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Homework Help: Abandon hope ye who enter here

  1. Apr 17, 2005 #1
    Hi. I'm a freshman and I royally suck at math. I know this isn't really a math forum, but I know physics involves math, which means you guys are probably wicked good. Anyway, I have a bonus question that will earn me a point or two on my next test, but my teacher says only one or two people a year actually solve it. (Because he won't give you points for being on the right track. It is either right, or it isn't)

    Anyway, it really is simple, which is why I feel stupid for having to ask it.

    He says prove p*q=.25

    So I have been trying at it, but like I said, I suck.

    So... p=1/q or.. see I just... it just doesn't make sense.
    Uhm... I don't even know where to start...

    Any help would be greatly appreciated.
  2. jcsd
  3. Apr 17, 2005 #2


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    I think your problem asks to prove that the maximum value of the product [itex] pq [/itex] subject to the constraint [itex] p+q=1 [/itex] is 0.25.

    Two ways i can think of

    1.Without calculus.
    2.Using Lagrange multipliers.

    Ring a bell?

  4. Apr 18, 2005 #3
    # 1. Rings a bell. He said it could be done using simple college algebra. And uh, what you said about the problem looks about right too.
  5. Apr 18, 2005 #4
    Error 404,

    If you're really in college and think that p+q=1 means that p=1/q, you don't deserve extra credit on this problem. If someone here helps you get the answer, and you turn it in as your own work, you'll be cheating!
  6. Apr 18, 2005 #5


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    It's actually correct to write [itex] p=1-q [/itex] rather than [itex] p=\frac{k}{q} [/itex].I'm sure the study of a quadratic (min,max,vertex coordinates) it's no mistery to u.

  7. Apr 18, 2005 #6


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    OUCH! :rofl:
    Yeah, I'm sure the guy knows the difference between addition and multiplication!

    So you've got p(1-p), right?

    = p - p^2

    p < 1 by definition.

    The squares of numbers that are less than one end up being less than the original numbers. Hmm...that doesn't tell us much except that the result is positive, which we knew already.

    well, maybe if we considered three cases

    p = 1/2 ==> p-p^2 = 1/4

    1 > p > 1/2 ==> ?

    0 < p < 1/2 ==> ?

    Just some preliminary thoughts guys, just thinking out loud. Don't actually know how to solve it yet (without differentiating, that is). Hopefully this helps you along, Error 404.
    Last edited: Apr 18, 2005
  8. Apr 18, 2005 #7
    Just try different values of p and q and see that the product of pq would be greatest when they are both equal.
    I dont know how you can proove this though... just show heaps of different values and the answers you get?
  9. Apr 18, 2005 #8


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    You can rigurously prove it by searching for the maximum of the product function:[itex] f(p)=p(1-p) [/itex]...

  10. Apr 18, 2005 #9


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    Searching for the maximum...without calculus?

  11. Apr 18, 2005 #10
    HINT: What will the function look like graphed?
  12. Apr 18, 2005 #11


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    Hehe...yeah ok. Of course.
  13. Apr 18, 2005 #12
    There's no other rigorous way than using calculus. Determine the value of p for which the derivative of [tex]p - p^2[/tex] is zero.
    A non-calculus way is to plot the function values on a graph. I wrote a short c++ program and gave me as values of p = 0.5 and pq = 0.25.
  14. Apr 18, 2005 #13
    error 404.

    is the "p + q = 1" given or is that of your own doing?
  15. Apr 18, 2005 #14
    That is given.

    Okay, I suck at math, but I am not that bad. I made a typo. I wasn't really paying attention.

    It isn't cheating, it is research. I don't think he expects us to figure it out on our own since he didn't really teach us to. I think he expects us to earn the points by putting forth the effort to solve it.

    I am not sure what he wants. He gave us that example. He was like, "And don't turn in p=.5, q=.5 so pq=.25

    Argh, forget it. Looks like wayyy too much work for a couple of bonus points. Let's just hope I do well on the test today.
  16. Apr 18, 2005 #15
    Actually, it's very simple. Dextercioby told you everything you need to know. But, I will explain it even further.
    When you graph the function you will see that it has a maximum. Since the graph is symmetric through the line parallel to the f(p) axis passing to the maximum, the point in which the maximum is can be found if you find the points at which the function is zero. The point of maximum will be in the middle of their interval.
  17. Apr 18, 2005 #16
    or you could toss together 5 statements that can be deduced from the original constraints and call that a proof. seeing as its just a bonus question i think thats what hes looking for.
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