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Question Regarding System of Equations

  1. Jan 12, 2012 #1
    Hey Pf, I'm working on precalculus review and I have found myself stumped by a question.

    1. The problem statement, all variables and given/known data

    A rectangular field is enclosed by 600m of fencing. A second rectangular field, which is alongside a river, has the same area and is also enclosed by 600m of fencing. However, this second field has fencing on only three sides because there is no need for fencing along the riverbank. Create a system of quadratic equations to model the problem. (Answer: A = -x2+300x, A = -2x2+60x)

    2. Relevant equations


    3. The attempt at a solution

    I began by modeling the first field:
    P = 2x+2y
    600 = 2x+2y
    300 = x+y
    y = -x+300

    A = xy
    A = x(-x+300)
    A = -x2+300x

    Afterwards I modeled the second field:
    600 = 2x+y
    y = -2x+600

    A = xy
    A = x(-2x+600)
    A = -2x2+600x

    As you can see, the equations are different. I cannot see where I went wrong. Could someone please correct my folly? :)
  2. jcsd
  3. Jan 12, 2012 #2
    I can't see where you wrong, are you sure you read the answer right?
  4. Jan 12, 2012 #3
    Positive. It may be of use to note when I entered the equations I created into my calculator it yielded solutions of (0,0) and (0,300). Seeing as those solutions are illogical, that means it's either a bad question, or I made an error somewhere along the way.

    EDIT: I just used the same method of checking for the "correct" solution; it yielded values of (-240,-129600) and (0,0). Perhaps it's just a bad question...
  5. Jan 12, 2012 #4
    Well the area of the second one should be [itex]\geq[/itex] the first one. And if the answers are right, that doesn't always hold true.
  6. Jan 12, 2012 #5
    I'll skip it and move on I suppose. :)
  7. Jan 12, 2012 #6
    Yeah that might be a good idea.
  8. Jan 12, 2012 #7


    Staff: Mentor

    You can't use the same variables for the two fields.
    Let x1 and y1 be the width and length of the first field (the one with fences along all four sides).
    Let x2 and y2 be the width and length of the second field (the one with the river as one boundary).

    For the first field, A = x1*y1 = x1(300 - x1)
    For the second field, A = x2*y2 = x2(600 - 2x2)
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