Optimisation Problem (costs of running ship)

[f1player]

I'm not even sure how to start this problem

The daily running costs of a ship are $(c + dx^3), where c and d are constants and x km is the distance travelled in the day. Show that the cost of a voyage of 1000 km will be least when the fixed part of the cost is double the variable part.

I started like this: Cost per km = (c + dx^3)/(x)

Therefore the cost of voyage = (c + dx^3)/(x) * 1000

Is this right?

What's the next step?? I know you have to differentiate, but how do you differentiate that
:?
Thanks for any help

 

[JakeD]

Re: Optimisation Problem

I'm not even sure how to start this problem. You're on the right track.

The daily running costs of a ship are $(c + dx^3), where c and d are constants and x km is the distance travelled in the day. Show that the cost of a voyage of 1000 km will be least when the fixed part of the cost is double the variable part.

I started like this: Cost per km = (c + dx^3)/(x)

Therefore the cost of voyage = (c + dx^3)/(x) * 1000

Is this right? Yes.

What's the next step?? I know you have to differentiate, but how do you differentiate that

You could use the quotion rule on the cost of the voyage. Or you could simplify it like this:

\(\displaystyle \Large 1000 \frac{c + dx^3}{x} = 1000 (\frac{c}{x} + \frac{dx^3}{x}) = 1000(cx^{-1} + dx^2).\)

Can you take from here?

 

[f1player]

Yes i can now differentiate it and get c = 2dx^3

However, how would i go about proving that the cost is least when the fixed part is double the variable part

 

[JakeD]

Yes i can now differentiate it and get c = 2dx^3

However, how would i go about proving that the cost is least when the fixed part is double the variable part

It's right there in your equation c = 2dx^3. Just interpret the left and right sides.