SCSA WACE 2022-style practice: Calculator-assumed. A closed cylindrical can must hold 500 cm^3. (a) Show that the surface area is S(r)=2π r^2+1000r. (b) Find the radius that minimises the surface area. (c) Find the minimum surface area, to the nearest cm^2.

A constrained optimisation of surface area. (a) Volume V=π r^2h=500 gives h=500π r^2. Surface area S=2π r^2+2π r h=2π r^2+2π r·500π r^2=2π r^2+1000r. (b) S'(r)=4π r-1000r^2. Setting S'(r)=0: 4π r=1000r^2, so r^3=10004π=250π≈ 79.58, giving r≈ 4.30 cm. Since S''(r)=4π+2000r^3>0, this is a minimum. (c) S(4.30)=2π(4.30)^2+10004.30≈ 116.2+232.6≈ 349 cm^2. Markers reward eliminating h via the volume constraint, the derivative, the cube-root solution, the minimum justification, and the evaluated area.

§-Syllabus dot point

WAMath MethodsSyllabus dot point

How do we use differentiation to find the maximum or minimum value of a quantity in a real context?

Solve optimisation problems by modelling a quantity as a function of one variable and using the derivative to find and justify extreme values

WACE Year 12 Mathematics Methods Unit 3 optimisation: modelling a quantity with one variable using a constraint, solving f-prime equals zero, justifying the extremum, checking domain endpoints, with a worked SCSA-style problem.

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

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

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What this dot point is asking
The optimisation method
Worked optimisation
Checking the domain endpoints
Rate-based optimisation
Why a square or cube so often wins

What this dot point is asking

SCSA Unit 3 applies differentiation to optimisation: finding the largest or smallest possible value of a real quantity such as area, volume, cost or time. This dot point appears in extended-response form, most often in the calculator-assumed section, and marks are awarded for the full modelling process, not just the final number.

The optimisation method

The most marked step is justification: a value of $x$ that solves $f'(x)=0$ is only a candidate until you confirm it gives the required maximum or minimum.

Worked optimisation

Maximising the volume of an open box

Step 1: Write the volume function and state the domain

A square of side $12$ cm has equal squares of side $x$ cut from each corner; the sides are folded up to form an open box. After cutting, the base has side $(12-2x)$ and the height is $x$ , so the volume is:

V(x) = x(12-2x)^{2}, \qquad 0 < x < 6.

The domain $0 < x < 6$ comes from the physical requirement that both the height and the base side must be positive.

Step 2: Expand and differentiate

Expanding gives $V(x) = 144x-48x^{2}+4x^{3}$ , and differentiating:

V'(x) = 144-96x+12x^{2} = 12(x^{2}-8x+12) = 12(x-2)(x-6).

Step 3: Solve $V'(x)=0$ and discard invalid candidates

Setting $V'(x)=0$ gives $x=2$ and $x=6$ , but $x=6$ is excluded by the domain (the box would have zero base area). The only candidate is $x=2$ .

Step 4: Justify the maximum and find the volume

Since $V''(x)=-96+24x$ and $V''(2)=-48<0$ , the point $x=2$ is a local maximum. The maximum volume is:

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

Final answer: Maximum volume is $128$ cm $^{3}$ when $x=2$ cm.

Checking the domain endpoints

In a restricted-domain problem the optimum can occur at an endpoint rather than at a stationary point. After finding stationary values, evaluate the function at the endpoints of the physical domain and compare. In the box problem the endpoints $x=0$ and $x=6$ both give $V=0$ , so the interior stationary point is indeed the maximum, but this comparison must be made, not assumed.

Rate-based optimisation

Some optimisation problems minimise a rate or a cost expressed as a function. The method is identical: form the function, differentiate, solve and justify. For instance, minimising the surface area of a cylinder of fixed volume uses the volume constraint $V=\pi r^{2}h$ to write the surface area as a function of $r$ alone.

Minimising cost with a constraint

Cheapest fencing for a fixed area

A rectangular paddock must enclose $800$ m $^{2}$ . Fencing costs the same per metre on all four sides. Find the dimensions that minimise the total perimeter.

Let the sides be $x$ and $y$ with constraint $xy=800$ , so $y=\dfrac{800}{x}$ . The perimeter is

P=2x+2y=2x+\frac{1600}{x},\qquad x>0.

Differentiate and solve.

P'(x)=2-\frac{1600}{x^{2}}=0\;\Rightarrow\;x^{2}=800\;\Rightarrow\;x=\sqrt{800}\approx 28.3.

Since

P''(x)=\dfrac{3200}{x^{3}}>0

, this is a minimum. Then

y=\dfrac{800}{28.3}\approx 28.3

, so the cheapest paddock is a square of side about

28.3

m. Markers reward eliminating

y

, the derivative, the positive root, and the minimum justification.

Why a square or cube so often wins

Many SCSA optimisation answers turn out to be square or cube shaped. This is not a coincidence: for a fixed area the perimeter is least when the rectangle is a square, and for a fixed volume the surface area of a box is least when it is a cube. Recognising this gives a quick sanity check on your answer, though you must still show the calculus to earn the marks.

Exam-style practice questions

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

WACE 20228 marksCalculator-assumed. A closed cylindrical can must hold

500

^{3}

. (a) Show that the surface area is

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

. (b) Find the radius that minimises the surface area. (c) Find the minimum surface area, to the nearest cm

^{2}

Show worked answer →

A constrained optimisation of surface area.

(a) Volume $V=\pi r^{2}h=500$ gives $h=\dfrac{500}{\pi r^{2}}$ . Surface area $S=2\pi r^{2}+2\pi r h=2\pi r^{2}+2\pi r\cdot\dfrac{500}{\pi r^{2}}=2\pi r^{2}+\dfrac{1000}{r}$ .

(b) $S'(r)=4\pi r-\dfrac{1000}{r^{2}}$ . Setting $S'(r)=0$ : $4\pi r=\dfrac{1000}{r^{2}}$ , so $r^{3}=\dfrac{1000}{4\pi}=\dfrac{250}{\pi}\approx 79.58$ , giving $r\approx 4.30$ cm. Since $S''(r)=4\pi+\dfrac{2000}{r^{3}}>0$ , this is a minimum.

Markers reward eliminating $h$ via the volume constraint, the derivative, the cube-root solution, the minimum justification, and the evaluated area.

WACE 20246 marksCalculator-free. A rectangle is inscribed under the curve

y=12-x^{2}

with its base on the

x

-axis and two upper corners on the curve, symmetric about the

y

-axis. (a) Write the area as a function of

x

, the half-width. (b) Find the

x

that maximises the area.

Show worked answer →

An inscribed-rectangle optimisation.

(a) By symmetry the width is $2x$ and the height is $y=12-x^{2}$ , so $A(x)=2x(12-x^{2})=24x-2x^{3}$ for $0<x<\sqrt{12}$ .

(b) $A'(x)=24-6x^{2}$ . Setting $A'(x)=0$ : $x^{2}=4$ , so $x=2$ (taking the positive root). Since $A''(x)=-12x<0$ at $x=2$ , this is a maximum. The maximum area is $A(2)=24(2)-2(8)=48-16=32$ .

Markers reward the area model using symmetry, the derivative, $x=2$ , and the maximum justification.