differentiate $S = 2\pi r^2 + 1000 r^{-1}$ :

\frac{dS}{dr} = 4\pi r - \frac{1000}{r^2}.

Set $\dfrac{dS}{dr} = 0$ : $4\pi r = \dfrac{1000}{r^2}$ , so $4\pi r^3 = 1000$ and $r^3 = \dfrac{1000}{4\pi} = \dfrac{250}{\pi}$ . Thus $r = \sqrt[3]{\dfrac{250}{\pi}} \approx 4.3\text{ cm}$ .

Confirm a minimum: $\dfrac{d^2S}{dr^2} = 4\pi + \dfrac{2000}{r^3} > 0$ for all $r > 0$ , so it is a minimum.

Minimum area: $S \approx 2\pi(4.3)^2 + \dfrac{1000}{4.3} \approx 116.2 + 232.6 \approx 348.7\text{ cm}^2$ .

Marks: one for $\dfrac{dS}{dr}$ , one for solving to find $r$ , one for the second-derivative justification, and two for the numerical $r$ and $S$ to one decimal place.

Practise this

Sources & how we know this

Stage 2 Mathematical Methods subject outline (2025 onwards) — SACE Board
Stage 2 Mathematical Methods: external assessment — SACE Board