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Math Help - calculus 1 help please! word problem (cost of production/derivative)

  1. #1
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    Exclamation calculus 1 help please! word problem (cost of production/derivative)

    " this project is an application of derivatives to cost and fuel efficiency for your vehicle"

    the average yearly mileage for my car to the nearest thousand is 7000 miles (from question #1)...

    in question #2 i had to state the average price per gallon that i pay for gas which came to be $2.87/gallon

    now i have to write the function C(x) which is the annual cost of fuel for my vehicle as a function of the miles per gallon x. (do not use a specific number for miles per gallon for x here) show work

    help me with this last question ..i dnt know what to do!!!

    (my car has a fuel capacity of 20 mpg...and i use 23 miles (from my house to school) as a reference to get the 7000 miles in question #1 maybe this can help ..i dnt know!!)
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  2. #2
    Member sinewave85's Avatar
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    Quote Originally Posted by natty210 View Post
    " this project is an application of derivatives to cost and fuel efficiency for your vehicle"

    the average yearly mileage for my car to the nearest thousand is 7000 miles (from question #1)...

    in question #2 i had to state the average price per gallon that i pay for gas which came to be $2.87/gallon

    now i have to write the function C(x) which is the annual cost of fuel for my vehicle as a function of the miles per gallon x. (do not use a specific number for miles per gallon for x here) show work

    help me with this last question ..i dnt know what to do!!!

    (my car has a fuel capacity of 20 mpg...and i use 23 miles (from my house to school) as a reference to get the 7000 miles in question #1 maybe this can help ..i dnt know!!)
    Well, just looking at what you have, a function for cost in terms of x = miles per gallon is:

    C(x) = \frac{(7000\ miles)(2.87\ dollars/gallon)}{x\ miles/gallon}

    I left the labels because they show how the "miles" and "gallons" cancel out and the quantity you want -- "dollars" -- remains. Something my physics teacher is big about, and it helps show whether or not you have set up an equation correctly.

    This is the equation you want for math class:

    C(x) = \frac{(7000)(2.87)}{x}
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    Question

    i got the same answer but what caught my attention was that when i took the derivative of that equation ill get (-7000/x^2)(2.87)!! and i have to do this chart where x is gonna be 10, 15, 20,25 etc.. (the miles) and i am gonna have C(x) and dC/dx where im gonna plug in 10, 15 etc..for x.
    and when i pug in this numbers for x on the derivative formula i get negative numbers!! is this right?? cause it dsnt make sence that the cost is going to be negative!... or did i do the derivative wrong
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    Member sinewave85's Avatar
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    Quote Originally Posted by natty210 View Post
    i got the same answer but what caught my attention was that when i took the derivative of that equation ill get (-7000/x^2)(2.87)!! and i have to do this chart where x is gonna be 10, 15, 20,25 etc.. (the miles) and i am gonna have C(x) and dC/dx where im gonna plug in 10, 15 etc..for x.
    and when i pug in this numbers for x on the derivative formula i get negative numbers!! is this right?? cause it dsnt make sence that the cost is going to be negative!... or did i do the derivative wrong
    I don't see why you need to take the derivative of the equation. Unless there was more to it than what you typed out, we have satisfied the assignment: "Write the function C(x) which is the annual cost of fuel for my vehicle as a function of the miles per gallon x." That is not a related rate problem and there is no need to differentiate the equation.

    Now if you had something like, for instance, "Assume that efficency declines by 0.1 mpg per year, and find the rate at which fuel cost is increasing anually," that would suggest that you diffentiate the equation with regard to time and give you an equation that described the relationship between the rate of change of the efficiency and the rate of change of the cost.

    But as is, this is fairly straightforward: C(10) = 2009, C(15) = 1339.33, C(20) = 1004.5, C(25) = 803.6, etc. As fuel effiecency increases, the annual cost decreases. Your graph is a convex up, downward sloping cuve with a vertical asymptotoe at x = 0 and a horizontal asymptote at y = 0.
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