A ferry operator wishes to minimise the cost per unit distance travelled of operating the ferry. When the ferry is travelling at speed v km $h^{-1}$, the cost in fuel per hour is $\dfrac {v^3}{10}$ dollars. The other costs of running the ferry (wages, maintenance, etc.) are $675 per hour. At what speed should the ferry be operated in order to minimise the cost per kilometre?
So for this question, I think that we first have to find a function for 'cost per unit distance'. So if C is the cost, t is the time in units of hours and d is the distance, then $\dfrac {dC}{dd}$ would be the 'cost per unit distance'. From the question, 'cost in fuel per hour' is given by $\dfrac{dv}{dt}$ = $\dfrac {v^3}{10}$, and 'other costs' is $\dfrac{dC}{dt}$ = $675$. So $\dfrac{dC}{dd}$ = $\dfrac {dv}{dt}$ $\times$ $\dfrac {dC}{dt}$, therefore $\dfrac {dC}{dd}$ = $\dfrac {v^3}{10}$ $\times$ $675$. Then to minimise the speed, find the second derivative of 'cost per unit distance travelled' and find the speed. But I don't know if my setting of the given values are correct.
As we are working in km and hours, we may as well take the unit distance as $1$ km. When we are traveling at speed $v$, it takes $\frac 1v$ hour to travel $1$ km. The cost for fuel is then $\frac 1v \cdot \frac {v^3}{10}=\frac {v^2}{10}$ per km. The cost for crew is $\frac 1v \cdot 675=\frac {675}v$ per km. The total cost is $\frac {v^2}{10}+\frac {675}v$. Now take the derivative, set to zero...