Results 1 to 4 of 4

Math Help - Help with this problem please

  1. #1
    Newbie
    Joined
    Oct 2007
    Posts
    3

    Help with this problem please

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

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

    ================================================== ========
    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 x_A and x_B so that there are some range of values of Y_A and Y_B to make both P_A and P_B positive?
    2) What is the restriction (i.e. the general rule) for x_A and x_B such that no matter what combination of Y_A and Y_B is chosen (but still sums to 100), P_A and P_B is always negative?
    3) How can you prove your answers to questions 1) and 2) ?
    Last edited by tongzilla; October 21st 2007 at 02:03 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member angel.white's Avatar
    Joined
    Oct 2007
    Posts
    723
    Awards
    1
    Quote Originally Posted by tongzilla View Post
    \\ \mbox{Give that,} \\ x_A , x_B >1 \\ \mbox{ and } \\ Y_A,Y_B\geqslant0 \mbox{ (but strictly positive for at least one)}
    Y_A + Y_B = 100

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

    Y_A=Y_B=50
    Which satisfies Y_A + Y_B = 100

    Then x_A, x_B > 2

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

    And the math will be the same for the other equation as well.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2007
    Posts
    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 x_A and x_B so that there are some range of values of Y_A and Y_B to make both P_A and P_B positive?
    2) What is the restriction (i.e. the general rule) for x_A and x_B such that no matter what combination of Y_A and Y_B is chosen (but still sums to 100), P_A and P_B is always negative?
    3) How can you prove your answers to questions 1) and 2) ?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Oct 2007
    Posts
    3
    Quote Originally Posted by angel.white View Post
    I think you could do it like this:

    Y_A=Y_B=50
    Which satisfies Y_A + Y_B = 100

    Then x_A, x_B > 2

    Which satisfies:
    P_A=Y_A(x_A -1) - Y_B
    because Y_A = Y_B
    P_A=Y_A(x_A -1) - Y_A
    P_A=Y_A[(x_A -1) - 1]
    P_A=Y_A(x_A -2)
    And because we define x_A>2 we know that x_A-2>0 which means that Y_A(x_A-2)>0, which means that 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?
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum