2025 RVHS P1 Q13

2025 RVHS P1 Q13

12 marks

A cafe roasts its own coffee beans and packages them to be sold. The amount of roasted coffee beans remaining in the cafe at time \(t\) days is denoted by \(x\) kg. The cafe produces roasted coffee beans at a fixed rate of \(10\) kg/day, and sells the roasted coffee beans at a rate proportional to \(x^2\) kg/day.

  1. Given that there is no change in the amount of roasted coffee beans remaining in the cafe when there is \(5\) kg of beans remaining, show that \(\frac{\mathrm{d}x}{\mathrm{d}t} = 10 - \frac{2}{5}x^2\). Find the general solution of this differential equation. (You do not have to make \(x\) the subject.)[3]

Due to improvement in roasting capabilities of the cafe, it is now able to produce roasted coffee beans at the rate of \(3x\) kg/day, and sells the roasted coffee beans at a rate proportional to \(x^2\) kg/day.

  1. Given that amount of roasted coffee beans is increasing at the rate of \(6\) kg/day when the amount of roasted coffee beans remaining in the cafe is \(6\) kg, write down the modified differential equation relating \(\frac{\mathrm{d}x}{\mathrm{d}t}\) and \(x\).[2]
  2. Given that the initial amount of roasted coffee beans in the cafe is \(13.5\) kg, solve the differential equation in part (b), expressing \(x\) in terms of \(t\). Deduce the amount of roasted coffee beans remaining in the cafe in the long run, and sketch the curve of \(x\) against \(t\).

    [6]
  3. Based on the curve in part (c), explain in context why this might be favourable for the cafe owner.

    [1]

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Answer:(a) \(\frac1{10}\ln\left|\frac{x-5}{x+5}\right|=-\frac25t+C\); (b) \(\frac{dx}{dt}=3x-\frac13x^2\); (c) \(x=\frac{27e^{3t}}{3e^{3t}-1}\), tending to \(9\text{ kg}\).

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