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Thread: Efficiency problem (nautical miles per hour) in "Calculus Made Easy": y=0.3+0.001v^3

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    Efficiency problem (nautical miles per hour) in "Calculus Made Easy": y=0.3+0.001v^3

    I'm working my way through Thompson's Calculus Made Easy, and I'm stuck on a problem in Chapter 11:

    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 **** are together equal per hour to the cost of one ton of coal. What speed will make the total cost of a voyage of 1,000 nautical miles a minimum? And if coal costs ten dollars per ton, what will that minimum cost of the voyage amount to?

    OK, so I realize that we have to express the first equation in terms of cost (C). If the added expense of wages, interest, etc. per hour are equal to one ton of coal burned per hour, then I think that the cost would equal 2y or:
    C=0.6 + 0.002v^3
    Is this correct? And what should my next step be? In view of the 1,000 mile trip, should I change v^3 to (m/h)^3 and plug that in? I'm not sure how to proceed.
    Thanks!
    Have a Kafka Day

    Cogito ergo sum . . . cogito

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    Quote Originally Posted by Linty Fresh View Post
    I'm working my way through Thompson's Calculus Made Easy, and I'm stuck on a problem in Chapter 11:

    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 **** are together equal per hour to the cost of one ton of coal. What speed will make the total cost of a voyage of 1,000 nautical miles a minimum? And if coal costs ten dollars per ton, what will that minimum cost of the voyage amount to?

    OK, so I realize that we have to express the first equation in terms of cost (C). If the added expense of wages, interest, etc. per hour are equal to one ton of coal burned per hour, then I think that the cost would equal 2y or:
    C=0.6 + 0.002v^3
    Is this correct? And what should my next step be? In view of the 1,000 mile trip, should I change v^3 to (m/h)^3 and plug that in? I'm not sure how to proceed.
    Thanks!
    Like algebra, one of the first things to do is to assign letters to variables and unknowns. But, as you did, it is frequently good in differential calculus to ask what is to be varied in order to optimize what.

    You correctly saw that you want to minimize a cost function. Name it.

    [tex]c = \text { total cost of the voyage of 1000 nautical miles.}[/tex]

    But what are you being asked to vary? Why velocity obviously. So name it.

    [tex]v = \text { velocity of ship.}[/tex]

    So we know that we are looking to set up something like [tex]c = f(v).[/tex]

    The cost information is given in terms of hourly costs. And we are told that there is an hourly cost of coal and hourly cost of additional requirements like labor, etc. And to turn hourly costs into total costs, we need to know hours. That's three different variables. Name them.

    [tex]c_c = \text { cost of coal used per hour.}[/tex]

    [tex]c_a = \text { additional costs per hour.}[/tex]

    [tex]h = \text { duration of voyage in hours.}[/tex]

    [tex]\therefore \text {Equation 1: } c = h(c_c + c_a).[/tex]

    But that is not a function in v. Moreover, we have not said how to evaluate any of those variables. But the problem gives us some equations. One involves the quantity of coal used per hour. Name it.

    [tex]y = \text { tons of coal used per hour.}[/tex] And we have an equation for that.

    [tex]\text {Equation 2: }y = 0.3 + 0.002v^3.[/tex]

    At last we see v. But be careful of dimensional analysis in calculus because y is expressed in tons per hour, not miles cubed per hour cubed. And notice that y is not a monetary quantity at all.

    How do we change a physical quantity into a monetary quantity? By multiplying it by a price. But we do not know (initially) the price. So name the variable.

    [tex]p = \text { the price of one ton of coal.}[/tex]

    That gives us [tex]\text {Equation 3: } c_c = py = p(0.3 + 0.002v^3).[/tex]

    Now what about those other hourly costs. Are they equal to the costs of the coal burned per hour or the cost of one ton of coal? Just one ton.

    [tex]\text {Equation 4: } c_a = p.[/tex]

    Putting this together, we derive:

    [tex]\text {Equation 5: } c = h(c_t + c_a) = h\{p(0.3 + 0.002v^3) + p\} = h(1.3p + 0.002pv^3).[/tex]

    That is a function in three variables. Can we replace either h or p with an expression in v. Yes. We can go to what learned in first year algebra

    [tex]v = \dfrac{1000}{h} \implies h = \dfrac{1000}{v}.[/tex]

    And finally we get to:

    [tex]c = h(1.3p + 0.002pv^3) = \dfrac{1300p}{v} + 2pv^2.[/tex]

    Now what?
    Last edited by JeffM; 03-03-2018 at 06:02 PM.

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    Thanks, Jeff

    Hi Jeff
    Sorry it took so long to reply. Work kind of got in the way. Just wanted to say thank you for your help. That was a real eye-opener, and I now realize that my problem isn't with calculus. It's with algebra. I'm not bad at algebra, but I still need a little more practice in creating new functions from the original.

    Well, p=$10/ton, so just plug that into the equation. You now have C(v)=13,000/v + 20v^2 where C is the total cost of the trip. Derive that and set the derivative to zero, solve for y, and you have your velocity to make the cost a minimum. Is that right?
    Last edited by Linty Fresh; 03-26-2018 at 07:47 PM.
    Have a Kafka Day

    Cogito ergo sum . . . cogito

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