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Thread: need math help

  1. #1
    Storm's Avatar Poster BT Rep: +3
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    im kinda rusty with math, its been some time since i did this, so id like your help

    i need to find the top of the following function:

    Y(T)= (c*r^T - 10000)*( (r+0.02)^(50-T) )

    c is starting cash, and r is interest, both are known

    anyone know how to solve this one? really cant remember how its done...... thnx in advance

  2. The Drawing Room   -   #2
    Differentiation? Sorry too lazy to do it myself
    Differentiation of something to the power of x

    Product rule


    edit: then solve for derivative = 0, remmber to check is a maxima not minima
    Last edited by ilw; 01-08-2005 at 12:23 PM.

  3. The Drawing Room   -   #3
    Storm's Avatar Poster BT Rep: +3
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    yikes!!!!


    so i split it in (s=r+0.02)
    f(T)=(c*r^T - 10000) and
    g(T)=(r+0.02)^(50-T) = (s^50)*(s^-T)

    f'(T)= c (ln r)r^T
    g'(T)=(s^50)(ln s)(s^-T)

    so f(T)*g'(T)+f'(T)*g(T)=
    (c*r^T - 10000)(s^50)(ln s)(s^-T)+(c (ln r)r^T)(s^50-T)

    try equaling THAT to 0 makes me wish i took that graphic calculator back from my ex

    well anyway thanks for your help

  4. The Drawing Room   -   #4
    tesco's Avatar woowoo
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  5. The Drawing Room   -   #5
    Skiz's Avatar (_8(I)
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    *scratches head*



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  6. The Drawing Room   -   #6
    S!X's Avatar L33T Member BT Rep: +5
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    *no clue*

  7. The Drawing Room   -   #7
    cpt_azad's Avatar Colonel
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    umm at first i thought it was compound interest, then i looked at ilw's post and now i'm just

    Jeff Loomis: He's so good, he doesn't need to be dead to have a tribute.

  8. The Drawing Room   -   #8
    It is possible and some bits do cancel out, I did it roughly and my final answer was nasty, but plausible i think.

    I noticed i have an extra -ve sign in my g'(x) from taking the derivative of (50-T), if u don't get where this comes from, try following the stages in that 'something to the power of x' thing i posted before, with the actual function we have.
    ie g'(T)=-(s^50)(ln s)(s^-T)

    and i put the s to the power (50-T) back together ie
    g'(T)=-ln(s) * s^(50-T)


    setting the derivative to 0 and multiplying out the (c*r^T - 10000) I get

    [c * ln(r) * r^T * s^(50-T)] - [c * ln(s) * s^(50-T) * r^T] + [10000* ln(s) * s^(50-T)] = 0

    take the 10000 ln(s)... term across, cancel the s^(50-T) terms and do some jiggery pokery with brackets and logarithms and i got:

    r^T = (10000 * ln(s)) / (c * ln(s/r) )

    and this is equivalent to:

    T = ln [(10000 * ln(s)) / (c * ln(s/r) )] / [ln(r)]

    very messy answer, but an answer nevertheless
    btw i've most likely made a mistake or 2 somewhere. I'm confident up to the g' stage though.

  9. The Drawing Room   -   #9
    Storm's Avatar Poster BT Rep: +3
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    ok, tried to do the same as you ilw, and i think your right.........

    (c*r^T-10000)(-(ln s)*s^(50-T))+c(ln r)r^T*s^(50-T)=0
    multiply out
    -c*r^T*(ln s)*s^(50-T)-10000(ln s)*s^(50-T)+c(ln r)r^T*s^(50-T)=0
    divide by s^(50-T)
    -c*r^T*(ln s)-10000(ln s)+c(ln r)r^T=0
    sort
    (c(ln r)-c(ln s0))r^T=10000(ln s)
    c(ln (r/s))r^T=10000(ln s)
    move
    r^T=10000(ln s)/c(ln r/s)
    take the ln of the function
    T=(ln (10000(ln s)/c(ln r/s)))/(ln r)

    THANKS FOR THE HELP MAN!!!!!!

    Quote Originally Posted by cpt_azad
    umm at first i thought it was compound interest, then i looked at ilw's post and now i'm just
    it is compound interest, but the interest increases after you pay 10 000

  10. The Drawing Room   -   #10
    DanB's Avatar Smoke weed everyday
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    wtf

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