2025 Paper 1, Q19
A farmer wishes to construct the shortest possible fence to enclose a triangular area in the corner of a paddock. There are two existing fences, the western fence and the southern fence.
The new fence must pass through the old homestead gate, as shown.
Determine the length of the shortest possible fence.
Verifying that the fence length is a minimum is not required.
Every possible fence through the gate, one after another
The gate is fixed, so there is exactly one fence for each angle. Swing the fence around the gate and the total length falls, touches 8 km, and climbs again.
θ sweeps 35° to 75° and back, pausing at the best angle
The gate splits the fence into two right triangles
$A$ on the western fence, $H$ the gate, $B$ on the southern fence. The fence we want is $AB=AH+HB$.
The fence is one straight line, so the angle at $H$ is the same $\theta$. Then $\cos\theta=\dfrac{1}{AH}$, so $AH=\dfrac{1}{\cos\theta}$.
The angle at $B$ is $\theta$ and the opposite side is $\sqrt{27}$. Then $\sin\theta=\dfrac{\sqrt{27}}{HB}$, so $HB=\dfrac{\sqrt{27}}{\sin\theta}$.
That single line is the only insight mark in the question, “an expression for the total length in terms of $\theta$”. Everything after it is Unit 3 differentiation.
Two right-angled triangles that share an angle
The gate splits the fence into $AH=\frac{1}{\cos\theta}$ and $HB=\frac{\sqrt{27}}{\sin\theta}$. Add them and you have a function of one variable. The length curve also has a minimum.
1 km east of the west fence, $\sqrt{27}$ km north of the south fence
$L(\theta)=\frac{1}{\cos\theta}+\frac{\sqrt{27}}{\sin\theta}$
What is the variable?
QCAA marking guide · 5 marks
Let the new fence be $AB$ and the gate be $H$. The two offsets are the adjacent and opposite sides of two right triangles:
$$AH=\frac{1}{\cos\theta},\quad HB=\frac{\sqrt{27}}{\sin\theta},\quad AB=\frac{1}{\cos\theta}+\frac{\sqrt{27}}{\sin\theta}$$
Marker: correctly determines an expression for the total length of the new fence in terms of the angle $\theta$. Equivalent forms accepted, e.g. $(\cos\theta)^{-1}+\sqrt{27}(\sin\theta)^{-1}$.
$$\frac{d(AB)}{d\theta}=\frac{\sin\theta}{\cos^2\theta}-\frac{\sqrt{27}\cos\theta}{\sin^2\theta}$$
Marker: determines the derivative of the length expression. Write the terms as $(\cos\theta)^{-1}$ and $(\sin\theta)^{-1}$ and it is two chain rules, no quotient rule needed. FT marks allowed.
$$\frac{\sin\theta}{\cos^2\theta}=\frac{\sqrt{27}\cos\theta}{\sin^2\theta}\ \Rightarrow\ \sin^3\theta=\sqrt{27}\cos^3\theta\ \Rightarrow\ \tan^3\theta=\sqrt{27}$$
$$\tan^3\theta=\left(\sqrt{3}\right)^3\ \Rightarrow\ \tan\theta=\sqrt{3}\ \Rightarrow\ \theta=60^\circ$$
Marker: determines the angle $\theta$ corresponding to the minimum length. The question says verification is not required, so no second-derivative test is needed here.
$$AB=\frac{1}{\cos 60^\circ}+\frac{\sqrt{27}}{\sin 60^\circ}=2+\sqrt{27}\times\frac{2}{\sqrt{3}}=2+6$$
The shortest fence would be 8 km long
Marker: determines the minimum length of the new fence. Exact values matter on a technology-free paper: $\cos 60^\circ=\frac12$ and $\sin 60^\circ=\frac{\sqrt3}{2}$ turn $\sqrt{27}$ into 6.
Name the points, say what $\theta$ is, show the two right triangles, and finish with a sentence in kilometres. A fifth of the marks is for a solution someone else can follow.
Marker: shows logical organisation, a clear flow of the solution using appropriate mathematical terminology, symbols, conventions and representations.
The 8 km fence, and a shortcut you should know
Take the stretch of southern fence west of the gate as the variable instead. Similar triangles give the western leg, Pythagoras gives the length, and the same 8 km falls out. QCAA marks either model.
For a point $a$ from one wall and $b$ from the other, the shortest line through it is $\left(a^{2/3}+b^{2/3}\right)^{3/2}$. Here $\left(1+3\right)^{3/2}=8$. That is why the paper chose $\sqrt{27}$.
At $\theta=60^\circ$ the triangle has legs of 4 km along the south and $4\sqrt3\approx6.93$ km along the west, enclosing about 13.9 km². Note what is not optimised: the shortest fence does not enclose the smallest or the largest area. Only the fence itself is being minimised.
Nothing in the question tells you to build a function. The diagram gives you two distances and an angle, and that angle is your hint. If you write both parts of the fence in terms of it, the geometry problem becomes an optimisation with one variable. The second hidden step is confidence with the arithmetic. You have to recognise $\tan^3\theta=\sqrt{27}$ as $\left(\sqrt3\right)^3$, and work with $\sqrt{27}$ by hand. On a calculator paper this would only be worth two marks, which is why the QCAA put it on Paper 1. I would practise working with surds like this until it feels comfortable.
Question wording and marking-guide steps are from the 2025 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2025, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.