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A rectangular field is to be enclosed by a fence. Two fences parallel to one side of the field divide the field into three rectangular fields. If 2400m of fence are available find the dimensions giving the max area.

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  1. Because there are 2 dividers, the perimeter = 2l + 4w, where l is the length of the field, and w is the width of the field (and the dividers)

    2l + 4w = 2400

    let x = l
    2x + 4w = 2400
    4w = 2400 - 2x
    w = 600 - 0.5x

    Now you can find the area.
    A = l*w
    A = x(600 - 0.5x)

    Now you need to optimize area. A calculator is the easiest way, although if you are in Calculus you would likely have to show more work.

    If you can use a graphing calculator, enter the equation in and find the maximum value on the graph.

    I will demonstrate the calculus way below:
    A = 600x - 0.5x^2
    dA/dt = 600 - x

    Find where dA/dt = 0

    600 - x = 0
    x = 600

    dA/dt changes signs from + to - at x = 600, so there is a relative maximum at x = 600

    Because x = length, the length is 600
    w = 600 - 0.5x
    w = 600 - 300
    w = 300

    The dimensions are 300m * 600m

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  2. let the width of the whole rectangle be x m (there will be 4 of these)
    let the length be y m
    then 4x + 2y = 2400
    2x + y = 1200
    y = 1200 - 2x

    Area = xy
    = x(1200-2x)
    = -2x^2 + 1200x

    Now, I don't know if you are studying Calculus.
    If you do, then
    d(Area)/dx = -4x + 1200
    = 0 for a max area
    x = 300

    then y = 600
    and the max area is (300(600) = 180000

    If you don't know Calculus, complete the square on the above quadratic
    you should end up with
    Area = -2(x-300)^2 + 180000

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  3. Hey, we think alike, lol

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  4. Interesting...I did not think to complete the square to find the maximum area though. Based on steven's other questions I was not sure whether he was in Calculus.

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  5. a farmer has a rectangle field with an area of 3/4 square like the field is 1/2 mule wide what is the length of the field

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