Calculus

a rancher has 1200 fet of fencing to enclose two adjacent cattle pens. What dimensions should be used so that the enclosed area is maximized?

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  1. You don't describe the diagram, but I am familiar with this type of question.
    Assuming there is a common width between the two fields
    let the width be x ft
    let the length of the combined field be y
    so
    3x + 2y = 1200
    y = 600 - (3/2)x

    area = xy
    = x(600 - (3/2)x)
    = 600x - (3/2)x^2
    d(area)/dx = 600 - 3x
    = 0 for a max of area
    3x = 600
    x = 200
    y = 600-(3/2)(200) = 300

    state the conclusion

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