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Math Help - Average Rate of Change

  1. #1
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    Average Rate of Change

    I need some help with this problem:
    A company introduces a new product for which the number of units sold S is given by the equation below, where t is the time in months.


    I need to:
    a) Find the average rate of change of S(t) during the first year (rounded to 1 decimal place).

    And:
    b) During what month does S'(t) equal the average rate of change during the first year?

    I'm not too sure what to do for the first one. Some people have told me to find the derivative of the problem then plug 12 in for t, but that's not right. I could probably find B after I find A.
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  2. #2
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    The average rate of change on an closed interval [a,b] is given by
    \frac{S(b)-S(a)}{b-a}

    So in your case a=0, b=12. And the average rate of change is given by
    \frac{S(12)-S(0)}{12-0}

    For the second part, just use the mean value theorem
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  3. #3
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by BeSweeet View Post
    I need some help with this problem:
    A company introduces a new product for which the number of units sold S is given by the equation below, where t is the time in months.


    I need to:
    a) Find the average rate of change of S(t) during the first year (rounded to 1 decimal place).

    And:
    b) During what month does S'(t) equal the average rate of change during the first year?

    I'm not too sure what to do for the first one. Some people have told me to find the derivative of the problem then plug 12 in for t, but that's not right. I could probably find B after I find A.
    The average rate of change is given by the slope of the line drawn through 2 points on a graph. So, the points in question here are, (0,S(0)), and (12,S(12)). Can you find the slope?
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  4. #4
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    Hmm... So I got 15.2 for A, which is incorrect. I plugged 12 in for t and got 182.8, then dividing that by 12 gave me 15.2. Plugging 0 in for [t] doesn't seem possible, since 9/0 is undefined...
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  5. #5
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    Actually I did that wrong. So I got 189.4736 for s(12), then 56.25 for s(0). I subtracted S(12) and S(0), and got 133.223684211. I then divided that by 12 and got 11.1, which is incorrect.
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  6. #6
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    Bump...

    Ok, so I got 182.8125 for S(12) and 56.25 for S(0). Subtracting the two and dividing by 12 gives me 10.6 (rounding to one decimal).

    So I took the derivative of the original problem and set it equal to 10.6. I got two answers; -11.9799 and 3.97993.

    This is the last submission I have for this problem, so I need to make sure this is right ahead of time. Is part A correct? For part B, I just have to choose the month from a drop-down menu.
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  7. #7
    Member Haven's Avatar
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    Part A)
    S(12) = 182.8125, S(0) = 56.25
    S_{average} = \frac{S(12)-S(0)}{12-0} = \frac{126.5625}{12} = 10.546875
    Part B
    \frac{dS(t)}{dt} = \frac{675}{(4+t)^2}
    By MVT:
    \frac{675}{(4+t)^2} = 10.546875
    Solve for t
    t = \{-12,4\}
    Since there is no such thing a negative month, we use t=4
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  8. #8
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    Quote Originally Posted by Haven View Post
    Part A)
    S(12) = 182.8125, S(0) = 56.25
    S_{average} = \frac{S(12)-S(0)}{12-0} = \frac{126.5625}{12} = 10.546875
    Part B
    \frac{dS(t)}{dt} = \frac{675}{(4+t)^2}
    By MVT:
    \frac{675}{(4+t)^2} = 10.546875
    Solve for t
    t = \{-12,4\}
    Since there is no such thing a negative month, we use t=4
    I figured out part A, that was sorta simple (after seeing what others have posted). I guess I got the answer wrong, because I put 10.6 (rounding up...), but I guess they were looking for 10.5.

    So anyway, I sort of understood part B. Although the correct month was May, I didn't understand how they got that.
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