Mathematical Methods — Question 19

QCAA 2025, Paper 1 · 5 marks

Q19 · 2025 · Technology-freeComplex unfamiliar5 marks

QUESTION 19 (5 marks)

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.

Diagram of the western and southern fences, old homestead gate and proposed new fence

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WORKED SOLUTION

QCAA guide · typeset solution

5 marks

ANSWER

8

km.

Worked solution

Method 1:

Let the new fence be

AB

and the old homestead gate be

H

AH=\frac1{\cos\theta},\qquad HB=\frac{\sqrt{27}}{\sin\theta}.

Hence

AB=\frac1{\cos\theta}+\frac{\sqrt{27}}{\sin\theta}.

Differentiating,

\frac{d(AB)}{d\theta}=\frac{\sin\theta}{(\cos\theta)^2}-\frac{\sqrt{27}\cos\theta}{(\sin\theta)^2}.

For minimum length, set the derivative equal to zero:

(\sin\theta)^3-\sqrt{27}(\cos\theta)^3=0,

(\tan\theta)^3=\sqrt{27}=(\sqrt3)^3,

\tan\theta=\sqrt3

and

\theta=60^\circ

. Then

AB=\frac1{\cos60^\circ}+\frac{\sqrt{27}}{\sin60^\circ}=8.

The shortest fence would be 8 km long.

Method 2:

Let

H=(1,\sqrt{27})

. With the horizontal and vertical extensions labelled

x

and

y

\tan\theta=\frac y1=\frac{\sqrt{27}}x,

y=\frac{\sqrt{27}}x

. Thus

OB=1+x,

OA=\sqrt{27}+y=\sqrt{27}+\frac{\sqrt{27}}x,

and

AB=\sqrt{(1+x)^2+\left(\sqrt{27}+\frac{\sqrt{27}}x\right)^2}.

Differentiating,

\frac{d(AB)}{dx}=\frac{(1+x)-\frac{\sqrt{27}}{x^2}\left(\sqrt{27}+\frac{\sqrt{27}}x\right)}{\sqrt{(1+x)^2+\left(\sqrt{27}+\frac{\sqrt{27}}x\right)^2}}.

For minimum length, set the derivative equal to zero. This gives

x^4+x^3-27x-27=0,

(x+1)(x^3-27)=0,

x=-1

(invalid) or

x=3

. Substitution gives

AB=8.

The shortest fence would be 8 km long.

correctly determines an expression for the total length of the new fence in terms of the angle

\theta

or a single variable

[1 mark]

determines the derivative of the length expression of the new fence

[1 mark]

determines the angle

\theta

or variable value corresponding to the minimum length

[1 mark]

determines the minimum length of the new fence

[1 mark]

shows logical organisation

[1 mark]

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Mathematical Methods — Question 19

QUESTION 19 (5 marks)

QCAA guide · typeset solution

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