intermediate algebra

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A farmer with 3000 feet wants to enclose a rectangular plot that borders on a straight highway. If the farmer does not fence the side along the highway, what is the largest area that can be enclosed?

  • intermediate algebra -

    Let x be the side length parallel to the highway. The side lengths perpendicular to the highway must be
    (1/2) (3000 -x)
    The area is
    A = (x/2)*(3000-x) = 3000x -x^2/2
    When there is a maximum,
    dA/dx = 0 = 3000 - 2x
    x = 1500 feet

    I used calculus but you can try completeing the square or try various values of x until you get a maximum area.

    The enclosed area will be 1500 x 750 = 1,125,000 ft^2

  • intermediate algebra -

    where did you get the 750? I have a similar problem here:

    a farmer wants to build a rectangular pen using a side of a barn and 60ft of fence. find the dimensions and area of the largest such pen

  • intermediate algebra -

    Let l = measure of the parallel side of the highway in meters
    w = measure of the perpendicular side of the highway in meters

    l + w = 3000
    l = 3000 - w

    length = 3000 - w
    width = w

    A= lw
    **since we will only use 1 side of the length, we will use:
    A= [(3000-w)/2]w 0r w[(3000-w)/2]
    = -(w^2)/2 + 1500w
    **complete the square
    = -1/2 (w^2 - 3000w + 225,000) + 1,125,000
    = -1/2 (w-1500)^2 + 1,125,000
    w=1500
    A max.= 1,125,000
    **substitution
    l=(3000-w)/2
    =(3000-1500)/2
    =750
    dimensions: 1500 x 750

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