2024 H2 Mathematics Paper 2 Question 2

Differentiation II: Maxima, Minima, Rates of Change

Answers

$23.43 m^{2} .$

Since the total perimeter is $20 m,$

\begin{aligned} 3 a + 2 b & = 20 \\ 2 b & = 20 - 3 a \\ b & = \frac{20 - 3 a}{2} \end{aligned}

\begin{aligned} A & = a b + \frac{1}{2} a^{2} \sin 60^{\circ} \\ = a b + \frac{\sqrt{3} a^{2}}{4} \\ = a (\frac{20 - 3 a}{2}) + \frac{\sqrt{3} a^{2}}{4} \\ = \frac{20 a - 3 a^{2}}{2} + \frac{\sqrt{3} a^{2}}{4} \\ \frac{d A}{d a} & = 10 - 3 a + \frac{\sqrt{3} a}{2} \end{aligned}

At maximum $A,$

\begin{matrix} \frac{d A}{d a} = 0 \\ 10 - 3 a + \frac{\sqrt{3} a}{2} = 0 \\ \frac{\sqrt{3} a}{2} - 3 a = - 10 \\ a (\frac{\sqrt{3}}{2} - 3) = - 10 \\ a = 4.6861 \end{matrix}

\begin{aligned} A & = \frac{20 (4.6861) - 3 ({4.6861}^{2})}{2} + \frac{\sqrt{3} ({4.6861}^{2})}{4} \\ = 23.43 m^{2} (4 s.f.) ∎ \end{aligned}

\begin{aligned} \frac{d^{2} A}{d a^{2}} & = - 3 + \frac{\sqrt{3}}{2} \\ = - 2.1340 \\ < 0 \end{aligned}

Hence $A = 23.43$ is a maximum $∎$