| Date | November 2017 | Marks available | 1 | Reference code | 17N.1.sl.TZ0.15 |
| Level | SL only | Paper | 1 | Time zone | TZ0 |
| Command term | Write down | Question number | 15 | Adapted from | N/A |
Question
Maria owns a cheese factory. The amount of cheese, in kilograms, Maria sells in one week, \(Q\), is given by
\(Q = 882 - 45p\),
where \(p\) is the price of a kilogram of cheese in euros (EUR).
Maria earns \((p - 6.80){\text{ EUR}}\) for each kilogram of cheese sold.
To calculate her weekly profit \(W\), in EUR, Maria multiplies the amount of cheese she sells by the amount she earns per kilogram.
Write down how many kilograms of cheese Maria sells in one week if the price of a kilogram of cheese is 8 EUR.
Find how much Maria earns in one week, from selling cheese, if the price of a kilogram of cheese is 8 EUR.
Write down an expression for \(W\) in terms of \(p\).
Find the price, \(p\), that will give Maria the highest weekly profit.
Markscheme
522 (kg) (A1) (C1)
[1 mark]
\(522(8 - 6.80)\) or equivalent (M1)
Note: Award (M1) for multiplying their answer to part (a) by \((8 - 6.80)\).
626 (EUR) (626.40) (A1)(ft) (C2)
Note: Follow through from part (a).
[2 marks]
\((W = ){\text{ }}(882 - 45p)(p - 6.80)\) (A1)
OR
\((W = ) - 45{p^2} + 1188p - 5997.6\) (A1) (C1)
[1 mark]
sketch of \(W\) with some indication of the maximum (M1)
OR
\( - 90p + 1188 = 0\) (M1)
Note: Award (M1) for equating the correct derivative of their part (c) to zero.
OR
\((p = ){\text{ }}\frac{{ - 1188}}{{2 \times ( - 45)}}\) (M1)
Note: Award (M1) for correct substitution into the formula for axis of symmetry.
\((p = ){\text{ }}13.2{\text{ (EUR)}}\) (A1)(ft) (C2)
Note: Follow through from their part (c), if the value of \(p\) is such that \(6.80 < p < 19.6\).
[2 marks]
