2020 H2 Mathematics Paper 2 Question 4

Differentiation II: Maxima, Minima, Rates of Change

Answers

{H^2 = 225 - 15a}

Maximum volume

{= 144 \sqrt{5} \textrm{ cm}^3}

{a=15}

The net forms a square with sides

{15\sqrt{2} \textrm{ cm}}

\begin{gather*} a + 2h = 30 \\ h = \frac{30-a}{2} \end{gather*}

\begin{align*} H^2 &= h^2 - \frac{a}{2}^2 \\ &= \frac{(30-a)^2 - a^2 }{4} \\ &= 225 - 15a \; \blacksquare \end{align*}

Let

{V}

denote the volume of the pyramid

\begin{align*} V &= \frac{1}{3} a^2 H \\ V^2 &= \frac{1}{9} a^4 H^2 \\ &= \frac{1}{9} a^4 (225-15a) \\ &= \frac{1}{9} (225a^4 - 15a^5) \end{align*}

Differentiating w.r.t.

{a,}

2V \frac{\mathrm{d}V}{\mathrm{d}a} = \frac{1}{9} (900a^3 - 75a^4)

At maximum

{V, \displaystyle \frac{\mathrm{d}V}{\mathrm{d}a} = 0}

\begin{gather*} 900a^3 - 75a^4 = 0 \\ a = 12 \; \blacksquare \end{gather*}

\begin{align*} & \textrm{Maximum volume} \\ &= \frac{1}{3} (12)^2 \sqrt{225 - 15(12)} \\ &= 144\sqrt{5} \textrm{ cm}^3 \; \blacksquare \end{align*}

(iiia)

Let

{A}

denote the total surface area of the four triangular faces of the pyramid

\begin{align*} A &= 4 \cdot \frac{1}{2} a h \\ &= 2a \left( \frac{30-a}{2} \right) \\ &= 30a - a^2 \end{align*}

\frac{\mathrm{d}A}{\mathrm{d}a} = 30 - 2a

At maximum

{A, \displaystyle \frac{\mathrm{d}A}{\mathrm{d}a} = 0}

\begin{gather*} 30-2a = 0 \\ a = 15 \; \blacksquare \end{gather*}

(iiib)

The net forms a square with sides

{15\sqrt{2} \textrm{ cm}}

{\blacksquare}