1. ## Help with this problem please

$\displaystyle \\ \mbox{Give that,} \\ x_A , x_B >1 \\ \mbox{ and } \\ Y_A,Y_B\geqslant0 \mbox{ (but strictly positive for at least one)}$
$\displaystyle Y_A + Y_B = 100$

$\displaystyle P_A=Y_A(x_A -1) - Y_B$
$\displaystyle P_B=Y_B(x_B -1) - Y_A$
Problem:
$\displaystyle x_A \mbox{ and } x_B \mbox{ is given }$
$\displaystyle \mbox{Find values of }Y_A, Y_B \mbox{ such that:}$
1) Both $\displaystyle P_A$ and $\displaystyle P_B$ is always greater than zero. Are there any restictions needed on $\displaystyle x$ for this to be true?

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EDIT (21 OCT 07):

Let me clarify what my actual question is. Sorry for not making this clear enough early on.

The question I am asking is this:
1) What restrictions (i.e. the general rule) are needed on $\displaystyle x_A$ and $\displaystyle x_B$ so that there are some range of values of $\displaystyle Y_A$ and $\displaystyle Y_B$ to make both $\displaystyle P_A$ and $\displaystyle P_B$ positive?
2) What is the restriction (i.e. the general rule) for $\displaystyle x_A$ and $\displaystyle x_B$ such that no matter what combination of $\displaystyle Y_A$ and $\displaystyle Y_B$ is chosen (but still sums to 100), $\displaystyle P_A$ and $\displaystyle P_B$ is always negative?
3) How can you prove your answers to questions 1) and 2) ?

2. Originally Posted by tongzilla
$\displaystyle \\ \mbox{Give that,} \\ x_A , x_B >1 \\ \mbox{ and } \\ Y_A,Y_B\geqslant0 \mbox{ (but strictly positive for at least one)}$
$\displaystyle Y_A + Y_B = 100$

$\displaystyle P_A=Y_A(x_A -1) - Y_B$
$\displaystyle P_B=Y_B(x_B -1) - Y_A$
Problem:
$\displaystyle x_A \mbox{ and } x_B \mbox{ is given }$
$\displaystyle \mbox{Find values of }Y_A, Y_B \mbox{ such that:}$
1) $\displaystyle P_A$ and $\displaystyle P_B$ is always greater than zero. Are there any restictions needed on $\displaystyle x$ for this to be true?
I think you could do it like this:

$\displaystyle Y_A=Y_B=50$
Which satisfies $\displaystyle Y_A + Y_B = 100$

Then $\displaystyle x_A, x_B > 2$

Which satisfies:
$\displaystyle P_A=Y_A(x_A -1) - Y_B$
because Y_A = Y_B
$\displaystyle P_A=Y_A(x_A -1) - Y_A$
$\displaystyle P_A=Y_A[(x_A -1) - 1]$
$\displaystyle P_A=Y_A(x_A -2)$
And because we define $\displaystyle x_A>2$ we know that $\displaystyle x_A-2>0$ which means that $\displaystyle Y_A(x_A-2)>0$, which means that $\displaystyle P_A > 0$

And the math will be the same for the other equation as well.

3. EDIT (21 OCT 07):

Let me clarify what my actual question is. Sorry for not making this clear enough early on.

The question I am asking is this:
1) What restrictions (i.e. the general rule) are needed on $\displaystyle x_A$ and $\displaystyle x_B$ so that there are some range of values of $\displaystyle Y_A$ and $\displaystyle Y_B$ to make both $\displaystyle P_A$ and $\displaystyle P_B$ positive?
2) What is the restriction (i.e. the general rule) for $\displaystyle x_A$ and $\displaystyle x_B$ such that no matter what combination of $\displaystyle Y_A$ and $\displaystyle Y_B$ is chosen (but still sums to 100), $\displaystyle P_A$ and $\displaystyle P_B$ is always negative?
3) How can you prove your answers to questions 1) and 2) ?

4. Originally Posted by angel.white
I think you could do it like this:

$\displaystyle Y_A=Y_B=50$
Which satisfies $\displaystyle Y_A + Y_B = 100$

Then $\displaystyle x_A, x_B > 2$

Which satisfies:
$\displaystyle P_A=Y_A(x_A -1) - Y_B$
because Y_A = Y_B
$\displaystyle P_A=Y_A(x_A -1) - Y_A$
$\displaystyle P_A=Y_A[(x_A -1) - 1]$
$\displaystyle P_A=Y_A(x_A -2)$
And because we define $\displaystyle x_A>2$ we know that $\displaystyle x_A-2>0$ which means that $\displaystyle Y_A(x_A-2)>0$, which means that $\displaystyle P_A > 0$

And the math will be the same for the other equation as well.
So is x_A and x_B > 2 the one and only restriction needed? Does it only work for Y_A and Y_B =50 or it also works for other values?