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March 30, 2017

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A farmer decides to enclose a rectangular garden, using the side of a barn as one side of the rectangle. What is the maximum area that the farmer can enclose with 80 feet of fence? What should the dimensions of the garden be to give the area?

  • word problem - ,

    He will need one length and two widths
    let the length by y ft
    let the width be x ft

    2x + y = 80
    y = 80-2x

    area = xy
    = x(80-2x)
    = -2x^2 + 80x

    The x value to produce the maximum area is the x of the vertex of this parabola.

    If you know Calculus
    d(area)/dx = -4x + 80= 0 for a max of area
    x = 20
    then y = 80-2x = 40
    the field should be 20 by 40 , with 40 being the one side.

    non-Calculus:
    the x of the vertex of y = ax^2 + bx +c is -b/(2a)
    which in this case is -80/(2(-2)) = 20
    continue as above.

    or you can complete the square:
    area = -2(x^2 - 40x + 400 - 400 )
    = -2(x-20)^2 + 800

    max of area is 800 when x = 20 and y = 40

  • word problem - ,

    The farmer wants to use the 500 m to an enclosure divided into two equal areas. What is the total maximum area that can is achieved with the 500 m fence. use differential calculus to the solution.

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