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  1. #1
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    recursive formula help

    Here is the problem: A company maintains 50,000 plants and sells 10% annually. They plant 4000 new plants each year. Write a recursive formula and find the number of plants being maintained in the 20th year of the business.

    Any help?
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    Quote Originally Posted by iknowyou View Post
    Here is the problem: A company maintains 50,000 plants and sells 10% annually. They plant 4000 new plants each year. Write a recursive formula and find the number of plants being maintained in the 20th year of the business.

    Any help?
    The question is a little unclear, I think by "maintains" they mean "starts out with." Also it's unclear whether the new plants are included when calculating that 10% get sold. I'll assume not. Additionally, it's not clear whether 50,000 should be the number of plants for year zero or for year one! I'll assume one.

    So we can say

    f(1) = 50000

    for all integers n > 1: f(n) = 0.9 * f(n-1) + 4000

    I get f(20) approximately equals 41350.85
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  3. #3
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    Quote Originally Posted by iknowyou View Post
    Here is the problem: A company maintains 50,000 plants and sells 10% annually. They plant 4000 new plants each year. Write a recursive formula and find the number of plants being maintained in the 20th year of the business.

    Any help?
    Sorry- for some reason I was thinking "sold" rather than "planted" 4000 new plants. Where I had "-4000" I have changed to "+4000".

    Just write down exactly what is said. If in year n, the company maintains A_n plants, sells 10% of them, 0.1A_n, and [b]plants 4000, then in year n+1, the company will have A_{n+1}= A_n- 0.1A_n+ 4000= 0.9A_n+ 4000. As undefined says, the "50,000 plants" must be some initial value. It wouldn't make sense to say the maintain 50,000 plants every year and then ask how many they maintain in the twentieth year!

    Given that A_0= 50000 and that A_{n+1}= 0.9A_n+ 4000, can you find A_{20}?
    Last edited by HallsofIvy; September 18th 2010 at 04:43 AM. Reason: Wrong equation
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  4. #4
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    Just write down exactly what is said. If in year n, the company maintains A_n plants, sells 10% of them, 0.1A_n, and sells 4000, then in year n+1, the company will have A_{n+1}= A_n- 0.1A_n- 4000= 0.9A_n- 4000. As undefined says, the "50,000 plants" must be some initial value. It wouldn't make sense to say the maintain 50,000 plants every year and then ask how many they maintain in the twentieth year!

    Given that A_0= 50000 and that A_{n+1}= 0.9A_n- 4000, can you find A_{20}?
    Why did you subtract 4000 instead of add? Was that a typo or did I misinterpret what it means to plant new plants?
    Last edited by undefined; September 17th 2010 at 03:35 AM. Reason: typo
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  5. #5
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    Hello, iknowyou!

    undefined is right . . . The wording is not clear.
    I'll make my own interpretation and solve it.


    A company starts with 50,000 plants and sells 10% annually.
    They plant 4000 new plants each year.
    (a) Write a recursive formula for the number of plants in year \,n
    (b) Find the number of plants being maintained in the 20th year.

    Each year they have 90% of the previous year's plants ... plus 4000 new plants.


    (a) We have: . \boxed{f(n) \;=\;\tfrac{9}{10}f(n\!-\11) + 4000} .[1]

    Write the equation for the "next year":
    . . . . . . . . . . f(n\!+\!1) \;=\;\frac{9}{10}f(n) + 4000 .[2]


    Subtract [2] - [1]: . f(n\!+\!1)-f(n) \;=\;\frac{9}{10}f(n) - \frac{9}{10}f(n\!-\!1)

    and we have: . f(n\!+\!1) - \frac{19}{10}f(n) + \frac{9}{10}f(n\!-\!1) \;=\;0


    Let f(n) \,=\,X^n

    We have: . X^{n+1} - \frac{19}{10}X^n + \frac{9}{10}X^{n-1} \;=\;0

    Divide by X^{n-1}\!:\;\;X^2 - \frac{19}{10}X + \frac{9}{10} \;=\;0

    Factor: . (X - 1)\left(X - \frac{9}{10}\right) \;=\;0

    Hence: . X \,=\,1,\;X \,=\,\frac{9}{10}


    The generating function is of the form: . f(n) \;=\;A + B\left(\frac{9}{10}\right)^n

    We know that: . f(0) = 50,\!000,\;\;f(1) = 49,\!000

    We have: . \begin{Bmatrix}f(0) = 50,\!000\!: & A + B &=& 50,\1000 & [3] \\ f(1) = 49,\!000\!: & A + \frac{9}{10}B &=& 49,\!000 & [4] \end{Bmatrix}

    Subtract [3] - [4]: . \frac{1}{10}B \:=\:1,\!000 \quad\Rightarrow\quad B \:=\:10,\!000

    Substitute into [3]: . A + 10,\!000 \:=\:50,\!000 \quad\Rightarrow\quad A \:=\:40,\!000


    The generating function is: . f(n) \;=\;40,\!000 + 10,\!000\left(\frac{9}{10}\right)^n


    (b) In the 20th year: . f(20) \;=\;40,\1000 + 10,\!000\left(\frac{9}{10}\right)^{20} \;\approx\;\boxed{41,216}

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  6. #6
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    Yes, that was a typo. I have changed it to +4000.
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