What is not eliminating one variable in optimisation?

You cannot differentiate a function of two variables in Maths Advanced. Use the constraint to reduce to one.

NSWMaths AdvancedSyllabus dot point

How can the derivative be used to analyse curves and solve optimisation and rate problems?

Use the first and second derivatives to find stationary points, points of inflection, and to solve optimisation and related rates problems

A focused answer to the HSC Maths Advanced dot point on applications of differentiation. Stationary points, concavity and inflection, maxima and minima word problems, and related rates with worked examples.

Generated by Claude Opus 4.814 min answerUpdated 2026-06-20

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

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What this dot point is asking

NESA wants you to use derivatives to analyse functions and model real-world problems. The first derivative locates stationary points (where the tangent is horizontal); the second derivative tells you about concavity and identifies maxima, minima and inflection points. You then apply this to optimisation problems (largest area, least cost) and related rates (how two changing quantities are linked).

The answer

A stationary point of $f(x)$ occurs where $f'(x) = 0$ . To classify it, use one of two tests.

First derivative test. Check the sign of $f'(x)$ just before and just after the stationary point.

Sign change from positive to negative: local maximum.
Sign change from negative to positive: local minimum.
No sign change: horizontal point of inflection.

Second derivative test. Evaluate $f''(x)$ at the stationary point.

$f''(x) < 0$ : local maximum (curve is concave down).
$f''(x) > 0$ : local minimum (curve is concave up).
$f''(x) = 0$ : test is inconclusive, fall back on the first derivative test.

Concavity and points of inflection

A point of inflection is where the concavity changes. Find candidates by solving $f''(x) = 0$ , then confirm the concavity actually changes by checking the sign of $f''$ either side.

Sketching a curve, stage by stage

A curve sketch is a procedure, and markers want the features labelled, not a freehand squiggle. Work through it in order for $y = x^3 - 3x^2$ , a typical HSC cubic.

Stage 1, axes and intercepts. Find where the curve meets the axes. The $y$ -intercept is $f(0) = 0$ . For the $x$ -intercepts, solve $x^3 - 3x^2 = x^2(x - 3) = 0$ , giving $x = 0$ (a double root) and $x = 3$ . Mark these on the axes.

Stage 2, locate the stationary points. Differentiate and solve $f'(x) = 0$ . Here $f'(x) = 3x^2 - 6x = 3x(x - 2)$ , so the stationary points are at $x = 0$ and $x = 2$ . Their coordinates are $(0, 0)$ and $(2, -4)$ . Plot them with a short horizontal tangent to show the zero gradient.

Stage 3, classify and find the inflection. Use the second derivative $f''(x) = 6x - 6$ . At $x = 0$ , $f''(0) = -6 < 0$ , so $(0, 0)$ is a local maximum; at $x = 2$ , $f''(2) = 6 > 0$ , so $(2, -4)$ is a local minimum. Solving $f''(x) = 0$ gives $x = 1$ , and the concavity changes there, so $(1, -2)$ is a point of inflection.

Stage 4, join the features. Draw a smooth curve rising to the maximum at the origin, falling through the inflection at $(1, -2)$ to the minimum at $(2, -4)$ , then rising through the $x$ -intercept at $x = 3$ . Label every feature; that labelling is what earns the marks.

Tangents and normals

Many application questions ask for the equation of the tangent or the normal at a point. The tangent at $x = a$ has gradient $m = f'(a)$ and equation $y - f(a) = f'(a)(x - a)$ . The normal is perpendicular, so its gradient is $-\frac{1}{f'(a)}$ (provided $f'(a) \neq 0$ ). For $y = x^2$ at the point $P(1, 1)$ , $f'(x) = 2x$ so the tangent gradient is $2$ and the normal gradient is $-\tfrac{1}{2}$ ; the two lines meet at right angles at $P$ .

Optimisation word problems

Follow a standard recipe.

Draw a diagram and label variables.
Write the quantity to optimise (area, volume, cost) as a function.
Use the constraint to reduce to one variable.
Differentiate and set to zero to find critical values.
Confirm maximum or minimum using $f''$ or a sign table.
Check endpoints if the domain is restricted.
State the final answer with units.

Related rates

When two quantities $x$ and $y$ both change with time and are linked by an equation, differentiate the equation with respect to $t$ (implicit differentiation). Substitute the given rate and the instantaneous value to solve for the unknown rate.

How exam questions ask about applications of differentiation

"Find the stationary points and determine their nature." Solve $f'(x) = 0$ , then classify each with $f''$ or a sign table. "Nature" is the explicit instruction to classify.
"Find the coordinates of any points of inflection." Solve $f''(x) = 0$ and confirm the concavity changes. Give full coordinates, not just the $x$ -value.
"Sketch the curve, showing all important features." Mark intercepts, stationary points (with their nature) and inflections, then draw a smooth curve through them.
"Find the maximum / minimum value", or any "largest / least / cheapest" word problem. An optimisation question: reduce to one variable, differentiate, solve, justify, and state the answer with units.
"At what rate is ... changing?" with two linked quantities. A related rates question: differentiate the link with respect to time and substitute the instantaneous values.
"Find the equation of the tangent / normal at ...". Differentiate, evaluate the gradient at the point, and write the line equation; the normal gradient is the negative reciprocal.

Optimisation: least metal for a fixed volume

A closed cylindrical can must hold $V = 500$ cm $^3$ . Find the radius $r$ and height $h$ that minimise the surface area (the amount of metal used).

Write the quantity you want to minimise: A closed cylinder has two circular ends and a curved side, so its surface area is

S = \underbrace{2\pi r^2}_{\text{two ends}} + \underbrace{2\pi r h}_{\text{curved side}}.

This has two variables ( $r$ and $h$ ), so we cannot differentiate yet. First we need $S$ in terms of one variable.

Use the fixed volume to eliminate $h$ : The volume is fixed, which links $r$ and $h$ :

\pi r^2 h = 500 \quad\Rightarrow\quad h = \frac{500}{\pi r^2}.

Substituting that into $S$ removes $h$ :

S(r) = 2\pi r^2 + 2\pi r\cdot\frac{500}{\pi r^2} = 2\pi r^2 + \frac{1000}{r}.

Now $S$ depends on $r$ alone, so calculus applies.

Differentiate and solve $S'(r) = 0$ : A minimum can only occur where the gradient is zero:

S'(r) = 4\pi r - \frac{1000}{r^2} = 0 \quad\Rightarrow\quad r^3 = \frac{250}{\pi} \quad\Rightarrow\quad r = \sqrt[3]{\tfrac{250}{\pi}} \approx 4.30\text{ cm}.

Confirm it is a minimum, not a maximum: A stationary point could be either, so test the concavity with the second derivative:

S''(r) = 4\pi + \frac{2000}{r^3}.

This is positive for every $r > 0$ , so the curve is concave up, and the surface area is least here. (That makes sense: a tall thin can and a short wide can both waste metal, so the best sits between them.)

Find $h$ and state the answer:

h = \frac{500}{\pi r^2} \approx 8.60\text{ cm}.

The can uses the least metal when $r \approx 4.30$ cm and $h \approx 8.60$ cm.

Related rates: a sliding ladder

A $5$ m ladder leans against a vertical wall. Its base is pulled away from the wall at $0.2$ m/s. How fast is the top sliding down when the base is $3$ m from the wall?

Name the changing quantities and link them: Let $x$ be the base's distance from the wall and $y$ the height of the top, both changing with time $t$ . The ladder length never changes, so Pythagoras links them at every instant:

x^2 + y^2 = 5^2 = 25.

Differentiate the link with respect to time: Since $x$ and $y$ both depend on $t$ , differentiate implicitly with respect to $t$ :

2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \quad\Rightarrow\quad \frac{dy}{dt} = -\frac{x}{y}\cdot\frac{dx}{dt}.

This is the "related rates" equation: it ties the rate we know ( $\frac{dx}{dt}$ ) to the rate we want ( $\frac{dy}{dt}$ ).

Find the missing value at this instant: We have $x = 3$ and $\frac{dx}{dt} = 0.2$ , but we still need $y$ when $x = 3$ . From the link:

y = \sqrt{25 - 3^2} = \sqrt{16} = 4\text{ m}.

Substitute, then read the sign:

\frac{dy}{dt} = -\frac{3}{4}\times 0.2 = -0.15\text{ m/s}.

The negative sign means $y$ is decreasing: the top slides down the wall at $0.15$ m/s.

Common traps

Forgetting to classify the stationary point: Setting $f'(x) = 0$ only finds candidates. You must justify whether each is a maximum, minimum or horizontal inflection.
Assuming $f''(x) = 0$ guarantees an inflection: The concavity must actually change. For example, $f(x) = x^4$ has $f''(0) = 0$ but no inflection there.
Not eliminating one variable in optimisation: You cannot differentiate a function of two variables in Maths Advanced. Use the constraint to reduce to one.
Ignoring units and domain: A negative radius is meaningless. Always check the feasible domain and state units in the final answer.
Confusing $\frac{dy}{dx}$ with $\frac{dy}{dt}$ in related rates: When time is the underlying variable, every quantity must be differentiated with respect to $t$ .

Exam-style practice questions

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

2021 HSC Q144 marksA rectangular paddock is to be fenced along three sides (one side borders a river and needs no fence). If 200 m of fencing is available, find the maximum area that can be enclosed.

Show worked answer →

Let $x$ be the side perpendicular to the river and $y$ the side parallel to it.

Constraint: $2x + y = 200$ , so $y = 200 - 2x$ .

Area: $A(x) = x y = x(200 - 2x) = 200 x - 2 x^2$ .

Differentiate: $A'(x) = 200 - 4x$ . Set $A'(x) = 0$ : $x = 50$ .

Second derivative: $A''(x) = -4 < 0$ , so this is a maximum.

$y = 200 - 100 = 100$ . Maximum area $= 50 \times 100 = 5000$ m $^2$ .

Markers reward defining variables, writing the constraint, expressing area in one variable, and confirming a maximum using $A''$ or sign analysis.

2019 HSC Q133 marksWater is poured into a conical container at a constant rate of 4 cm

^3

/s. The cone has its vertex pointing down, with a base radius of 6 cm and a height of 12 cm. Find the rate at which the water level is rising when the water is 3 cm deep.

Show worked answer →

By similar triangles, $\frac{r}{h} = \frac{6}{12} = \frac{1}{2}$ , so $r = \frac{h}{2}$ .

Volume of water: $V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \left(\frac{h}{2}\right)^2 h = \frac{\pi h^3}{12}$ .

Differentiate with respect to $t$ : $\frac{dV}{dt} = \frac{\pi h^2}{4} \cdot \frac{dh}{dt}$ .

Substitute $\frac{dV}{dt} = 4$ and $h = 3$ : $4 = \frac{9 \pi}{4} \cdot \frac{dh}{dt}$ .

$\frac{dh}{dt} = \frac{16}{9 \pi} \approx 0.566$ cm/s.

Markers expect the similar-triangle relation, volume in one variable, implicit differentiation in $t$ , and a final answer with units.

What this dot point is asking

The answer

Concavity and points of inflection

Sketching a curve, stage by stage

Tangents and normals

Optimisation word problems

Related rates

How exam questions ask about applications of differentiation

Exam-style practice questions

Related dot points