Hi folks,

this is not a university level question, but it is calculus so I hope it's ok here.

A ship is to make a voyage of 200 km at a constant speed. When the speed of the ship is v km/h the cost is $ v^2 + \frac{4000}{v} $ USD per hour.

Find the speed at which the ship should travel so that the cost of the voyage is a minimum.

So, we have a cost function in terms of velocity, so my first try was to differentiate the cost with respect to the velocity

$ \frac{d (cost)}{dv} = 2v - 4000v^-2 $ i.e. $v^3 = 2000$ therefore v = 12.6 km/h

$ \frac {d^2 (cost)}{dv^2}$ is > 0 so the first differential is a minimum as required.

I was concerned that I did not used the distance information (200km) and even more concerned that the answer was wrong anyway.

So, I decided to try velocity = distance / time i.e. $v = \frac{200}{t}$ where t is the time in hours and substitute this in the cost function

cost = $ (\frac{200}{t})^2 + 20.t $

To minimise the cost $ \frac{d (cost)}{dt} = -80,000 t^-3 + 20.t = 0 $ this gives $ t^3 = 4000$ so t = 15.9 hours

therefore velocity = 200 / 15.9 = 12.6 km/h

you will see that this is the same wrong answer.