Express the volume of the parcel in terms of $l$ and $w$.

Show that $l = \frac{{150}}{w}$.

The parcel is tied up using a length of string that fits exactly around the parcel, as shown in the following diagram.

Show that the length of string, $S$ cm, required to tie up the parcel can be written as

$$S = 40 + 4w + \frac{{300}}{w},{\text{ }}0 < w \leqslant 20.$$

The parcel is tied up using a length of string that fits exactly around the parcel, as shown in the following diagram.

Draw the graph of $S$ for $0 < w \leqslant 20$ and $0 < S \leqslant 500$, clearly showing the local minimum point. Use a scale of $2$ cm to represent $5$ units on the horizontal axis $w$ (cm), and a scale of $2$ cm to represent $100$ units on the vertical axis $S$ (cm).

The parcel is tied up using a length of string that fits exactly around the parcel, as shown in the following diagram.

Find $\frac{{{\text{d}}S}}{{{\text{d}}w}}$.

The parcel is tied up using a length of string that fits exactly around the parcel, as shown in the following diagram.

Find the value of $w$ for which $S$ is a minimum.

The parcel is tied up using a length of string that fits exactly around the parcel, as shown in the following diagram.

Write down the value, $l$, of the parcel for which the length of string is a minimum.

The parcel is tied up using a length of string that fits exactly around the parcel, as shown in the following diagram.

Find the minimum length of string required to tie up the parcel.

$20lw$   OR   $V = 20lw$     (A1)

[1 mark]

$3000 = 20lw$     (M1)

Note: Award (M1) for equating their answer to part (a) to $3000$.

$l = \frac{{3000}}{{20w}}$     (M1)

Note: Award (M1) for rearranging equation to make $l$ subject of the formula. The above equation must be seen to award (M1).

OR

$150 = lw$     (M1)

Note: Award (M1) for division by $20$ on both sides. The above equation must be seen to award (M1).

$l = \frac{{150}}{w}$     (AG)

[2 marks]

$S = 2l + 4w + 2(20)$     (M1)

Note: Award (M1) for setting up a correct expression for $S$.

$2\left( {\frac{{150}}{w}} \right) + 4w + 2(20)$     (M1)

Notes: Award (M1) for correct substitution into the expression for $S$. The above expression must be seen to award (M1).

$= 40 + 4w + \frac{{300}}{w}$     (AG)

[2 marks]

     (A1)(A1)(A1)(A1)

Note: Award (A1) for correct scales, window and labels on axes, (A1) for approximately correct shape, (A1) for minimum point in approximately correct position, (A1) for asymptotic behaviour at $w = 0$.

     Axes must be drawn with a ruler and labeled $w$ and $S$.

     For a smooth curve (with approximately correct shape) there should be one continuous thin line, no part of which is straight and no (one-to-many) mappings of $w$.

     The $S$-axis must be an asymptote. The curve must not touch the $S$-axis nor must the curve approach the asymptote then deviate away later.

[4 marks]

$4 - \frac{{300}}{{{w^2}}}$     (A1)(A1)(A1)

Notes: Award (A1) for $4$, (A1) for $-300$, (A1) for $\frac{1}{{{w^2}}}$ or ${w^{ - 2}}$. If extra terms present, award at most (A1)(A1)(A0).

[3 marks]

$4 - \frac{{300}}{{{w^2}}} = 0$   OR   $\frac{{300}}{{{w^2}}} = 4$   OR   $\frac{{{\text{d}}S}}{{{\text{d}}w}} = 0$     (M1)

Note: Award (M1) for equating their derivative to zero.

$w = 8.66{\text{ }}\left( {\sqrt {75} ,{\text{ 8.66025}} \ldots } \right)$     (A1)(ft)(G2)

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

[2 marks]

$17.3 \left( {\frac{{150}}{{\sqrt {75} }},{\text{ 17.3205}} \ldots } \right)$     (A1)(ft)

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

[1 mark]

$40 + 4\sqrt {75}  + \frac{{300}}{{\sqrt {75} }}$     (M1)

Note: Award (M1) for substitution of their answer to part (f) into the expression for $S$.

$= 110{\text{ (cm) }}\left( {40 + 40\sqrt 3 ,{\text{ 109.282}} \ldots } \right)$     (A1)(ft)(G2)

Note: Do not accept $109$.

     Follow through from their answers to parts (f) and (g).

[2 marks]