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.