1. the same wrong answer twice!

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.

2. Re: the same wrong answer twice!

why do you think 12.6 km/h is wrong?

3. Re: the same wrong answer twice!

You are rounding. Perhaps the correct answer is not supposed to be rounded? Try $10\sqrt[3]{2}$

4. Re: the same wrong answer twice!

The question came from a Maths book and the answer at the back is 20 km/h. I guess the book could be wrong. No errata is published so I cannot tell.
But if you guys think I am on the right track, that's great!

5. Re: the same wrong answer twice!

Yes, I believe your book is wrong:

minimize v^2+4000/v - Wolfram|Alpha Results

You can also plug in the result you found and the book's result to see that the cost is less at 12.6 than it is at 20 km/h.

6. Re: the same wrong answer twice!

You got the same wrong answer by both methods because you are solving the wrong problem! You, and the others, are minimizing v^2+ 4000/v, the cost per hour, but you should be minimizing the cost of the entire trip. At v km/h you will go 200 km in 200/v hours. Since the cost is v^2+ 400/v per hour, the cost of going 200 km will be (200/v)(v^2+ 400/v)= 200v+ 80000/v^2. The derivative of that is 200- 160000/v^3= 0 so 200v^3= 160000, v^3= 8000, v= 20 km/h.