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 4.76 min answerUpdated 2026-05-29

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What this dot point is asking
The optimisation method
Worked optimisation
Checking the domain endpoints
Rate-based optimisation

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

Maximum volume of a box from a square sheet

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. Find $x$ that maximises the volume.

Write the volume. The base is $(12-2x)$ by $(12-2x)$ and the height is $x$ , so

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

Expand and differentiate.

V(x) = x(144-48x+4x^{2}) = 144x-48x^{2}+4x^{3},

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

Solve $V'(x)=0$ . The candidates are $x=2$ and $x=6$ , but $x=6$ is excluded by the domain.

Justify $x=2$ . $V''(x)=-96+24x$ , so $V''(2)=-48<0$ , confirming a maximum.

Find the volume. $V(2)=2(12-4)^{2}=2\times 64=128$ cm $^{3}$ .

The maximum volume is $128$ cm $^{3}$ when $x=2$ cm. Markers reward the correct volume function, the restricted domain, the justification and the interpreted answer.

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.