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System of 4 equations.

  1. Jun 30, 2010 #1
    Hello, I'm just wondering if there's a better way to solve systems such as these:

    variables are x,y,z,a and I need to solve for x, y, and z.

    y^2z^3 + a4x=0

    2xyz^3 + a6y = 0

    3xy^2z^2 + a12z = 0

    2x^2 +3y^2 + 6z^2 -6 = 0


    I'm aware of matrix methods to solve systems of linear equations. But for those you nead to have linear equations, correct? And here my variables are multiplied together and to higher powers.

    I was thinking of definine something like y^2z^3 = b or something and just calling those terms variables, and then creating the necessary new equations. But I don't know, that seems like a lot of work.

    Anyway, is there any other way to solve such systems other than back elimination of varibales and back substitution? because I have to do something like this on an exam and that method takes a lot of time, and I always make mistakes. what algorithm (if any) do you use to solve a simlar system??


    thanks for your help.
     
  2. jcsd
  3. Jun 30, 2010 #2

    Mark44

    Staff: Mentor

    What do a4x, a6y, and a12z mean?
     
  4. Jun 30, 2010 #3
    sorry, that was kind of unclear.

    it should probably be written as:

    y^2z^3 + 4xa=0

    2xyz^3 + 6ya = 0

    3xy^2z^2 + 12za = 0

    2x^2 +3y^2 + 6z^2 -6 = 0



    I define a as just another variable. in the question I'm trying to do it's the lagrange multiplier but I don't know how to make a lamda, I'm trying to find the extrema, the last equation is the restriction.
     
  5. Jun 30, 2010 #4

    Mark44

    Staff: Mentor

    Try solving for a ([itex]\lambda[/itex]) in the first three equations, starting with the second and third.

    From the second equation, y(2xz3 + 6a) = 0 ==> y = 0 or a = -(1/3)xz3.
    Do a similar thing for the third equation and the first. The goal is to get a number of critical points that you can substitute into the objective function.
     
  6. Jun 30, 2010 #5

    Mark44

    Staff: Mentor

    This might work if you had multiple equations with y^2z^3, but you don't, so this is not a good idea.

    BTW, this does not seem to be a precalc-level problem, so you should have posted it in the Calculus & Beyond section.
     
  7. Jun 30, 2010 #6
    Hmm, yeah ok. that seems like a good way. I did it and it gave me the same results a bit faster. The way I initally tried was just solving for lamda from Eqn. 1 and then plugging lamda into Eqns 2 and 3.

    I have a question though, you say it's 0 if y is 0 or if that expression is 0.
    Does that mean later we have to check if y=0 is part of some solution?

    thanks for your help
     
  8. Jun 30, 2010 #7
    actually, maybe I should just post my entire question, because all of it is a bit fuzzy for me.

    the question is find the max of

    {xy^2z^3 : 2x^2 +3Y^2 + 6z^3 = 6, x> or = 0, y > or = 0, z> or = o}


    So, basically I set up the equations above and muddled around with a lot of arithmatic mistakes and got this:

    y= +/- 2/sqrt(15)
    x= +/- sqrt(3)/sqrt(15)
    z= +/- 2sqrt(3)/sqrt(15)


    Then I said the max was when you have either a combination of the three coordinates where x and z are their positive options, or either both their negative options,

    and the minimum is when you have three coordinates where x and z do not have the same sign.


    does this seem somewhat correct?
     
    Last edited: Jun 30, 2010
  9. Jun 30, 2010 #8

    Mark44

    Staff: Mentor

    Yes.

    Since you have conditions that x >= 0, y >= 0, and z >= 0, you can eliminate the three negative values above. As already noted, though, some of the variables can be zero.
    No, due to the non-negativity constraints.
     
  10. Jun 30, 2010 #9
    Oops, sorry.
    I'm kind of new to these forums...


    Anyway,

    Yeah, obviously no negative solutions. Sorry, that was just an oversight on my part. Also I only need the max, so there's no minimum solution needed either.

    But I'm still confused with the solution of the system of equations.

    I can factor out a z from equation 3, as you did the y from eqn 2, and then I get that z=0 along with y=0 are possible solutions.

    if I plug y=0 and x=0 into eqn 4, then I will get a different value for x. hmmm. Well I'm just not sure how to check these solutions. basically I'll get 4 different solution sets??, one for x,y,z >0 one for x,z>0, y=0 and one for x,y>0 z=0 and one for x>0, z,y=0


    That's a lot of equations to solve....
     
  11. Jun 30, 2010 #10
    oh, btw, thanks for helping me out. I appreciate it. I know my questions are kind of dumb but I really want to understand this problem.
     
  12. Jun 30, 2010 #11

    Mark44

    Staff: Mentor

    If I have arrived at something like this:

    x = 0 or y = 7
    AND
    y = 0 or z = 3

    I have four things to check: x = 0 AND y = 0, x = 0 AND z = 3, y = 7 AND y = 0 (can't happen), y = 7 and z = 3.

    If you add another pair of values, this doubles the number of things you need to check to 8. Each time you have to pick one from the first group, one from the second group, AND one from the third group.

    I haven't worked this problem, so I'm speaking as generally as I can from that perspective.
     
  13. Jun 30, 2010 #12
    Ok, I think I understand in general. It seems just like logic.
    But I'm not sure how to apply this to the problem.

    when faced with this:

    From the second equation, y(2xz3 + 6a) = 0 ==> y = 0 or a = -(1/3)xz3.

    I first assumed the second case, and did the same for a similar situation with z.

    This led me to the solution set:


    y= 2/sqrt(15)
    x= sqrt(3)/sqrt(15)
    z= 2sqrt(3)/sqrt(15)


    now for instance, if I choose that actually it's y=0 and a=/= 0

    then I substitute y = 0 into the system of equations and get this

    0 + a2x = 0
    0 + 0 = 0
    0 + a12z = 0
    2x^2 +6z^2 - 6 = 0


    which gives me the new solution set:

    x = 6*sqrt(1/13)
    y=0
    z=sqrt(1/13)


    So now you're saying I need to consider

    y= 2/sqrt(15) or y=0

    AND

    x= sqrt(3)/sqrt(15) Or x = 6*sqrt(1/13)



    AND

    z= 2sqrt(3)/sqrt(15) or z=sqrt(1/13)


    now I have to do the same for z=0
    and I'll get an alternate solution set... and so I'll have

    x1 or x2 or x3
    AND
    y1 or y2 or y3
    And
    z1 or z2 or z3

    This seems like a huge amount of work... just to solve an system. Now I have to test every possible permutation in the constraint equation to see if it's allowed/ a maximum??

    the thing is, that when I did it my way the first time. I just solved for lamda from eqn 1 and substituted back into the other two equations and rearanged, I never had to consider if y=0 and z=0. how could one method of solving give different results?
     
  14. Jun 30, 2010 #13

    Mark44

    Staff: Mentor

    Yes, exactly like logic.
    Then it's possible you weren't getting all of the possible critical points. Since the goal is to get every one of these points and evaluate the objective function at each of them, if you're missing some, it's possible to overlook the true maximum point.

    From the 2nd equation I got y = 0 OR a = -(1/3)xz^3
    From the 3rd equation I get z = 0 OR a = -(1/4)xy^2z

    There are four combinations of the pairs of equations above, and you need to investigate all four. As already stated, if you end up with another pair of equations, you'll need to look at 8 different combinations of three equations, with one from each pair.
     
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