QLDMath MethodsSyllabus dot point

Topic 3: Introduction to differential calculus

Use the derivative to find stationary points of a polynomial function and classify them, and apply differentiation to simple optimisation problems

Q: Year 11 SAC HSC: A farmer has $40$ m of fence to enclose a rectangular paddock against a long straight river (no fence needed along the river). Find the dimensions that maximise the enclosed area.

Let the side perpendicular to the river be $x$ and the side parallel be $y$. Fence uses $2x + y = 40$, so $y = 40 - 2x$. Area: $A(x) = x y = x(40 - 2x) = 40x - 2x^2$. Differentiate: $A'(x) = 40 - 4x$. Stationary point: $40 - 4x = 0$, so $x = 10$ m. Classify: $A'(9) = 4 > 0$, $A'(11) = -4 < 0$. Sign change positive-to-negative confirms a maximum. Dimensions: $x = 10$ m, $y = 40 - 20 = 20$ m. Maximum area: $A = 200$ m$^2$. Markers reward the explicit modelling step (defining variables, writing the constraint, building the area function), the sign test, and the final dimensions plus area.

A focused answer to the QCE Math Methods Unit 2 dot point on stationary points. Locates stationary points by solving $f'(x) = 0$, classifies them as maxima, minima or stationary points of inflection using the first-derivative sign test, and works the QCAA-style optimisation problem (maximising the area of a fenced rectangle).

Generated by Claude OpusReviewed by Better Tuition Academy6 min answerUpdated 2026-05-19

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What this dot point is asking

QCAA wants you to locate stationary points of polynomial functions (where $f'(x) = 0$ ), classify them as local maxima, local minima or stationary points of inflection, and apply the technique to simple worded optimisation problems.

Stationary points

A stationary point is a point on the curve where $f'(x) = 0$ , that is, the tangent line is horizontal. Stationary points come in three types:

Local maximum: $f'$ changes from positive (rising) to negative (falling) as $x$ passes through the point.
Local minimum: $f'$ changes from negative (falling) to positive (rising).
Stationary point of inflection: $f'$ does not change sign (positive on both sides, or negative on both sides).

Finding stationary points

Procedure:

Differentiate the function to get $f'(x)$ .
Solve $f'(x) = 0$ for $x$ to find the $x$ -coordinates of stationary points.
Substitute each $x$ -value back into $f(x)$ to find the $y$ -coordinate.

Classifying with the first-derivative sign test

Pick a point just to the left and just to the right of the stationary point, and compute the sign of $f'$ at each.

Sign of $f'$ before	Sign of $f'$ after	Type
+	-	Local maximum
-	+	Local minimum
+	+	Stationary point of inflection (rising)
-	-	Stationary point of inflection (falling)

Optimisation procedure

For a worded problem asking for a maximum or minimum:

Identify the quantity to optimise (area, volume, cost).
Express it as a function of a single variable (use the constraint to eliminate other variables).
Differentiate; set $f'(x) = 0$ ; solve for $x$ .
Verify with a sign test (or check endpoints of the feasible region).
State the answer in words with units.

Worked example

Find the stationary points of $f(x) = x^3 - 3x^2 + 4$ and classify them.

$f'(x) = 3x^2 - 6x = 3x(x - 2)$ .

Stationary points where $f'(x) = 0$ : $x = 0$ and $x = 2$ .

$y$ -coordinates: $f(0) = 4$ , $f(2) = 8 - 12 + 4 = 0$ .

Sign test for $x = 0$ : $f'(-1) = 3 + 6 = 9 > 0$ . $f'(1) = 3 - 6 = -3 < 0$ . Sign change $+$ to $-$ , so local max at $(0, 4)$ .

Sign test for $x = 2$ : $f'(1) = -3 < 0$ . $f'(3) = 27 - 18 = 9 > 0$ . Sign change $-$ to $+$ , so local min at $(2, 0)$ .

Common traps

Forgetting the $y$ -coordinate. A stationary point is a point, so report both $x$ and $y$ .

Skipping the sign test. Setting $f'(x) = 0$ finds candidate stationary points. Without a classification, you cannot tell which are maxima, minima, or points of inflection.

Treating local extrema as global. A local maximum is the largest value in a neighbourhood. The global maximum might be at an endpoint of the feasible region.

Forgetting the domain in optimisation problems. A rectangle with negative or zero side length is not physically meaningful. Enforce the practical domain (e.g. $0 < x < 20$ for a $40$ m fence).

In one sentence

Stationary points of $f$ are where $f'(x) = 0$ , found by setting the derivative to zero and solving for $x$ , then classified by the first-derivative sign test as local maxima (sign $+$ to $-$ ), local minima ( $-$ to $+$ ) or stationary points of inflection (no change), which lets you solve worded optimisation problems by setting up a single-variable function and locating its extremum.

Past exam questions, worked

Real questions from past QCAA papers on this dot point, with our answer explainer.

Year 11 SAC5 marksA farmer has $40$ m of fence to enclose a rectangular paddock against a long straight river (no fence needed along the river). Find the dimensions that maximise the enclosed area.

Show worked answer →

Let the side perpendicular to the river be $x$ and the side parallel be $y$ . Fence uses $2x + y = 40$ , so $y = 40 - 2x$ .

Area: $A(x) = x y = x(40 - 2x) = 40x - 2x^2$ .

Differentiate: $A'(x) = 40 - 4x$ .

Stationary point: $40 - 4x = 0$ , so $x = 10$ m.

Classify: $A'(9) = 4 > 0$ , $A'(11) = -4 < 0$ . Sign change positive-to-negative confirms a maximum.

Dimensions: $x = 10$ m, $y = 40 - 20 = 20$ m.

Maximum area: $A = 200$ m $^2$ .

Markers reward the explicit modelling step (defining variables, writing the constraint, building the area function), the sign test, and the final dimensions plus area.