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 Subject: "Help no one else seems to be able to give me..." Previous Topic | Next Topic
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guest
Jan-23-07, 10:52 PM (EST)

"Help no one else seems to be able to give me..."

 I had a question, and my parents, math teachers, etc. can't help me. Maybe someone here can. I've never done any sort of difficult math (I'm in 9th grade, Algebra 2) and no one can help me with this. I'm not even sure how to notate what I'm asking, but I'll try:z=x^(y-x)This is the equation. What I want to know is, for a specified value y, what value of x will make z the biggest? What I want is an equation that, in terms of y, returns the maximum value of z, with x. So, I'm asking, with a set y, what x will make z the biggest. Sorry if you're confused... Thats the only way I know how to explain it.

alexb
Charter Member
1953 posts
Jan-23-07, 10:57 PM (EST)

1. "RE: Help no one else seems to be able to give me..."
In response to message #0

 (I am pretty sure) this is impossible. The equation you get is implicit and transcendental. There may of cause be an unobvious property of this equation that I miss, but that somehow simplifies for the extreme points. Personally, I do not believe you can do that.

JJ
guest
Jan-24-07, 05:01 AM (EST)

2. "RE: Help no one else seems to be able to give me..."
In response to message #1

 z = x^(y-x) = exp((y-x)ln(x))dz/dx = z*(((y - x)/x) - ln(x)) = 0(y - x)/x) - ln(x) = 0y = x x*ln(x) = x*(1 ln(x)) = x*ln(e*x)Let Y = e*y and X = e*xthen : Y = X*ln(X)The roots of this kind of equation cannot be expressed in terms of elementary function, but in term of the special function named the "Lambert's W function". See :https://mathworld.wolfram.com/LambertW-Function.htmlThe result is : X = exp(W(Y))Finally : x = exp(W(e*y)-1). Remark : In case of some particular value of y, eventually this special function may be reduced to an elementary function.

guest
Jan-24-07, 07:33 AM (EST)

3. "RE: Help no one else seems to be able to give me..."
In response to message #2

 On the first comment, it has to be possible. Think about it: x^(10-x) has to have a maximum value. With it notated (x,x^(10-x)):1,12,2563,21874,40965,31256,12967,3438,649,910,1With 0

Ricardo
guest
Jan-27-07, 07:44 AM (EST)

4. "RE: Help no one else seems to be able to give me..."
In response to message #3

 That was a nice question Brad!What these guys above are saying is that even thought, as you argued, there is a function that does that, in general there is no "nice" equation for it.The "no nice function equation" is in fact a common occurrence in mathematics. The 'solution' for that is to find a simpler function that approximates the "ugly" function or to make some computer program/algorithm that gives good approximations of it.These "approximation methods" have lots of nice ideas actually, and you are going to see some of them if you do a calculus course.Ricardo

Kh4o\$
guest
Feb-05-07, 11:19 PM (EST)

5. "RE: Help no one else seems to be able to give me..."
In response to message #3

 >On the first comment, it has to be possible. Think about >it: >x^(10-x) has to have a maximum value. With it notated >(x,x^(10-x)): >1,1 >2,256 >3,2187 >4,4096 >5,3125 >6,1296 >7,343 >8,64 >9,9 >10,1 >With 0 Maybe this will help. X cannot be negative. With my >original function, >x^(y-x), 0=with this now... >Bye >Brad Hmm, the solution above only applied for integral, Which is not stated in the problem. In addition, it's done by trial and error ??? =.=' And the solution provided by JJ is way to complex for me ( a grade 12 student, lol). That being said, this kind of problem is way too ambiguity ( or you'd prefer: general)for high school or even first year university, so you'd better not fooling around with it. Come back when you have a specific number for y :P. Regards.