Date | May 2014 | Marks available | 2 | Reference code | 14M.2.sl.TZ2.5 |
Level | SL only | Paper | 2 | Time zone | TZ2 |
Command term | Draw | Question number | 5 | Adapted from | N/A |
Question
A parcel is in the shape of a rectangular prism, as shown in the diagram. It has a length l cm, width w cm and height of 20 cm.
The total volume of the parcel is 3000 cm3.
Express the volume of the parcel in terms of l and w.
Show that l=150w.
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+300w, 0<w⩽
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.
Markscheme
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]