2008 H2 Mathematics Paper 1 Question 7

Differentiation II: Maxima, Minima, Rates of Change

Answers

{x = \frac{120}{4+5\pi} = 6.09}

{y = \frac{60+60\pi}{4+5\pi} = 12.6}

\begin{gather*} 3(x+2y) + 9\Bigg( \pi \left(\frac{x}{2}\right) \Bigg) = 180 \\ 3x + 6y + \frac{9}{2}\pi x = 180 \\ 6y = 180-3x-\frac{9}{2}\pi x \\ y = 30 - \frac{1}{2}x - \frac{3}{4}\pi x \end{gather*}

Let

{A}

denote the area of the flower bed

\begin{align*} A &= xy + \frac{1}{2} \pi \left(\frac{x}{2}\right)^2 \\ &= x \left( 30 - \frac{1}{2}x - \frac{3}{4}\pi x \right) + \frac{1}{8} \pi x^2 \\ &= 30x - \frac{1}{2}x^2 - \frac{5}{8} \pi x^2 \end{align*}

\frac{\mathrm{d}A}{\mathrm{d}x} = 30 - x - \frac{5}{4} \pi x

At maximum

A, \frac{\mathrm{d}A}{\mathrm{d}x} = 0

30 - x - \frac{5}{4} \pi x = 0

\begin{align*} x &= 30 \div \left( 1 + \frac{5}{4} \pi \right) \\ &= \frac{120}{4+5\pi} \; \blacksquare \end{align*}

\begin{align*} y &= 30 - \frac{1}{2} \left( \frac{120}{4+5\pi} \right) - \frac{3}{4}\pi \left( \frac{120}{4+5\pi} \right) \\ &= 30 - \frac{60}{4+5\pi} - \frac{90\pi}{4+5\pi} \\ &= \frac{120+150\pi-60-90\pi}{4+5\pi} \\ &= \frac{60+60\pi}{4+5\pi} \; \blacksquare \end{align*}

\begin{align*} \frac{\mathrm{d}^{2}A}{\mathrm{d}x^{2}} &= -1 - \frac{5}{4}\pi \\ &< 0 \end{align*}

Hence, when

{x=\frac{120}{4+5\pi}}

and

{y=\frac{60+60\pi}{4+5\pi}}

, the area of the flower-bed is a maximum

{\blacksquare}