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Math Help - Word Problem

  1. #1
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    Word Problem

    Hello everybody,

    Just needed some help on this problem.

    Write the Function C(x) which is the annual cost of fuel for your vehicle as a function of the miles per gallon X. (Do not use a specific number for miles per gallon for X here)

    cost of fuel = 3.98
    car gets 27.5 on average.

    So far all I can think of is C(x) = 3.98x then I'm lost. Any help would be appreciated. Thanks.
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by Afterme View Post
    Hello everybody,

    Just needed some help on this problem.

    Write the Function C(x) which is the annual cost of fuel for your vehicle as a function of the miles per gallon X. (Do not use a specific number for miles per gallon for X here)

    cost of fuel = 3.98
    car gets 27.5 on average.

    So far all I can think of is C(x) = 3.98x then I'm lost. Any help would be appreciated. Thanks.
    Let x be the number of miles driven then

    \left( \frac{x \mbox{ miles}}{1}\right)\left( \frac{1 \mbox{ gallon}}{27.5 \mbox{ miles }}\right)\left( \frac{3.98 \mbox{ dollars}}{\mbox{ gallon} }\right) \approx 0.14472x \mbox{ dollars}

    So the cost of driving the car x miles in dollars is

    C(x) \approx 0.14472x

    I hope this helps.

    Good luck.
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  3. #3
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    Sorry but that doesn't seem really clear to me

    How would I even write a function from this?
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  4. #4
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    Quote Originally Posted by Afterme View Post
    Sorry but that doesn't seem really clear to me

    How would I even write a function from this?

    What part do you have a question about?

    The function is C(x) \approx 0.14472x

    Please be specific and I will try to answer your questions.

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  5. #5
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    Well the next problem is

    Find the derivative function dC
    dx

    then the next problem is

    Complete the following table:

    x 10 15 20 25 30 35 40
    c
    dC/dx

    I guess my question is. The derivative of .14472x is .14472 so how would I even work that into the next problem which is that table?

    Also the first problem says: write the function C(x) which is the annual cost of fuel for your vehicle. So through the function C(x) = .14472x how would I even find the annual cost of fuel? The average miles I drive a year is 14,000 so would that have any play in this equation?
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  6. #6
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    Quote Originally Posted by Afterme View Post
    Well the next problem is

    Find the derivative function dC
    dx

    then the next problem is

    Complete the following table:

    x 10 15 20 25 30 35 40
    c
    dC/dx

    I guess my question is. The derivative of .14472x is .14472 so how would I even work that into the next problem which is that table?

    Also the first problem says: write the function C(x) which is the annual cost of fuel for your vehicle. So through the function C(x) = .14472x how would I even find the annual cost of fuel? The average miles I drive a year is 14,000 so would that have any play in this equation?
    Okay so I have a reading problem. Sorry you want to implut mpg and get the cost of driving 14000 miles.

    C(x)=\left( \frac{14000 \mbox{ miles }}{1}\right) \left( \frac{3.98 \mbox{ dollars }}{1 \mbox{ gallon }}\right)\left( \frac{1}{x \frac{miles}{gallon}}\right)

    Simplifying we get

    C(x)=\frac{5572}{x}\mbox{dollars}

    Where x is mpg for your car.

    Again sorry about my reading problem good luck.
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  7. #7
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    If I found the derivative of that function C(x)= 55720/x my answer would come out to

    -55720/x^2 is that correct? Just need a double check.

    Also how would I start on that table posted above?

    btw thanks for your help. =)


    Also I started plugging in the numbers for the table above and the dC/dx values are coming out negative. What does this mean?
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  8. #8
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    -55720/x^2 is that correct? Just need a double check.
    exactly right. Good job

    Also I started plugging in the numbers for the table above and the dC/dx values are coming out negative. What does this mean?
    This means that when you increase the miles per gallon, it costs less. Whenever the derivative is negative it means that increasing your input decreases the value of the function.
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