SACE Board SACE 2021-style practice: Calculator-assumed. A closed cylindrical can has volume 500 cm^3. Its surface area is S = 2π r^2 + 1000r, 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.

Differentiate S = 2π r^2 + 1000 r^-1: $(dS)/(dr) = 4π r - (1000)/(r^2).$ Set dSdr = 0: 4π r = 1000r^2, so 4π r^3 = 1000 and r^3 = 10004π = 250π. Thus r = [3]250π ≈ 4.3 cm. Confirm a minimum: d^2Sdr^2 = 4π + 2000r^3 > 0 for all r > 0, so it is a minimum. Minimum area: S ≈ 2π(4.3)^2 + 10004.3 ≈ 116.2 + 232.6 ≈ 348.7 cm^2. Marks: one for dSdr, one for solving to find r, one for the second-derivative justification, and two for the numerical r and S to one decimal place.

§-Syllabus dot point

SAMath MethodsSyllabus dot point

How do we use calculus to find the maximum or minimum value of a real-world quantity?

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

A step-by-step method for solving optimisation problems: build the function, use a constraint to reduce to one variable, differentiate, solve for stationary points, and justify the maximum or minimum.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section

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

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.

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.

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

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 practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20224 marksCalculator-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 20215 marksCalculator-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.