Calc.

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A farmer has 600m of fence and wants to enclose a rectangular field beside a river. Determine the dimensions of the fence field in which the maximum area is enclosed. (Fencing s required on only three sides: those that aren't next to the river.)

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  1. A=L*W
    P=L*2W three sides
    600m=L*2W
    L=600-2W
    substitute in the area
    A=(600-2W)*W
    A=600W-2W^2
    convert the eq. to vertex form y=a(x-h)^2+k
    A=-2W^2+600W
    complete the square
    A=-2(W^2-300W+22500)+22500*2
    A=(W-150)^45000
    w=150
    substitute calculate....
    L=600-2w
    L=600-300 =300
    dimension
    150x300

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  2. A=L*W
    P=L*2W three sides
    600m=L*2W
    L=600-2W
    substitute in the area
    A=(600-2W)*W
    A=600W-2W^2
    convert the eq. to vertex form y=a(x-h)^2+k
    A=-2W^2+600W
    complete the square
    A=-2(W^2-300W+22500)+22500*2
    A=(W-150)^45000
    w=150
    substitute calculate....
    L=600-2w
    L=600-300 =300
    dimension
    150x300

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  3. A family wants to fence a rectangular play area alongside the wall of their house. The wall of their house bounds one side of the play area. If they want the play area to be exactly 400ft2, what is the least amount of fencing needed to make this? Round your answer to the nearest two decimal places.

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