2012 H2 Mathematics Paper 1 Question 10

Differentiation II: Maxima, Minima, Rates of Change

Answers

(i)

r=3k5π3{r=\sqrt[3]{\frac{3k}{5\pi}}}
h=3k5π3{h=\sqrt[3]{\frac{3k}{5\pi}}}

(ii)

r=3.04{r=3.04}
h=4.88{h = 4.88}

Full solutions

(i)

Volume=kπr2h+23πr3=k\begin{align*} \textrm{Volume} &= k \\ \pi r^2 h + \frac{2}{3}\pi r^3 &= k \end{align*}
h=k−23πr3πr2\begin{equation} h = \frac{k - \frac{2}{3}\pi r^3}{\pi r^2} \end{equation}
Let A{A} denote the external surface area
A=2πrh+2πr2+πr2=2πrh+3πr2\begin{equation}\begin{split} \qquad A &= 2 \pi r h + 2 \pi r^2 + \pi r^2 \\ &= 2 \pi r h + 3 \pi r^2 \end{split}\end{equation}
Substituting (1){(1)} into (2),{(2),}
A=2k−43πr3r+3πr2=2kr−43πr2+3πr2=2kr+53πr2\begin{align*} A &= \frac{2k-\frac{4}{3}\pi r^3}{r} + 3 \pi r^2 \\ &= \frac{2k}{r} - \frac{4}{3}\pi r^2 + 3 \pi r^2 \\ &= \frac{2k}{r} + \frac{5}{3} \pi r^2 \end{align*}
dAdx=−2kr2+103πr\frac{\mathrm{d}A}{\mathrm{d}x} = -\frac{2k}{r^2} + \frac{10}{3} \pi r
At minimum A,dAdx=0,{\displaystyle A, \frac{\mathrm{d}A}{\mathrm{d}x}=0,}
−2kr2+103πr=0103πr=2kr2r3=3k5πr=3k5π3  ■\begin{gather*} -\frac{2k}{r^2} + \frac{10}{3} \pi r = 0 \\ \frac{10}{3} \pi r = \frac{2k}{r^2} \\ r^3 = \frac{3k}{5\pi} \\ r = \sqrt[3]{\frac{3k}{5\pi}} \; \blacksquare \end{gather*}
h=k−23πr3πr2=k−23π3k5ππr2=k−25kπr2=3k5πr2=3k5π÷(3k5π3)2=(3k5π)13=3k5π3  ■\begin{align*} h &= \frac{k - \frac{2}{3}\pi r^3}{\pi r^2} \\ &= \frac{k - \frac{2}{3}\pi \frac{3k}{5\pi}}{\pi r^2} \\ &= \frac{k - \frac{2}{5}k }{\pi r^2} \\ &= \frac{3k}{5\pi r^2} \\ &= \frac{3k}{5\pi} \div \left( \sqrt[3]{\frac{3k}{5\pi}} \right)^2 \\ &= \left( \frac{3k}{5\pi} \right)^{\frac{1}{3}} \\ &= \sqrt[3]{\frac{3k}{5\pi}} \; \blacksquare \end{align*}
d2Adx2=4kr3+103π\frac{\mathrm{d}^{2}A}{\mathrm{d}x^{2}} = \frac{4k}{r^3} + \frac{10}{3}\pi
When x=3k5π3,{x = \sqrt[3]{\frac{3k}{5\pi}}, } d2Adx2>0{\frac{\mathrm{d}^{2}A}{\mathrm{d}x^{2}}>0} so the A{A} is a minimum ■{\blacksquare}

(ii)

A=2kr+53πr22(200)r+53πr2=180400r+53πr2−180=0\begin{gather*} A = \frac{2k}{r} + \frac{5}{3}\pi r^2 \\ \frac{2(200)}{r} + \frac{5}{3}\pi r^2 = 180 \\ \frac{400}{r} + \frac{5}{3}\pi r^2 - 180 = 0 \end{gather*}
Solving with a GC,
r=−6.7587 (NA)orr=3.0372 (5 sf)orr=3.7215 (5 sf)\begin{align*} r &= -6.7587 \textrm{ (NA)} \quad &\textrm{or} \\ r &= 3.0372 \textrm{ (5 sf)} \quad &\textrm{or} \\ r &= 3.7215 \textrm{ (5 sf)} \\ \end{align*}
Since r>0,{r>0,} there are two possible values of r  ■{r \; \blacksquare}
If r=3.7215,{r = 3.7215,}
h=200−23πr3πr2=2.1156\begin{align*} h &= \frac{200 - \frac{2}{3}\pi r^3}{\pi r^2} \\ &= 2.1156 \end{align*}
This is not valid as r<h{r < h}
If r=3.0372,{r = 3.0372,}
h=200−23πr3πr2=4.88 (3 sf)  ■r=3.04 (3 sf)  ■\begin{align*} h &= \frac{200 - \frac{2}{3}\pi r^3}{\pi r^2} \\ &= 4.88 \textrm{ (3 sf)} \; \blacksquare \\ r &= 3.04 \textrm{ (3 sf)} \; \blacksquare \end{align*}