What is the Optimal Speed for Minimizing Coal Consumption on a Ship?

[54stickers]

I cannot seem the crack this problem. The solution may be trivial, but normal methods of max/min do not seem to work. Thanks for your help.

Homework Statement

Suppose it to be known that consumption of coal by a certain ship may be represented by the formula
y = 0.3 + .001v³
Where y is the tons of coal burned per hour and v is the speed expressed in nautical miles per hour. if coal costs $10 per ton
what speed will make the cost of a voyage of 1000 nautical miles a minimum?

 

[Mark44]

I cannot seem the crack this problem. The solution may be trivial, but normal methods of max/min do not seem to work. Thanks for your help.

Homework Statement

Suppose it to be known that consumption of coal by a certain ship may be represented by the formula
y = 0.3 + .001v³
Where y is the tons of coal burned per hour and v is the speed expressed in nautical miles per hour. if coal costs $10 per ton
what speed will make the cost of a voyage of 1000 nautical miles a minimum?

What have you tried? If you did the obvious (but incorrect) thing and simply differentiated y with respect to v, that won't get you anywhere.

You need to tie in the given information that the trip is 1000 naut. miles. The distance formula comes into play here: d = vt.

BTW, welcome to Physics Forums!

 

[54stickers]

Thanks for the welcome!
I did the dy/dv. I also tried to use the distance formula, but nothing would coalesce. Just looking for some fresh starting points.

Thanks.

 

[Mark44]

vt = 1000

If the ship travels for t hours to cover the 1000 n.m., it will burn ty tons of coal, since y is the number of tons of coal used per hour. IOW, it will burn .3t + .001tv³ tons of coal.

Your goal is to find the speed that minimizes the cost, which will be the same speed that minimizes the amount of coal used.

 

[54stickers]

vt = 1000

If the ship travels for t hours to cover the 1000 n.m., it will burn ty tons of coal, since y is the number of tons of coal used per hour. IOW, it will burn .3t + .001tv³ tons of coal.

Your goal is to find the speed that minimizes the cost, which will be the same speed that minimizes the amount of coal used.

Thanks! I was doing all of the substitutions into the dy/dv function.

 

[54stickers]

After a few days of trying this problem over and over, I still come up with incorrect numbers (I checked the back of the book, and it says 8.66 is the n.m/h)

My apologies, for it may be hard to read(I couldn't get TeX to work properly). This is what I did:

y = .3t + .001tv³ t = 1000/v

y = .3(1000)/v + .001(1000/v)v^3

= 300/v + v^2

= (300 + v^3)/v

dy/dv = [v(3v^2) - (300 + v^3)]/v^2

(2v^3 - 300)/v^2 = 0

2v³ = 300

v = cube root of 150 = 5.31329285

 

[Mark44]

I get the same as you did using a slightly different technique. Textbooks have been known to have incorrect answers.

You can check by comparing your answer to the book's answer. If your speed produces a smaller value for the tons of coal used than the book's speed, you know the book's answer is wrong.

 

[54stickers]

Thanks! That is most helpful.

The end result comes out much less than the book's answer with 5.3..etc

 

[Mark44]

Well, there you go...