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March 30, 2017

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A dog kennel is to be constructed alongside a house using 60 m of fencing. Find the dimensions that will yield the maximum area.

How do you solve this problem?

  • algebra 2 - ,

    I assume you are implying that we don't need a fence along the house.

    let the length be y m (the single side)
    let the width be x m

    so we know that 2x + y = 60
    or y = 60 - 2x

    Area = xy
    = x(60-2x)
    = -2x^2 + 60x

    At this point I don't if you know Calculus or not.
    If you do, then
    d(Area)/dx = -4x + 60
    = 0 for a max of area
    -4x + 60 = 0
    x = 15
    then y = 30

    if you don't take Calculus, we have to complete the square
    area = -2(x^2 - 30x)
    = -2(x^2 - 30x + 225 - 225)
    = -2((x-15)^2 - 225)
    = -2(x-15)^2 + 450

    so the area is a maximum when x = 15
    and then y = 60-30 = 30

    by either method,
    the width has to be 15 m, and the length has to be 30 m for a maximum area of 15(30) or 450 m^2

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