Kingsholm Rovers set the price of a home match ticket at \(\pounds x\). The finance team models the profit made on the match, \(\pounds P\), by \[P = -2x^2 ...

The profit model \(P = -2x^2 + bx + c\) is a downward parabola in the ticket price \(x\) pounds. Two facts are supplied: the vertex (stationary point) is at \(x = 15\), and the profit there is 250 pounds. Each fact yields one equation, which is enough to pin down the two unknown constants.

1. Show that b = 60 [2 marks]
   Differentiate the model with respect to \(x\): \[\frac{dP}{dx} = -4x + b\] At a stationary point the gradient is zero, and this happens at \(x = 15\): \[-4(15) + b = 0 \quad \Rightarrow \quad -60 + b = 0 \quad \Rightarrow \quad b = 60\] as required. (The same value follows from the symmetry formula \(x = -\dfrac{b}{2a}\) with \(a = -2\).)
2. Find the value of c [2 marks]
   Substitute \(x = 15\) and \(b = 60\) into the model and set the profit to 250: \[P(15) = -2(15)^2 + 60(15) + c = -450 + 900 + c = 450 + c\] \[450 + c = 250 \quad \Rightarrow \quad c = -200\] So the model is \(P = -2x^2 + 60x - 200\). The negative constant makes sense: at a price of zero the club would make a loss of 200 pounds, its fixed costs for staging the match.
3. Solve P = 0 [3 marks]
   \[-2x^2 + 60x - 200 = 0\] Dividing through by \(-2\) simplifies the arithmetic: \[x^2 - 30x + 100 = 0 \tag{1 mark}\] This does not factorise, so use the quadratic formula with \(a = 1\), \(b = -30\), \(c = 100\): \[x = \frac{30 \pm \sqrt{(-30)^2 - 4(1)(100)}}{2} = \frac{30 \pm \sqrt{900 - 400}}{2} = \frac{30 \pm \sqrt{500}}{2} \tag{1 mark}\] Since \(\sqrt{500} = 22.3606\ldots\), \[x = \frac{30 + 22.3606}{2} = 26.18\ldots \quad \text{or} \quad x = \frac{30 - 22.3606}{2} = 3.8196\ldots\] To 3 significant figures the ticket prices are £3.82 and £26.2.
4. Interpretation [2 marks]
   These two prices are the break-even prices: at either of them the model predicts a profit of exactly zero. Because the parabola opens downwards, \(P\) is positive only between the roots, so the club makes a profit only for prices strictly between about £3.82 and £26.20. Charge less and the ticket income fails to cover the fixed costs; charge more and (in the model) too few supporters come. The most profitable price is the vertex, £15, giving £250.

Examination reminder: for a negative quadratic, the region of positive values lies between the roots, so break-even points define the whole permitted range of prices.