The number of Dragon heads??

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September 21st 2011, 11:04 PM
darandoma

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?
September 21st 2011, 11:53 PM
CaptainBlack

Re: The number of Dragon heads??

Quote:

Originally Posted by darandoma

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
September 22nd 2011, 12:55 PM
darandoma

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 :/
September 22nd 2011, 07:28 PM
CaptainBlack

Re: The number of Dragon heads??

Quote:

Originally Posted by darandoma

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
September 22nd 2011, 11:01 PM
darandoma

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
September 22nd 2011, 11:53 PM
CaptainBlack

Re: The number of Dragon heads??

Quote:

Originally Posted by darandoma

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.

CB
September 23rd 2011, 07:32 AM
HallsofIvy

Re: The number of Dragon heads??

Quote:

Originally Posted by darandoma

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.