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Math Help - The number of Dragon heads??

  1. #1
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    Cool The number of Dragon heads??

    The thinking power of a multi-headed dragon depends on how many heads it has. The thinking power of a group of dragons is the product of the number of heads on the individual dragons. A particular group has 100 heads available, how many dragons with what number of heads will maximise the thinking power of the group?
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    Re: The number of Dragon heads??

    Quote Originally Posted by darandoma View Post
    The thinking power of a multi-headed dragon depends on how many heads it has. The thinking power of a group of dragons is the product of the number of heads on the individual dragons. A particular group has 100 heads available, how many dragons with what number of heads will maximise the thinking power of the group?
    Well, what have you tried so far. You could start by assuming all the dragons have the same number of heads and see where that takes you.

    CB
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    Re: The number of Dragon heads??

    all I can do is x = number of dragons, n = number of heads ...

    thinking power is : n^(100/x) but I am really comfused :/
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    Re: The number of Dragon heads??

    Quote Originally Posted by darandoma View Post
    all I can do is x = number of dragons, n = number of heads ...

    thinking power is : n^(100/x) but I am really comfused :/
    If n is the number of heads then 100/n is the number of dragons and the thinking power is n^{100/n}

    CB
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    Re: The number of Dragon heads??

    Yeh I just got that and found that 3^32 x "2^2 is the highest, is that what you got
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    Re: The number of Dragon heads??

    Quote Originally Posted by darandoma View Post
    Yeh I just got that and found that 3^32 x "2^2 is the highest, is that what you got
    Yes, though I has 32 with 3 heads and one with 4 heads, but that gives the same as 32 with 3 and 2 with 2.

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    Re: The number of Dragon heads??

    Quote Originally Posted by darandoma View Post
    The thinking power of a multi-headed dragon depends on how many heads it has. The thinking power of a group of dragons is the product of the number of heads on the individual dragons. A particular group has 100 heads available, how many dragons with what number of heads will maximise the thinking power of the group?
    Suppose there were only two dragons, one with n heads, the other with 100- n. Then the "thinking power" would be n(100- n)= 100n- n^2. The maximum of that will come where 100- 2n= 0 or n= 50 and would be 50^2= 2500. If there were three dragons, with n_1, n_2 and 100- n_1- n_2 heads, the "thinking power" is n_1(n_2)(100- n_1- n_2)= 100n_1n_2- n_1^2n_2- n_1n_2^2. The maximum of that will come where 100n_2- 2n_1n_2+ n_2^2= n_2(100- 2n_1- n_2)= 0 and 100n_1- n_1^2- 2n_1n_2= n_1(100- n_1- 2n_2)= 0. Obviously, neither n_1= 0 nor n_2= 0 is acceptable so we must have 100- 2n_1- n_2= 0 and 100- n_1- 2n_2= 0. From the first equation, n_2= 100- 2n_1. Putting that into the second equation, 100- n_1- 2(100- 2n_1)= 3n_1- 100= 0 so that n_1= 100/3, n_2= 100-2n_1= 100/3 and, of course, the third dragon has 100- 100/3- 100/3= 100/3 heads. The "thinking power" is now \frac{100}{3}\frac{100}{3}\frac{100}{3}= \frac{1000000}{27} which is about 37037. Do you see the point? For a given number of dragons, the maximum "thinking power" occurs when all dragons have the same number of heads (try to show that algebraically). If there are n dragons, each having 100/n heads, then their "thinking power" is (100/n)^n.

    Here, n must be an integer but if we think of this as a continuous variable, y= (100/x)^x we can take the derivative and set it equal to 0. Specifically, ln(y)= xln(100/x)= xln(100)- xln(x) so \frac{y'}{y}= ln(100)- ln(x)- 1 and so y'= (ln(100)- ln(x)-1)(100/x)^x. Setting that equal to 0 we must have either (100/x)^x= 0, which is impossible, or ln(100)- ln(x)- 1= 0 so that ln(x)= ln(100)- 1= ln(100)- ln(e)= ln(100/e) so that x= 100/e= 36.788[tex] which rounds to 37 dragons. Now, 37 does not divide 100 evenly- it is about 2.7 so we cannot have our "ideal" of all dragons with the same number of heads. I would suggest allocating the 100 heads to 37 dragons, giving some 2 heads, others 3 heads.

    That is the as solving the pair of equations x+ y= 37, 2x+ 3y= 100 where x is the number of dragons with 2 heads, y the number with 3 heads.
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