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Minimization Problem

  1. Oct 11, 2009 #1
    1. The problem statement, all variables and given/known data

    The line joining P and Q crosses two parallel lines that are 8 units apart (thus, if I drew a vertical line from the top line to the bottom line, that would equal eight). The point R is 10 units from point P (as shown). How far from point Q should the point S be chosen so that the sum of the areas of the two triangles is at a minimum?

    Calculus4-10.jpg

    2. Relevant equations

    [tex]Area = \frac{1}{2}bh[/tex]

    3. The attempt at a solution

    Okay ... not so bad, I think I know some of the steps.

    [tex]Area of 2 Triangles = \frac{1}{2}bh + \frac{1}{2}(b_{1})(h_{1})[/tex]

    b1 and h1 are just the base/height of the other trianglw (the one with points S and Q).

    So should I substitute 10 in for b and then do implicit differentiation? And how do I include the "8" that's here? Have something like height x and height 8-x for the two triangles?
     
  2. jcsd
  3. Oct 12, 2009 #2

    lanedance

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    you need to find the areas of the triangles in terms of the distance QS = s, before you can differentiate with respect to s to find your minima

    are you sure this is everything in the question 7 how its was given? unless it comes out as a really nice result (unlikely), I think you need to know the position of Q relative to R & P first
     
  4. Oct 12, 2009 #3
    Yes, I coped it directly from my book.

    So when doing this problem, how can I first find the areas in terms of QS? Should I just assign a variable to QS?
     
  5. Oct 12, 2009 #4

    lanedance

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    yeah, so find a way to write the area in terms of s = QS,

    expanding qhat you previously wrote, including the dependence on s
    [tex]A_{total}(s) = A_1(s) + A_2(s) = \frac{1}{2}b_1(s)h_1(s) + \frac{1}{2}b_{2}h_{2}(s)[/tex]
    only the base PR = b2, is independent of s

    however unless i'm missing something I still can't see how you can do this without something pinning the location of Q
     
  6. Oct 12, 2009 #5

    Office_Shredder

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    I think you have enough. Notice the two triangles will be similar
     
  7. Oct 12, 2009 #6

    lanedance

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    knew i was missing something easy - nice, (can't believe i missed it)
    so DMOC, its all about ratios of length
     
    Last edited: Oct 12, 2009
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