| Date | May 2013 | Marks available | 2 | Reference code | 13M.2.hl.TZ1.7 |
| Level | HL only | Paper | 2 | Time zone | TZ1 |
| Command term | Find | Question number | 7 | Adapted from | N/A |
Question
An electricity station is on the edge of a straight coastline. A lighthouse is located in the sea 200 m from the electricity station. The angle between the coastline and the line joining the lighthouse with the electricity station is 60°. A cable needs to be laid connecting the lighthouse to the electricity station. It is decided to lay the cable in a straight line to the coast and then along the coast to the electricity station. The length of cable laid along the coastline is x metres. This information is illustrated in the diagram below.
The cost of laying the cable along the sea bed is US$80 per metre, and the cost of laying it on land is US$20 per metre.
Find, in terms of x, an expression for the cost of laying the cable.
Find the value of x, to the nearest metre, such that this cost is minimized.
Markscheme
let the distance the cable is laid along the seabed be y
\({y^2} = {x^2} + {200^2} - 2 \times x \times 200\cos 60^\circ \) (M1)
(or equivalent method)
\({y^2} = {x^2} - 200x + 40000\) (A1)
cost = C = 80y + 20x (M1)
\(C = 80{({x^2} - 200x + 40000)^{\frac{1}{2}}} + 20x\) A1
[4 marks]
\(x = 55.2786 \ldots = 55\) (m to the nearest metre) (A1)A1
\(\left( {x = 100 - \sqrt {2000} } \right)\)
[2 marks]
Examiners report
Some surprising misconceptions were evident here, using right angled trigonometry in non right angled triangles etc. Those that used the cosine rule, usually managed to obtain the correct answer to part (a).
Some surprising misconceptions were evident here, using right angled trigonometry in non right angled triangles etc. Many students attempted to find the value of the minimum algebraically instead of the simple calculator solution.
