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Math Help - Geometric progression question

  1. #1
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    Geometric progression question

    Question:

    A chicken farmer has 1000 chickens that cost $0.50 per chicken per week to rear at the start of each week. On the last day of every week, he sells exactly k chickens, where k is a positive integer and 1000 is divisible by k, to a restaurant. In the first week, the price of each chicken is $ 12. After each week, the price of the remaining chickens drops by 5 % the existing price. The sale continues until the farmer has sold all of his 1000 chickens.

    Show that, when he has sold all his chickens, the total cost of rearing the chickens is $ (250/k)(1000 + k)

    # There are actually 2 parts to the question, but I would like to try the second part myself...so please help me with the first part. Thank you very much!
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  2. #2
    Moo
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    Hello,

    Because at the end of each week he sells k chickens, there are \frac{1000}{k} weeks of selling (and rearing).

    THE REARING
    1st week : 1000x0.5
    2nd week : (1000-k)x0.5
    3rd week : (1000-2k)x0.5
    .
    .
    .
    \left(\frac{1000}{k}-1\right) th week : \left(1000-\left(\frac{1000}{k}-1\right)k\right) \times 0.5 because so far, he has sold \left(\frac{1000}{k}-1\right)k chickens.
    \frac{1000}{k} th week : \left(1000-\left(\frac{1000}{k}\right)k\right) \times 0.5=0


    So the total is :

    1000 \times 0.5+(1000-k) \times 0.5+(1000-2k) \times 0.5+ \dots + \left(1000-\left(\frac{1000}{k}-1\right)k\right) \times 0.5

    Oh yeah, this is awful...

    Factoring by 0.5, we get :

    S=0.5 \times \left(1000+(1000-{\color{red}k})+(1000-{\color{red}2k})+ \dots +\left(1000-{\color{red}\left(\frac{1000}{k}-1\right)k}\right)\right)


    You can see that 1000 minus the red terms belong to an arithmetic sequence of general term : a_n=1000-nk, starting at n=0, finishing at n= \frac{1000}{k}-1, and of progression -k
    ---> there are \frac{1000}{k} terms.

    So :

    S=0.5 \times \left(\frac{n}{2} (2a_1+(n-1)r) \right)
    This is the formula you should know. But we're starting at a_0, so it's different

    S=0.5 \times \left(\frac{\color{red}n+1}{2} (2a_{\color{red}0}+{\color{red}n}r)\right)

    a_0=1000, n=\frac{1000}{k}-1, r=-k


    ---> S=\frac{0.5}{2} \times \left(\frac{1000}{k} \left(2000-\left(\frac{1000}{k}-1\right)k\right)\right)

    \begin{aligned} S&=0.25 \times \frac{1000}{k} \times \left(2000-1000+k\right) \\<br />
&=\boxed{\frac{250}{k} \times (1000+k)} \end{aligned}





    If something is unclear, tell me.. And if you think there is an easier way.. I'd agree with you
    Last edited by Moo; June 13th 2008 at 08:09 AM.
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  3. #3
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    Hello, Tangera!

    A chicken farmer has 1000 chickens that cost $0.50 per chicken
    per week to rear at the start of each week.
    On the last day of every week, he sells exactly k chickens,
    where k is an integer and 1000 is divisible by k, to a restaurant.

    Show that, when he has sold all his chickens,
    . . . the total cost of rearing the chickens is: . \frac{250}{k}(1000 + k) dollars.
    Let n = number of weeks that he sold chickens.
    . . Then: . n \,=\,\frac{1000}{k} .[1]


    Let's chart out his cost . . .

    \begin{array}{ccc}\text{Week} & \text{Chickens} & \text{Cost (\$)} \\ \hline \\[-3mm]<br />
1 & 1000 & \frac{1}{2}(1000) \\ \\[-3mm]<br />
2 & 1000-k & \frac{1}{2}(1000-k) \\ \\[-3mm]<br />
3 & 1000-2k & \frac{1}{2}(1000-2k) \\ \\[-3mm]<br />
4 & 1000-3k & \frac{1}{2}(1000-3k) \\<br />
\vdots & \vdots & \vdots \\<br />
n & 1000-(n-1)k & \frac{1}{2}(1000-[n-1]k) \\ \hline\end{array}


    Total cost: . \frac{1}{2}(1000) + \frac{1}{2}(1000-k) + \frac{1}{2}(1000-2k) + \frac{1}{2}(1000-3k) + \hdots + \frac{1}{2}(1000 - [n-1]k)

    . . = \;\frac{1}{2}\bigg[1000 + (1000-k) + (1000-2k) + (1000-3k) + \hdots + (1000-[n-1]k)\bigg]

    . . = \;\frac{1}{2}\bigg[\underbrace{1000 + 1000 + \hdots + 1000}_{n\text{ terms}} - (k + 2k + 3k + \hdots + [n-1]k)\bigg]

    . . = \;\frac{1}{2}\bigg[1000n - k(1 + 2 + 3 + \hdots + [n-1])\bigg] \;= \;\frac{1}{2}\bigg[1000n - k\,\frac{n(n-1)}{2}\bigg]

    . . =\;\frac{n}{4}\bigg[2000-k(n-1)\bigg] \;=\;\frac{n}{4}\bigg[2000 -kn + k\bigg]


    Substitute [1]: . \frac{1000}{4k}\bigg[2000 - k\!\cdot\!\frac{1000}{k} + k\bigg] \;=\;\boxed{\frac{250}{k}(1000 + k)}


    Edit: Nice work, Moo!
    .
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  4. #4
    Moo
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    Hi Soroban,

    Your table is impressive... the forum doesn't even want to show that gigantic work of yours with the latex code
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  5. #5
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    Quote Originally Posted by Moo View Post
    Hello,

    Because at the end of each week he sells k chickens, there are \frac{1000}{k} weeks of selling (and rearing).

    THE REARING
    1st week : 1000x0.5
    2nd week : (1000-k)x0.5
    3rd week : (1000-2k)x0.5
    .
    .
    .
    \left(\frac{1000}{k}-1\right) th week : \left(1000-\left(\frac{1000}{k}-1\right)k\right) \times 0.5 because so far, he has sold \left(\frac{1000}{k}-1\right)k chickens.
    \frac{1000}{k} th week : \left(1000-\left(\frac{1000}{k}\right)k\right) \times 0.5=0


    So the total is :

    1000 \times 0.5+(1000-k) \times 0.5+(1000-2k) \times 0.5+ \dots + \left(1000-\left(\frac{1000}{k}-1\right)k\right) \times 0.5
    Hello Moo! Could you explain why is it that he would have sold \left(\frac{1000}{k}-1\right)k chickens for the \left(\frac{1000}{k}-1\right) th week? I thought at the start of that week he would only have sold \left(\frac{1000}{k}-2\right)k chickens...?
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  6. #6
    Moo
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    Because you're right

    Couting is not what I prefer... Nice remark
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