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5 June, 05:17

A homeowner has 60 feet of fencing material to enclose a rectangular area for his pets to play in. One side of the house will be used as a side of the play area. What deminsions should be used in order to maximize the play area?

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  1. 5 June, 05:28
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    Let the length of the side that would be a house be l. Given that the total length of the fencing material is 60 and the other side is x, then:

    2x+l=60

    ⇒l=60-2x

    The area of the house will therefore be:

    A=length*width

    A=x (60-2x)

    A=60x-2x²

    differentiating A w. r. t x we get

    dA/dx=60-4x

    equating the above to zero and solving for x we get

    60-4x=0

    hence

    4x=60

    ⇒x=60/4

    x=15

    this implies that one of the sides of the area is 15 ft and the other is 30 ft
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