Edexcel A level Maths (9MA0) · Pure mathematics › Algebra and functions
Practise Modelling with functions. 4 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Worked solution: Since \((x - 12)^2 \ge 0\), the greatest value of \(P\) is 150, when \(x = 12\): a profit of £150 thousand.
Question 2Medium9 marks
A new museum opens. The monthly number of visitors, \(V\) thousand, \(t\) months after the museum opens is modelled by \[V = 12 + 8t - t^2, \quad t \ge 0\]
(a) Find, according to the model, the maximum monthly number of visitors and the value of \(t\) at which it occurs.[3]
(b) Show that the model predicts that the monthly number of visitors falls to zero when \(t = 4 + 2\sqrt{7}\).[3]
(c) Comment on the suitability of the model for predicting the number of visitors two years after the museum opens.[1]
(d) Give a reason why a model in which the monthly number of visitors falls to zero is unlikely to be realistic, and suggest how the model could be improved.[2]
Show the answer and mark scheme
(a)Answer: 28 000 visitors per month, when \(t = 4\)
M1 for completing the square, \(28 - (t - 4)^2\), or solving \(8 - 2t = 0\)
A1 for \(t = 4\)
A1 for 28 000 (visitors per month)
Worked solution: \(V = 12 + 8t - t^2 = 28 - (t - 4)^2\). The maximum is \(V = 28\) when \(t = 4\): 28 000 visitors per month, 4 months after opening.
A1* for \(\sqrt{28} = 2\sqrt{7}\) and rejecting the negative root, with a reason
Worked solution: \(V = 0 \Rightarrow (t - 4)^2 = 28 \Rightarrow t = 4 \pm \sqrt{28} = 4 \pm 2\sqrt{7}\). \(4 - 2\sqrt{7} \lt 0\) is outside the domain \(t \ge 0\), so \(t = 4 + 2\sqrt{7}\) (about 9.3 months).
(c)Answer: At \(t = 24\), \(V = -372\): a negative number of visitors is impossible, so the model is unsuitable (it is only valid up to about 9.3 months).
B1 for evaluating (or noting the sign of) \(V\) at \(t = 24\): negative, which is impossible, so the model is not suitable
Worked solution: When \(t = 24\), \(V = 12 + 192 - 576 = -372\). A negative number of visitors is impossible, so the model cannot be used two years after opening.
(d)Answer: A museum is likely to keep attracting some visitors (tourists, school groups), so numbers should level off rather than fall to zero. Use a model that tends to a positive constant, e.g. one of the form \(V = A + B\mathrm{e}^{-kt}\) after the peak.
B1 for a sensible reason, e.g. visitor numbers are likely to settle at a steady level (regular visitors, tourists) rather than fall to zero
B1 for a sensible refinement, e.g. a function with a positive horizontal asymptote (such as \(V = A + B\mathrm{e}^{-kt}\)), a periodic term for seasonal variation, or restricting the model to \(0 \le t \le 9\)
Worked solution: After the initial interest, a museum would still attract tourists and school groups, so the monthly number is likely to level off at a positive value. A better model would tend to a positive constant for large \(t\), for example \(V = A + B\mathrm{e}^{-kt}\) for the period after the peak (possibly with a seasonal term added).
Question 3Hard7 marks
The entrance to a road tunnel is modelled as a parabolic arch. The arch meets horizontal ground at two points 12 m apart, and its highest point is 6 m above the ground. The arch is modelled by the equation \(y = kx(12 - x), \quad 0 \le x \le 12\) where \(y\) m is the height of the arch above the ground at a horizontal distance \(x\) m from one end, and \(k\) is a constant.
(a) Show that \(k = \frac{1}{6}\).[2]
(b) A lorry is 2.5 m wide and 4.9 m high. It drives through the tunnel entrance with its centre directly below the highest point of the arch. Use the model to determine whether the lorry fits through the entrance.[3]
(c) State the greatest height of a lorry of width 2.5 m that could pass through the entrance in this way, according to the model.[1]
(d) Give one reason why a real lorry of this greatest height should not attempt to pass through the entrance.[1]
Show the answer and mark scheme
(a)Answer: The highest point is at \(x = 6\), so \(6 = k \times 6 \times 6\), giving \(k = \frac{1}{6}\).
M1 for using the symmetry of the arch: the highest point is at \(x = 6\), so \(6 = k \times 6 \times (12 - 6)\)
A1* for \(k = \frac{6}{36} = \frac{1}{6}\)
Worked solution: By symmetry the highest point is midway between the ends, at \(x = 6\). So \(6 = k \times 6 \times (12 - 6) = 36k\) and \(k = \frac{6}{36} = \frac{1}{6}\).
(b)Answer: At \(x = 4.75\) (and \(x = 7.25\)) the arch height is \(\frac{4.75 \times 7.25}{6} = 5.74\) m (3 s.f.), which is more than 4.9 m, so the lorry fits.
M1 for identifying the positions of the top corners of the lorry: \(x = 6 \pm 1.25\), i.e. \(x = 4.75\) or \(x = 7.25\)
A1 for the height of the arch there: \(\frac{4.75 \times 7.25}{6} = 5.74\) (awrt)
A1ft for a correct conclusion: \(5.74 \gt 4.9\), so the lorry fits
Worked solution: The lorry's top corners are 1.25 m either side of the centre line, at \(x = 6 - 1.25 = 4.75\) and \(x = 6 + 1.25 = 7.25\). There \(y = \frac{1}{6} \times 4.75 \times 7.25 = 5.74\) m (3 s.f.), which is greater than 4.9 m. So, according to the model, the lorry fits through the entrance.
(c)Answer: 5.74 m (3 s.f.)
B1ft for 5.74 m (awrt)
Worked solution: The greatest possible height is the height of the arch at \(x = 4.75\), i.e. \(5.74\) m (3 s.f.).
(d)Answer: e.g. there would be no clearance for movement or bouncing; the arch may not be exactly parabolic; the road may not be level
B1 for a sensible reason, e.g. a safety margin is needed, the lorry may sway or bounce, the arch is only approximately parabolic, or measurements are approximate
Worked solution: For example, the model gives the height exactly at the corners, leaving no clearance; in reality the lorry moves and the arch is only approximately parabolic, so a safety margin is needed.