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Math Help - Find the Formula

  1. #1
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    Find the Formula

    1 Apple = (3.00 + (n x 1.00) ) * P
    1 Date = (3.50 + (n x 1.20) ) * P
    1 Chilli = (3.25 + (n x 1.10) ) * P
    1 Mango = (2.50 + (n x 0.80) ) * P

    Where n = the number you already have.
    and P = your current amount / the maximum amount


    If I have 5 Apples, and the maximum amount is 10

    Cost of buying 1 apple =
    = 3.00 + (5 * 1.00) * 5/10
    = 8.00 * 0.5
    = 4.00


    Cost of buying 2 apples (6th and 7th)
    =[ 3.00 + (5 * 1.00) * 5/10 ] + [ 3.00 + (6 * 1.00) * 6/10 ]
    =(8.00 * 0.5) + (9.00 * 0.6)
    =9.40


    Cost of buying all 5 apples (6th, 7th, 8th, 9th, 10th)
    =(8 * 0.5) + (9 * 0.6) + (10 * 0.7) + (11 * 0.8) + (12 * 0.9)
    =4.00 + 5.40 + 7.00 + 8.80 + 10.80
    =36.00


    But how do I put that into a formula that will work for apples, dates, chillis and mangos When the values will be much larger.
    Example

    Cost of buying 999 apples when;
    Current apples = 1500
    Maximum apple = 2700



    Cost of buying 899 mangos when;
    Current mangos = 2601
    Maximum mango = 3500


    Edit:
    If a minimum value is needed, P can't drop below 0.1
    So.
    If Current/Maximum < 0.1 : P = 0.1
    Last edited by S1eepy; January 20th 2009 at 12:25 PM.
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  2. #2
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    Did I put this is the wrong place? or just no-one has any ideas?

    Im pretty stuck trying to figure it out
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  3. #3
    Junior Member ursa's Avatar
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    1 Apple = 3.00 + (n x 1.00) * P
    1 Date = 3.50 + (n x 1.20) * P
    1 Chilli = 3.25 + (n x 1.10) * P
    1 Mango = 2.50 + (n x 0.80) * P

    Where n = the number you already have.
    and P = your current amount / the maximum amount


    If I have 5 Apples, and the maximum amount is 10

    Cost of buying 1 apple =
    = 3.00 + (5 * 1.00) * 5/10
    = 8.00 * 0.5
    = 4.00


    Cost of buying 2 apples (6th and 7th)
    =[ 3.00 + (5 * 1.00) * 5/10 ] + [ 3.00 + (6 * 1.00) * 6/10 ]
    =(8.00 * 0.5) + (9.00 * 0.6)
    =9.40


    Cost of buying all 5 apples (6th, 7th, 8th, 9th, 10th)
    =(8 * 0.5) + (9 * 0.6) + (10 * 0.7) + (11 * 0.8) + (12 * 0.9)
    =4.00 + 5.40 + 7.00 + 8.80 + 10.80
    =36.00


    But how do I put that into a formula that will work for apples, dates, chillis and mangos When the values will be much larger.
    Example

    Cost of buying 999 apples when;
    Current apples = 1500
    Maximum apple = 2700



    Cost of buying 899 mangos when;
    Current mangos = 2601
    Maximum mango = 3500


    Edit:
    If a minimum value is needed, P can't drop below 0.1
    So.
    If Current/Maximum < 0.1 : P = 0.1
    hi
    first you have placed brackets wrongly, its very confusing
    1 Apple = 3.00 + (n x 1.00) * P
    it should be
    1 Apple = (3.00 + (n x 1.00) )* P
    But how do I put that into a formula that will work for apples, dates, chillis and mangos When the values will be much larger.
    you have to calculate by yourself
    If a minimum value is needed, P can't drop below 0.1
    So.
    If Current/Maximum < 0.1 : P = 0.1
    what this for?????what you need in this???
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  4. #4
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    Quote Originally Posted by ursa View Post
    hi
    first you have placed brackets wrongly, its very confusing

    it should be
    1 Apple = (3.00 + (n x 1.00) )* P


    you have to calculate by yourself


    what this for?????what you need in this???
    sorry about the brackets, my mistake. Changed it in original post

    I thought a minimum value might be needed because while I was trying to work it out, I tried to do a sum of all the values, but it didnt work, because the minimum P is 0.1


    Is there no way to create a formula for this then?
    i started working on this without the P multiplier, which was simple arithmetic progression, and I thought I was just so used to using that I couldn't find the way to introduce P into the equation.


    Before P;
    Sn = (n/2) * ( 2a + (n-1)d)


    But when I had to introduce P
    I tried (Sn) * (average of P)
    but that way doesn't take into account the larger P multiplied by the growing d, so doesnt give an accurate figure.

    then I tried;

    [(n * a) * average P] + (an arithmetic progression with a variable d)

    but I'm pretty sure you cant have d as a variable in an arithmetic series
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  5. #5
    Junior Member ursa's Avatar
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    sorry about the brackets, my mistake. Changed it in original post

    I thought a minimum value might be needed because while I was trying to work it out, I tried to do a sum of all the values, but it didnt work, because the minimum P is 0.1


    Is there no way to create a formula for this then?
    i started working on this without the P multiplier, which was simple arithmetic progression, and I thought I was just so used to using that I couldn't find the way to introduce P into the equation.


    Before P;
    Sn = (n/2) * ( 2a + (n-1)d)


    But when I had to introduce P
    I tried (Sn) * (average of P)
    but that way doesn't take into account the larger P multiplied by the growing d, so doesnt give an accurate figure.

    then I tried;

    [(n * a) * average P] + (an arithmetic progression with a variable d)

    but I'm pretty sure you cant have d as a variable in an arithmetic series
    if u need a formula then you can try this one, its a bit complicated
    and only for
    1 Apple = (3.00 + (n x 1.00) ) * P
    ( ((d^2 + d -2)n) + (((2d + 1)(n)(n+1))/2) + (((n)(n+1)(n+2))/6) )/m
    d->no. of apples in our hand
    n->no. of apple to be buy
    m->maximum amount
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  6. #6
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    Quote Originally Posted by ursa View Post
    if u need a formula then you can try this one, its a bit complicated
    and only for

    ( ((d^2 + d -2)n) + (((2d + 1)(n)(n+1))/2) + (((n)(n+1)(n+2))/6) )/m
    d->no. of apples in our hand
    n->no. of apple to be buy
    m->maximum amount
    Current = 5
    Want = 5 more
    top = 10

    ( ((5^2 + 5 -2)5) + (((2*5 + 1)(5)(5+1))/2) + (((5)(5+1)(5+2))/6) )/10

    ((28*5) + ((11 * 5 * 6)/2) + ((5*6*7)/6) /10

    (140 + 165 + 35) /10

    =34

    Should be 36?


    and I think what Ive been doing is wrong, seen as I seem to be going down a completely different line.
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  7. #7
    Junior Member ursa's Avatar
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    Current = 5
    Want = 5 more
    top = 10

    ( ((5^2 + 5 -2)5) + (((2*5 + 1)(5)(5+1))/2) + (((5)(5+1)(5+2))/6) )/10

    ((28*5) + ((11 * 5 * 6)/2) + ((5*6*7)/6) /10

    (140 + 165 + 35) /10

    =34

    Should be 36?


    and I think what Ive been doing is wrong, seen as I seem to be going down a completely different line.
    oh
    i m sorry
    i typed wrong formula
    actually it is

    ( ((d^2 + d -2)n) + (((2d + 1)(n)(n+1))/2) + (((n)(n+1)(2n+1))/6) )/m
    d->no. of apples in our hand
    n->no. of apple to be buy
    m->maximum amount

    now try.
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  8. #8
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    Current = 5
    Want = 5 more
    top = 10


    ( ((5^2 + 5 -2)5) + (((2*5 + 1)(5)(5+1))/2) + (((5)(5+1)(2*5+1))/6) )/10

    ((28*5) + (11*5*6/2) + (5*6*11/6) )/10

    (140 + 165 + 55) /10

    =36 cool =)
    Thanks for the help

    Just to test

    When
    Current = 1500
    Want = 200 more
    top = 2900
    Should cost 177,002.2069

    ( ((1500^2 + 1500 -2)200) + (((2*1500 + 1)(200)(200+1))/2) + (((200)(200+1)(2*200+1))/6) )/2900
    (450299600 + 60320100 + 2686700) / 2900
    =177002.20689

    Sweet, thanks again
    Apples, 3.00 1.00
    -----------

    hmm, so why doesnt it work for the others?

    Mango, 2.50 0.80
    Current = 5
    Want = 5 more
    top = 10


    Cost of buying 6th =
    = 2.50 + (5 * 0.80) * 5/10
    = 6.50 * 0.5
    = 3.25


    Cost of buying 2 mangos (6th and 7th)
    =[ 2.50 + (5 * 0.80) * 5/10 ] + [ 2.50 + (6 * 0.80) * 6/10 ]
    =(6.50 * 0.5) + (7.30 * 0.6)
    =7.63


    Cost of buying all 5 mangos (6th, 7th, 8th, 9th, 10th)
    =(6.5 * 0.5) + (7.3 * 0.6) + (8.1 * 0.7) + (8.9 * 0.8) + (9.7 * 0.9)
    =3.25+4.38+5.67+7.12+8.73
    =20.42



    ( ((d^2 + d -2)n) + (((2d + 1)(n)(n+1))/2) + (((n)(n+1)(2n+1))/6) )/m
    d->no. of apples in our hand
    n->no. of apple to be buy
    m->maximum amount

    *works on it*
    Last edited by S1eepy; January 20th 2009 at 02:57 PM.
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  9. #9
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    c->Current no. of apples in our hand
    n->No. of apple to be buy
    t->Total allowed
    a->original cost of food
    d->changing cost

    Ive got it to

    When n=2 [ a( c + (c+1) ) + d( c^2 + (c+1)^2 ) ] /t
    When n=3 [ a( c + (c+1) + (c+2) ) + d( c^2 + (c+1)^2 + (c+2)^2) ] /t
    When n=4 [ a( c + (c+1) + (c+2) + (c+3) ) + d( c^2 + (c+1)^2 + (c+2)^2 + (c+3)^2) ] /t

    So
    Sn = [ a( nc + ( n(n-1)/2 ) ) + d( n(c^2) + (n^2 - n) + ???? ) ] /t


    Im unsure how to finish the equation... so close..
    should be;
    +1 when n=2
    +5 when n=3
    +14 when n=4
    +30 when n=5
    +55 when n=6
    Last edited by S1eepy; January 24th 2009 at 06:45 AM.
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  10. #10
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    Sn = [ a( nc + ( n(n-1)/2 ) ) + d( n(c^2) + (n^2 - n) + ???? ) ] /t
    You appear to have done quite well so far. So well, in fact, that the only thing I think I need to tell you is that
    1^2+2^2+3^2+4^2...+n^2 = \frac{n(n+1)(2n+1)}{6}
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  11. #11
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    Quote Originally Posted by badgerigar View Post
    You appear to have done quite well so far. So well, in fact, that the only thing I think I need to tell you is that
    1^2+2^2+3^2+4^2...+n^2 = \frac{n(n+1)(2n+1)}{6}
    That gives;
    +1 when n=1
    +5 when n=2
    +14 when n=3
    +30 when n=4
    +55 when n=5

    I need
    +1 when n=2
    +5 when n=3
    +14 when n=4
    +30 when n=5
    +55 when n=6

    \frac{n(n-1)(2-1)}{6} works

    cheers for the help


    Now the formula is;
    [an(c+\frac{n-1}{2}) + dcn(c+n-1) + (\frac{n(n-1)(2-1)}{6})] /t

    Thanks for the help =D
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  12. #12
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    You probably just made a mistake typing this out, but since it appeared twice in your post I thought I had better point out that there is an n missing in this. It should read
    <br />
[an(c+\frac{n-1}{2}) + dcn(c+n-1) + (\frac{n(n-1)(2n-1)}{6})] /t
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