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!!
Best Answer
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 :
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.