The cost of running a cargo ship is \(£150\) a day for wages and other crew expenses and \(£\dfrac{8v^2}{3}\) a day for fuel and maintenance, where \(v\) is the steady speed in \(\quantity{}{km/h}\). Find in terms of \(v\) how many days a voyage of \(\quantity{2400}{km}\) takes and hence obtain a formula for the total cost of the whole voyage.

If \(v\) is the steady speed of the ship, it takes time \(t=\dfrac{s}{v}\) to travel a distance \(s\).

In our case, we have \(s=\quantity{2400}{km}\), and hence the voyage takes \(\dfrac{2400}{v}\) hours, or \(\dfrac{100}{v}\) days.

The cost per day is \(150+\dfrac{8v^2}{3}\) so the total cost of the voyage is \[\begin{equation*} \left(150+\dfrac{8v^2}{3}\right)\frac{100}{v} = \frac{15\,000}{v} + \frac{800v}{3}. \end{equation*}\]

Deduce by calculus methods the most economical speed and the least total cost of the whole voyage.

The most economical speed reduces the total cost of the voyage to a minimum. First we find the derivative of the cost function:

\[ \frac{d}{dv} \left( \frac{15\,000}{v} + \frac{800v}{3} \right) = -\frac{15\,000}{v^2}+ \frac{800}{3}. \]

Setting this equal to zero and solving for \(v\),

\[\begin{align*} -\frac{15\,000}{v^2}+ \frac{800}{3} &= 0\\ -15\,000 + \frac{800v^2}{3} &= 0\\ v^2 &= \frac{450}{8}\\ v &= \pm \sqrt{\frac{450}{8}} = \pm \frac{15}{2}. \end{align*}\]

The speed \(v\) has to be positive, so \(v = \quantity{\dfrac{15}{2}}{km/h}\) is the most economical speed.

At this speed the cost of the voyage is \(\dfrac{15\,000}{15/2} + \dfrac{800\times15/2}{3} = £4\,000\).

To check this is the minimum rather than the maximum cost, we can compare the cost at a different speed such as \(v=\quantity{5}{km/h}\). \[\begin{align*} \text{Cost} &= \frac{15\,000}{5} + \frac{800\times5}{3}\\ &= 3\,000+\frac{4\,000}{3}\\ &> 4\,000 \end{align*}\]

So the \(£4\,000\) is indeed the minimum cost.

Alternatively, we could evaluate the second derivative at \(v=\dfrac{15}{2}\).