Date | November 2008 | Marks available | 8 | Reference code | 08N.1.hl.TZ0.9 |
Level | HL only | Paper | 1 | Time zone | TZ0 |
Command term | Justify and Show that | Question number | 9 | Adapted from | N/A |
Question
A packaging company makes boxes for chocolates. An example of a box is shown below. This box is closed and the top and bottom of the box are identical regular hexagons of side x cm.
(a) Show that the area of each hexagon is \(\frac{{3\sqrt 3 {x^2}}}{2}{\text{c}}{{\text{m}}^2}\) .
(b) Given that the volume of the box is \({\text{90 c}}{{\text{m}}^2}\) , show that when \(x = \sqrt[3]{{20}}\) the total surface area of the box is a minimum, justifying that this value gives a minimum.
Markscheme
(a) Area of hexagon \( = 6 \times \frac{1}{2} \times x \times x \times \sin 60^\circ \) M1
\( = \frac{{3\sqrt 3 {x^2}}}{2}\) AG
(b) Let the height of the box be h
Volume \( = \frac{{3\sqrt 3 h{x^2}}}{2} = 90\) M1
Hence \(h = \frac{{60}}{{\sqrt 3 {x^2}}}\) A1
Surface area, \(A = 3\sqrt 3 {x^2} + 6hx\) M1
\( = 3\sqrt 3 {x^2} + \frac{{360}}{{\sqrt 3 }}{x^{ - 1}}\) A1
\(\frac{{{\text{d}}A}}{{{\text{d}}x}} = 6\sqrt 3 x - \frac{{360}}{{\sqrt 3 }}{x^{ - 2}}\) A1
\(\left( {\frac{{{\text{d}}A}}{{{\text{d}}x}} = 0} \right)\)
\(6\sqrt 3 {x^3} = \frac{{360}}{{\sqrt 3 }}\) M1
\({x^3} = 20\)
\(x = \sqrt[3]{{20}}\) AG
\(\frac{{{{\text{d}}^2}A}}{{{\text{d}}{x^2}}} = 6\sqrt 3 + \frac{{720{x^{ - 3}}}}{{\sqrt 3 }}\)
which is positive when \(x = \sqrt[3]{{20}}\), and hence gives a minimum value. R1
[8 marks]
Examiners report
There were a number of wholly correct answers seen and the best candidates tackled the question well. However, many candidates did not seem to understand what was expected in such a problem. It was disappointing that a significant number of candidates were unable to find the area of the hexagon.