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Optimisation problems: WACE Year 12 Mathematics Methods Unit 3

Syllabus dot point

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

8 min read

Quick answer

Express the quantity to optimise as a function of one variable using the constraint, differentiate and solve f′(x)=0f'(x)=0, justify whether each candidate is a maximum or minimum, check the domain endpoints, and interpret the answer in context.

Jump to a section
  1. What this dot point is asking
  2. The optimisation method
  3. Worked optimisation
  4. Checking the domain endpoints
  5. Rate-based optimisation
  6. Why a square or cube so often wins
  7. Exam-style questions

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 optimisation steps
  1. Define variables and write the quantity to optimise.
  2. Use the constraint to eliminate variables until the quantity is a function of one variable.
  3. State the domain that makes physical sense.
  4. Differentiate and solve f′(x)=0f'(x)=0 for the candidate points.
  5. Justify each candidate as a maximum or minimum (sign of f′f' or f′′f'').
  6. Check the domain endpoints, then interpret the result.

The most marked step is justification: a value of xx that solves f′(x)=0f'(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 1212 cm has equal squares of side xx cut from each corner; the sides are folded up to form an open box. After cutting, the base has side (12−2x)(12-2x) and the height is xx, so the volume is:

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

The domain 0<x<60 < 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−48x2+4x3V(x) = 144x-48x^{2}+4x^{3}, and differentiating:

V′(x)=144−96x+12x2=12(x2−8x+12)=12(x−2)(x−6).V'(x) = 144-96x+12x^{2} = 12(x^{2}-8x+12) = 12(x-2)(x-6).

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

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

Step 4: Justify the maximum and find the volume

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

V(2)=2(12−4)2=2×64=128 cm3.V(2)=2(12-4)^{2}=2\times 64=128 \text{ cm}^{3}.

Final answer: Maximum volume is 128128 cm3^{3} when x=2x=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=0x=0 and x=6x=6 both give V=0V=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=πr2hV=\pi r^{2}h to write the surface area as a function of rr alone.

Minimising cost with a constraint

Cheapest fencing for a fixed area

A rectangular paddock must enclose 800800 m2^{2}. Fencing costs the same per metre on all four sides. Find the dimensions that minimise the total perimeter.

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

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

Differentiate and solve.
P′(x)=2−1600x2=0  ⇒  x2=800  ⇒  x=800≈28.3.P'(x)=2-\frac{1600}{x^{2}}=0\;\Rightarrow\;x^{2}=800\;\Rightarrow\;x=\sqrt{800}\approx 28.3.

Since P′′(x)=3200x3>0P''(x)=\dfrac{3200}{x^{3}}>0, this is a minimum. Then y=80028.3≈28.3y=\dfrac{800}{28.3}\approx 28.3, so the cheapest paddock is a square of side about 28.328.3 m. Markers reward eliminating yy, 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.

Common errors
Not eliminating the second variable
You cannot differentiate until the quantity is a function of a single variable; use the constraint first.
Skipping justification
Solving f′(x)=0f'(x)=0 finds candidates only. State the sign of f′′f'' or use a sign table to confirm a maximum or minimum.
Ignoring the domain
A solution outside the physical domain (such as x=6x=6 above) must be discarded, and endpoints must be checked.
Answering with xx instead of the optimised quantity
Re-read the question: it often asks for the maximum volume or minimum cost, not the value of xx.

Exam-style questions

Questions in the style of SCSA 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.

WACE 2022-style8 marks
Calculator-assumed. A closed cylindrical can must hold 500500 cm3^{3}. (a) Show that the surface area is S(r)=2πr2+1000rS(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 cm2^{2}.
Show worked answer →

A constrained optimisation of surface area.

(a) Volume V=πr2h=500V=\pi r^{2}h=500 gives h=500πr2h=\dfrac{500}{\pi r^{2}}. Surface area S=2πr2+2πrh=2πr2+2πr⋅500πr2=2πr2+1000rS=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)=4πr−1000r2S'(r)=4\pi r-\dfrac{1000}{r^{2}}. Setting S′(r)=0S'(r)=0: 4πr=1000r24\pi r=\dfrac{1000}{r^{2}}, so r3=10004π=250π≈79.58r^{3}=\dfrac{1000}{4\pi}=\dfrac{250}{\pi}\approx 79.58, giving r≈4.30r\approx 4.30 cm. Since S′′(r)=4π+2000r3>0S''(r)=4\pi+\dfrac{2000}{r^{3}}>0, this is a minimum.

(c) S(4.30)=2π(4.30)2+10004.30≈116.2+232.6≈349S(4.30)=2\pi(4.30)^{2}+\dfrac{1000}{4.30}\approx 116.2+232.6\approx 349 cm2^{2}.

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

WACE 2024-style6 marks
Calculator-free. A rectangle is inscribed under the curve y=12−x2y=12-x^{2} with its base on the xx-axis and two upper corners on the curve, symmetric about the yy-axis. (a) Write the area as a function of xx, the half-width. (b) Find the xx that maximises the area.
Show worked answer →

An inscribed-rectangle optimisation.

(a) By symmetry the width is 2x2x and the height is y=12−x2y=12-x^{2}, so A(x)=2x(12−x2)=24x−2x3A(x)=2x(12-x^{2})=24x-2x^{3} for 0<x<120<x<\sqrt{12}.

(b) A′(x)=24−6x2A'(x)=24-6x^{2}. Setting A′(x)=0A'(x)=0: x2=4x^{2}=4, so x=2x=2 (taking the positive root). Since A′′(x)=−12x<0A''(x)=-12x<0 at x=2x=2, this is a maximum. The maximum area is A(2)=24(2)−2(8)=48−16=32A(2)=24(2)-2(8)=48-16=32.

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

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Sources & how we know this

Based on the official SCSA syllabus. Not yet individually reviewed by a person; always check SCSA and your school's materials for current rules.

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