Calculate
(i)     the area of the base of the wastepaper bin;
(ii)    the height, $h$ , of Nadia’s wastepaper bin;
(iii)   the total external surface area of the wastepaper bin.

State whether Nadia’s design is practical. Give a reason.

Write down an equation in $h$ and $r$ , using the given volume of the bin.

Show that the total external surface area, $A$ , of the bin is $A = \pi {r^2} + \frac{{16000}}{r}$ .

Write down $\frac{{{\text{d}}A}}{{{\text{d}}r}}$.

(i)     Find the value of $r$ that minimizes the total external surface area of the wastepaper bin.
(ii)    Calculate the value of $h$ corresponding to this value of $r$ .

Determine whether Merryn’s design is an improvement upon Nadia’s. Give a reason.

(i)     ${\text{Area}} = \pi {(5)^2}$     (M1)
$= 78.5{\text{ (c}}{{\text{m}}^2}{\text{)}}$ ($78.5398 \ldots$)     (A1)(G2)

Note: Accept $25\pi$ .

(ii)    $8000 = 78.5398 \ldots  \times h$     (M1)
$h = 102{\text{ (cm)}}$ ($101.859 \ldots$)     (A1)(ft)(G2)

Note: Follow through from their answer to part (a)(i).

(iii)   ${\text{Area}} = \pi {(5)^2} + 2\pi (5)(101.859 \ldots )$     (M1)(M1)

Note: Award (M1) for their substitution in curved surface area formula, (M1) for addition of their two areas.

$= 3280{\text{ (c}}{{\text{m}}^2}{\text{)}}$ ($3278.53 \ldots$)     (A1)(ft)(G2)

Note: Follow through from their answers to parts (a)(i) and (ii).

No, it is too tall/narrow.     (A1)(ft)(R1)

Note: Follow through from their value for $h$.

$8000 = \pi {r^2}h$     (A1)

$A = \pi {r^2} + 2\pi r\left( {\frac{{8000}}{{\pi {r^2}}}} \right)$     (A1)(M1)

Note: Award (A1) for correct rearrangement of their part (c), (M1) for substitution of their rearrangement into area formula.

$= \pi {r^2} + \frac{{16000}}{r}$     (AG)

$\frac{{{\text{d}}A}}{{{\text{d}}r}} = 2\pi r - 16000{r^{ - 2}}$     (A1)(A1)(A1)

Note: Award (A1) for $2\pi r$ , (A1) for $- 16000$ (A1) for ${r^{ - 2}}$ . If an extra term is present award at most (A1)(A1)(A0).

(i)     $\frac{{{\text{d}}A}}{{{\text{d}}r}} = 0$     (M1)
$2\pi {r^3} - 16000 = 0$    (M1)
$r = 13.7{\text{ cm}}$ ($13.6556 \ldots$)     (A1)(ft)

Note: Follow through from their part (e).

(ii)    $h = \frac{{8000}}{{\pi {{(13.65 \ldots )}^2}}}$     (M1)
$= 13.7{\text{ cm}}$ ($13.6556 \ldots$)     (A1)(ft)

Note: Accept $13.6$ if $13.7$ used.

Yes or No, accompanied by a consistent and sensible reason.     (A1)(R1)

Note: Award (A0)(R0) if no reason is given.