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Math Help - Quadratic Function Problem

  1. #1
    Raj
    Raj is offline
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    Sep 2007
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    Quadratic Function Problem

    A rectangular lot has a river on one side and is fenced off into 3 areas with 752m of fencing. Find the dimensions and its maximum area. Note: There is no fencing along the river



    I have:

    Perimeter = 752m
    3w + 2l = 752
    2l = 752 - 3w
     <br />
l = 376 - 3/2w<br />

    Area = l x w
    A = (376-3/2w)(w)
    A = [376w-(3/2w^2)]




    Thats what i have so far, don't know if its right or wrong. Would be gratefull if anyone could tell me if what i have so far looks right, and what/how to do next.
    Last edited by Raj; September 28th 2007 at 03:19 PM.
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  2. #2
    MHF Contributor
    Joined
    Apr 2005
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    Very good so far.
    Except that 3/2w is misleading. (3/2)w is better.
    So is (3/2)w^2.

    Now that you have your quadratic equation
    A = 376w -(3/2)w^2
    You transform that into
    A = a(w-b) +c
    so that you can get the vertex (b,c),
    where b is the value of w at the vertex.
    and c is the value of A at the vertex.

    A = 376w -(3/2)w^2
    A = -(3/2)[w^2 -(376w)/(3/2)]
    A = -(3/2)[w^2 -(376*2 /3)w +(376/3)^2] -(3/2)(-(376/3)^2)

    . Why don't we use decimal points?

    A = -(1.5)[(w -125.333)^2] +23,562.667 ----------(i)

    Parabola (i) is vertical, that opens downward because of the negative coefficient of w^2.
    Therefore the vertex is the highest point, a maximum A.

    Hence, for maximum A,
    w = 125.333 m. ------------------------------answer.
    L = 376 -(3/2)(125.333) = 188 m. -------------answer.
    A = 23,562.7 sq.m. ---------------------------answer.

    Check,
    A = w*L
    max A = 125.333 *188
    23,562.7 =? 125.333 *188
    23,562.7 =? 23,562.6
    Umm, the 0.1 sq.m. difference in area was stolen by the neighbor.
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