2024 H2 Maths Paper 2 Q2 — Worked Solution

Solution

6 marks

Approach

With two side-lengths $a$ and $b$ related by a fixed perimeter, we express the area as a function of $a$ alone, then use calculus to maximise it.

Let $AE = a$ and $ED = b$ . Since $\text{ABDE}$ is a rectangle, $BD = AE = a$ and $AB = ED = b$ . Since triangle $\text{BCD}$ is equilateral, $BC = CD = BD = a$ .

Using the perimeter of the pentagon,

AB + BC + CD + DE + EA = 20

= b + a + a + b + a

= 3a + 2b

\Rightarrow b = \frac{20 - 3a}{2}\quad\text{...(1)}

The area is the rectangle $\text{ABDE}$ plus the equilateral triangle $\text{BCD}$ :

A = ab + \frac{\sqrt{3}}{4}a^{2}

Substituting (1),

A = a\left( \frac{20 - 3a}{2} \right) + \frac{\sqrt{3}}{4}a^{2}

= 10a - \frac{3}{2}a^{2} + \frac{\sqrt{3}}{4}a^{2}

= 10a + \left( \frac{\sqrt{3} - 6}{4} \right)a^{2}

Differentiating with respect to $a$ ,

\frac{dA}{da} = 10 + \left( \frac{\sqrt{3} - 6}{2} \right)a

Setting $\frac{dA}{da} = 0$ ,

10 = \left( \frac{6 - \sqrt{3}}{2} \right)a

\Rightarrow a = \frac{20}{6 - \sqrt{3}}

= \frac{20\left( 6 + \sqrt{3} \right)}{33}

\approx 4.686

Since $\frac{d^{2}A}{da^{2}} = \frac{\sqrt{3} - 6}{2} < 0$ (as $6 > \sqrt{3}$ ) for all $a$ , $A$ is a maximum at this value of $a$ .

From (1), $b \approx \frac{20 - 3(4.686)}{2} \approx 2.971 > 0$ , so this value of $a$ lies in the valid range $0 < a < \frac{20}{3}$ .

The maximum area is

A_{max} = ab + \frac{\sqrt{3}}{4}a^{2}

= \frac{100}{6 - \sqrt{3}}

= \frac{100\left( 6 + \sqrt{3} \right)}{33}

\approx 23.4305

To check your answer: substituting

a \approx 4.686

b \approx 2.971

directly into

A = ab + \frac{\sqrt{3}}{4}a^{2}

gives

\text{≈}13.923 + 9.508 = 23.431

, matching the exact-form value above ✓

$\therefore$ the maximum possible area of the flower bed is $\frac{100\left( 6 + \sqrt{3} \right)}{33} \approx 23.43\ \text{m}^{2}$ (4 s.f.).

2024 H2 Maths Paper 2, Question 2

Solution

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