A corner shop is deciding whether to raise the price of a bestselling snack. Weekly revenue from the snack, in pounds, is modelled by \(R(x) = -x^2 + 30x\),...
Question 1 Report
A corner shop is deciding whether to raise the price of a bestselling snack. Weekly revenue from the snack, in pounds, is modelled by \(R(x) = -x^2 + 30x\), where \(x\) is the price increase in pence. The owner will only consider an increase within the range shown in the diagram below.
- Work out the revenue when \(x = 5\). (1)
- Form the equation satisfied by \(x\) when the revenue is \(£200\). (1)
- Solve this equation to find both possible values of \(x\). (3)
- Using the diagram, state which value of \(x\) the owner should choose, giving a reason. (1)
Answer Details
This question links a specific evaluation of the revenue model, a quadratic equation for a target revenue, and a real-world restriction shown on a diagram, all built from the same formula \(R(x) = -x^2 + 30x\).
(a) Substituting \(x = 5\): \(R(5) = -5^2 + 30(5) = -25 + 150 = £125\). [1 mark]
(b) Setting revenue to £200: \(-x^2 + 30x = 200\), which rearranges to \(x^2 - 30x + 200 = 0\). [1 mark]
(c) Factorising: two numbers that multiply to \(200\) and add to \(-30\) are \(-10\) and \(-20\), so \((x-10)(x-20) = 0\), giving \(x = 10\) or \(x = 20\). [3 marks]
(d) The diagram shows the owner will only accept a price increase up to \(15\) pence, so \(x = 20\) is outside this allowed range and must be rejected, leaving \(x = 10\) pence as the owner's choice. [1 mark]
As in the previous question, both roots give the same £200 revenue mathematically, but the diagram's shaded range (\(0\) to \(15\) pence) is what rules out the larger solution.
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