The price that maximises nothing
Solving a quadratic model correctly and then noticing what it does not license.
Scenario
A stall's weekly profit in pounds is modelled by P(x) = -2x^2 + 60x - 250, where x is the price charged per item in pounds.
The model was fitted from one summer of trading at prices between £6 and £22, and the stallholder is deciding what to charge next season.
Answers in order, explanations withheld, and a report on where the chain broke.
- 19.1Quadratic functionsStandard
At which prices does the model predict a profit of exactly zero?
- A£5 and £25
- B£10 and £20
- C£2 and £30
- D£15 only
Show the answer and reasoningHide the reasoning
£5 and £25
Setting P(x) = 0 gives -2x^2 + 60x - 250 = 0, or x^2 - 30x + 125 = 0, which factors as (x - 5)(x - 25). Check: P(5) = -50 + 300 - 250 = 0.
- 19.2Quadratic functionsStandard
Builds on: The zeros from question 1, whose midpoint is the vertex. The £12.50 choice is what averaging the wrong pair of numbers produces.
Which price does the model predict will produce the greatest profit?
- A£30
- B£15
- C£25
- D£12.50
Show the answer and reasoningHide the reasoning
£15
The parabola opens downwards, so its vertex is the maximum, and the vertex sits midway between the zeros: (5 + 25) / 2 = 15. Then P(15) = -450 + 900 - 250 = £200.
- 19.3Interpreting modelsChallenge
Builds on: Question 1 fixed the zeros at 5 and 25, which is what tells you £3 lies outside them — and the stimulus is what tells you it also lies outside the data.
What does the model say about charging £3?
- AProfit would be about £200.
- BProfit would be negative, and £3 is outside the range the model was fitted to.
- CProfit would be zero.
- DProfit would be at its highest, because the price is lowest.
Show the answer and reasoningHide the reasoning
Profit would be negative, and £3 is outside the range the model was fitted to.
P(3) = -18 + 180 - 250 = -£88, so the model predicts a loss. But £3 is below the £6 floor of the data the model was fitted from, so the prediction is an extrapolation and carries far less weight than the arithmetic suggests.
- 19.4Interpreting modelsChallenge
Builds on: Question 2 supplied the £15, and question 3 established that the model has a range it was fitted over. Together they make the difference between what was computed and what is being claimed.
The stallholder concludes: "The maths says charge £15." What is the best assessment?
- ACorrect: £15 is the maximum of the model and the model is the evidence.
- BIncorrect: the maximum is at £25, where profit is zero.
- CThe arithmetic is right, and the model describes one summer's trading rather than next season's.
- DThe model cannot be used to compare prices at all.
Show the answer and reasoningHide the reasoning
The arithmetic is right, and the model describes one summer's trading rather than next season's.
£15 really is the model's maximum, so nothing in the calculation is wrong. What the conclusion adds is that next season will behave like last one, which no amount of solving establishes. The model is usable — it just supports a weaker claim than "the maths says".