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I am currently stuck on this question. The question and my steps so far are given below:

(10) Suppose it to be known that consumption of coal by a certain steamer may be represented by the formula: $$y = 0.3 + 0.001v^3$$ where $y$ is the number of tons of coal burned per hour and v is the speed expressed in nautical miles per hour. The cost of wages, interest on capital, and depreciation of that ship are together equal, per hour, to the cost of 1 ton of coal. What speed will make the total cost of a voyage of 1000 nautical miles a minimum? And, if coal costs 10 dollars per ton, what will that minimum cost of the voyage amount to?

I am stuck on the first part. Here are my steps so far:

Let price of one ton of coal $= c$
wages + interest + depreciaition = T
According to the question: $T \times t = c$ If distance = 1000, $v = \frac{1000}{time}$

basically $y = \frac{tonnes}{time}$ so : $$\frac{tonnes}{time} = 0.3 + 0.001\frac{1000^3}{time^3}$$ I solved this equation but the wring answer for v. I also do not think I am doing it right.

Please do not give away the answer but guide me in the right direction.

Many thanks and stay safe!!

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  • $\begingroup$ So far all you've shown is how you restated the problem. What did you do next? And after that? And after that, etc., to finally reach the wrong answer for $v$ (which was what)? $\endgroup$
    – David K
    Dec 17, 2020 at 21:08
  • $\begingroup$ Note, however, that wages + interest + depreciation per hour equals the cost of one ton of coal. If $T$ is wages + interest + depreciation per hour, and $t>1,$ then $T\times t > c$ because $T\times t$ is the cost for more than just one hour. And if $T$ is not wages + interest + depreciation per hour, what is it? $\endgroup$
    – David K
    Dec 17, 2020 at 21:13
  • $\begingroup$ @DavidK Re your first comment, good point, but partly my fault. Apparently, without showing his work, the OP explored the idea in my (originally wrong) analysis and (naturally enough) came up with a wrong answer. See the comment that he left after my answer. $\endgroup$ Dec 17, 2020 at 21:13
  • $\begingroup$ @user2661923 My fault too for missing a lot of that until after I posted (I blame a browser glitch). But I think we're asking the same thing from OP: to get guidance to a better solution, show the bad one. Then we can give pointers to where it went wrong. $\endgroup$
    – David K
    Dec 17, 2020 at 21:16
  • $\begingroup$ @DavidK Except that the OP may well have solved the problem accurately, and merely not shown his work. Assuming that my analysis is now correct, and assuming that the OP correctly followed my mistaken analysis before, if he repeats the method under my corrected analysis, he may well get the intended answer. In that case, it is game over. $\endgroup$ Dec 17, 2020 at 21:18

1 Answer 1

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Edit

Well, this is embarassing. I went off the rails, and made a mistake in analysis.

I have corrected the mistake below.


Hint

Although your analysis was a good step in the right direction, you should capitalize on :

The cost of wages, interest on capital, and depreciation of that ship are together equal, per hour, to the cost of 1 ton of coal.


Edit

I misinterpreted the meaning of the above constraint. Further, there is an ambiguity here. Are you supposed to minimize the total of [wages + interest + depreciation], or are you supposed to minimize the total of $\{$ [wages + interest + depreciation] + coal used $\}?$

Under the first interpretation, you are supposed to minimize the time taken for the voyage. It is doubtful that this would be the intended solution.

Under the second interpretation, the cost per hour is $(1 + y)$, rather than $y$. I suspect that this is the intent of the problem. Contrast that with the mistake that I made below, which (wrongly, indirectly) presumes that the cost per hour is $y$, instead of $(1 + y).$


This means that you do not need to use any variables other than $y$ and $v$.

Distance = rate $\times$ time.

The time for the voyage will be $[(1000)/v],$ so the amount of coal consumed will be $[(1000/v] \times y.$

Take both the first and second derivatives of the function [in the sole variable $v$] that represents the amount of coal that will be consumed on the voyage.

You want the first derivative to be zero and the second derivative to be positive. Then, you will have identified the value for $v$ that represents the minimum value of the function. From this, you can routinely determine the time of the voyage.

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  • $\begingroup$ Is rate basically the velocity? Thanks $\endgroup$
    – Thingybob
    Dec 17, 2020 at 20:18
  • $\begingroup$ @AkashGopinath Yes, rate, velocity, and speed are all terms that (generally) refer to the same thing : distance divided by time. $\endgroup$ Dec 17, 2020 at 20:20
  • $\begingroup$ This method gives me an answer that is different from the textbook. Is there anything I am missing? Thanks $\endgroup$
    – Thingybob
    Dec 17, 2020 at 20:35
  • $\begingroup$ @Thingybob In order to answer that question, I need you to either overhaul your original query or add an Addendum to your original query that focuses exclusively on the method provided by my answer. When editing your query to show your work, be as careful/detailed as you can, going from one step to the next. Then, when you have completed editing your query, if no one else steps up, leave another comment for me and I will look at your work. You can either address the comment using the at sign followed by user2661923, or you can place the comment after my answer. $\endgroup$ Dec 17, 2020 at 20:41
  • $\begingroup$ @Thingybob My bad! I made a mistake in my analysis. Please read my edited answer. $\endgroup$ Dec 17, 2020 at 21:00

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