A phone company's monthly profit, in thousands of pounds, from adjusting its tariff price by \(x\) pounds is modelled by \(P=x^3-12x^2+36x+5\), valid for \(...
Question 1 Report
A phone company's monthly profit, in thousands of pounds, from adjusting its tariff price by \(x\) pounds is modelled by \(P=x^3-12x^2+36x+5\), valid for \(0 \leq x \leq 8\).
- Find \(\frac{dP}{dx}\). (1)
- Find the values of \(x\) at the stationary points of \(P\). (2)
- Determine the nature of each stationary point. (2)
- The company wants to choose a price increase that maximises profit. State, with a reason, which value of \(x\) it should choose, and the resulting profit. (2)
Answer Details
Stationary points of a cubic are found by setting its derivative to zero, and the second derivative distinguishes a local maximum from a local minimum at each one, which is exactly what is needed to choose the price change that maximises profit.
(a) Differentiating \(P=x^{3}-12x^{2}+36x+5\) term by term:
\[\dfrac{dP}{dx}=3x^{2}-24x+36\] [1 mark](b) Setting the derivative to zero:
\[3x^{2}-24x+36=0\]Dividing every term by 3:
\[x^{2}-8x+12=0\]Factorising:
\[(x-2)(x-6)=0\]giving \(x=2\) or \(x=6\). [2 marks]
(c) Differentiating again:
\[\dfrac{d^{2}P}{dx^{2}}=6x-24\]At \(x=2\): \(6(2)-24=-12 \lt 0\), so this is a maximum. At \(x=6\): \(6(6)-24=12 \gt 0\), so this is a minimum. [2 marks]
(d) Since maximising profit requires the maximum, the company should choose \(x=2\). Substituting into the original profit formula:
\[P(2)=2^{3}-12(2)^{2}+36(2)+5=8-48+72+5=37\]So the resulting profit is \(\pounds37000\) (since \(P\) is measured in thousands of pounds). [2 marks]
A cubic profit model typically has one local maximum and one local minimum; the second derivative test in part (c) is what tells the company which of the two candidate price changes, \(x=2\) or \(x=6\), is actually the profitable one to choose rather than the one that minimises profit.
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