# 1-2-42 - d d 300 = 400 a Since the streets are one-way we...

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Section 1.2, Problem 42 The key to this problem is the assumption that “the vehicles leaving the area during the hour were exactly the same as those entering it.” At the end of the hour, there are no cars left over, in any of the streets or intersections. That means that the number entering each intersection must equal the number leaving. This gives four equations (labeling the unknowns clockwise from top): a + 100 + b = 250 120 + 150 = b + c c + 300 = 320 +
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Unformatted text preview: d d + 300 = 400 + a Since the streets are one-way, we have additional restrictions: a ≥ , b ≥ , c ≥ , d ≥ . The solution has one free parameter, r: ≤ a =-100 + r ≤ b = 250-r ≤ c = 20 + r ≤ d = r The inequalities are all satisﬁed as long as r ≥ 100 and r ≤ 250. Plug this in to get the ranges of a , b , c and d : ≤ a ≤ 150 ≤ b ≤ 100 120 ≤ c ≤ 270 100 ≤ d ≤ 250 . 1...
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## This note was uploaded on 09/15/2011 for the course AMAT 344 taught by Professor Raghaven during the Spring '11 term at SUNY Albany.

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