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Math Help - Product Rule in a word problem

  1. #1
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    Product Rule in a word problem

    I have no clue how this problem relates to the Product Rule - it seems to relate more to average rates. I would really appreciate it if someone would help me understand this and walk me through the steps to get the answer:

    In this exercise we estimate the rate at which the total personal income is rising in a metropolitan area. In 1999, the population of this area was 939,200, and the population was increasing at roughly 9500 per year. The average annual income was $30,388 per capita, and this average was increasing at about $1300 per year (a little above the national average of about $1225 yearly). Use the Product Rule and these figures to estimate the rate at which total personal income was rising in the area in 1999.
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  2. #2
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    Hello, Maziana!

    We estimate the rate at which the total personal income is rising in a city.

    In 1999, the population of the city was 939,200, and increasing at 9500 per year.
    The average annual income was $30,388 per capita, and increasing at $1300 per year.

    Use the Product Rule and these figures to estimate the rate
    at which total personal income was rising in the area in 1999.

    Let x = number of years since 1999.

    After x years, the population is: . 939,\!200 + 9500x

    After x years, the average income is: . 30\!,388 + 1300x dollars.

    Hence, after x years, the total income is: . I \;=\;(939,\!200 + 9500x)(30,\!388 + 1300x)


    They are asking for \frac{dI}{dx}

    Differentiate I using the Product Rule.

    . . \frac{dI}{dx} \;=\;(939,\!200 + 9500x)\cdot1300 + 9500(30,\!388 + 1300x) \quad\hdots and simplify.

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  3. #3
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    Thank you!

    I = 1220960000 + 12350000x + 288686000 + 12350000x

    I = 1509646000 + 24700000x

    What now? I don't think I can continue:

    24700000x = 1509646000

    x = -61.12

    That can't be right, since the rate should be positive...
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  4. #4
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    Hello, Maziana!

    \frac{dI}{dx} \;=\; 1,\!220,\!960,\!000 + 12,\!350,\!000x + 288,\!686,\!000 + 12,\!350,\!000x

    . . \frac{dI}{dx} \;=\;1,\!509,\!646,\!000 + 24,\!700,\!000x

    What now? I don't think I can continue.

    Continue to where?

    You've answered the question.

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  5. #5
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    I am expected to continue and get a number as an answer. The rate of change in which income was rising in the year 1999 - the answer is not exactly but is probably close to $1.5 billion/year, as I have figured out from the answer in the back of my math book. I'm not sure how to get there... any ideas? Thanks so much.
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  6. #6
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    Not sure if you still need help but it's obvious that the intercept at which your derivative has increased to has reached 1.5 billion (1,509,646,000). That is your answer to this problem. It has, and will aproximately increase that amount each year supposedly. I am still unsure myself why it is that number, but it's because of the population I believe being included in the number.

    Think of it this way: instead of just OOOONE person's average income..

    ITS THE ENTIRE POPULATION. IN 1999.

    Hope that helps (this is what I got from my professor) and what I got from the product rule. I was having trouble with this too, so I went ahead and tried my new answer just like this (mine was different because of different variables. This is it:

    In this exercise we estimate the rate at which the total personal income is rising in a metropolitan area. In 1999, the population of this area was 963,700, and the population was increasing at roughly 10000 per year. The average annual income was $34,549 per capita, and this average was increasing at about $1400 per year (a little above the national average of about $1225 yearly). Use the Product Rule and these figures to estimate the rate at which total personal income was rising in the area in 1999.
    )

    I ended up with about 1.6 billion.

    It's
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