Two school sports day fundraising stalls model their profit, in pounds, as \(P_{1} = 3x^{2}-2x\) and \(P_{2} = 3x^{2}+4x-30\), where \(x\) is the number of ...
Question 1 Report
Two school sports day fundraising stalls model their profit, in pounds, as \(P_{1} = 3x^{2}-2x\) and \(P_{2} = 3x^{2}+4x-30\), where \(x\) is the number of raffle tickets sold, in tens.
- Find an expression for \(P_{2}-P_{1}\), fully simplified. (2)
- Work out the value of \(x\) for which the two stalls make equal profit. (2)
- State which stall is more profitable when \(x = 3\), showing your working. (2)
Answer Details
Subtracting one profit expression from the other cancels the matching \(3x^{2}\) terms, leaving a simple linear expression that can be set to zero (to find equal profit) or evaluated directly (to compare a specific case).
(a) \(P_{2}-P_{1} = (3x^{2}+4x-30)-(3x^{2}-2x)\). The \(3x^{2}\) terms cancel, leaving \(4x-(-2x)-30 = 4x+2x-30 = 6x-30\). [2 marks]
(b) Setting the difference to zero: \(6x-30=0\), so \(x=5\) (tens of tickets). [2 marks]
(c) At \(x=3\): \(6(3)-30 = -12\), which is negative, meaning \(P_{2}\lt P_{1}\). So Stall 1 is the more profitable stall when \(x=3\). [2 marks]
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