Review question

# What's the most economical speed for this ship's journey? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7280

## Solution

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 $v=\dfrac{15}{2}$.