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Homework Help: Finding the length of wire needed

  1. Mar 5, 2009 #1
    1. The problem statement, all variables and given/known data

    Two isolated farms are 12km apart on a straight country road that runs parallel to the main highway 20km away. The power company decides to run a wire from the highway to the junction box, and from there, wired of equal length to two houses. Where should the junction box be placed to minimize the length of wire needed?

    2. Relevant equations

    3. The attempt at a solution

    Where should I start with this?
  2. jcsd
  3. Mar 6, 2009 #2


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    try and write an equation for the length of wire neded in terms of the juntion box position, then minimise

    drawing a picture will help

    also use the symmetry of the problem to simplfy the possible positions
  4. Mar 6, 2009 #3
    I got nothing.
  5. Mar 6, 2009 #4


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    The shortest distance between two "farms" is a straight line. If the two farms were on opposite sides of the road, so that the straight line crosses it, this would be easy. What would the answer be then?

    Now use the "symmetry" that lanedance mentioned.
  6. Mar 6, 2009 #5
    I dont understand what you mean by using "symmetry" I also dont understand what your asking here.

    The two farms are 12km apart. Ill try and trace out a diagram here.

    FARM A


    FARM B

    So I tried to create 2 triangles between the farms and the junction, with one side of the triangle being 6km (since the question asks for equal length of wire.)
  7. Mar 6, 2009 #6


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    looking good, now try & write an equation for the total length of wire in terms of the junction box position

    you could do it a few ways, but using of the distance junction box to the road (ie the other side of the two triangles) as your variable might be a good place to start
  8. Mar 6, 2009 #7
    Im still lost as to where to start. What im attempting right now is that I have 1 triangle and a straight length of wire.

    I have the hypot. of the triangle = y, the base = x and the other side = 6. Hope that makes sense. I also have the length of wire = 20-x. If I can solve for one of these I can figure out the rest.

    DO you recommend starting with the 20-x function? If so how?
  9. Mar 6, 2009 #8


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    the wire if i understand right, will be the length from highway to junction box, call it HJ, then 2 times the hypotenuse, JF

    so length L(x)= HJ(x) + 2*JF(x)

    you've already told me
    HJ(x) = 20-x

    what is JF(x)?
  10. Mar 6, 2009 #9
    I dont know how we can figure that out. JF(x) = L - (20-x), I dont know where you can from there without knowing more of the variables.
  11. Mar 6, 2009 #10
    I did some calculations and figured out x = 10, which then translated into a total length of wire of 33.3 km.

    What do you guys think?
    Last edited: Mar 6, 2009
  12. Mar 7, 2009 #11


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    you need to show your calcs then

    i'm not going to do the problem then check your answer, but i will confirm whether yoru thinking sounds right
  13. Mar 7, 2009 #12
    Its diffcult without being able to show a diagram. Essentially I created 2 triangles, One triangle with the sides of 6 and t with a hypotenuse of x, and the other triangle with sides of 6, 20-t, and y. I created equations for x and y to single out the variable t. So then I did D = x + y then differentiated. From there I singled out t which you can find out 20-t and also x. 2 times x plus 20-t gives the answer.
  14. Mar 8, 2009 #13


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    Have you given any thought to what I said before?

  15. Mar 8, 2009 #14
    I dont get what that means. I would need a diagram to understand what that means.
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