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31 March, 15:24

Find the dimensions of a rectangle with perimeter 64 m whose area is as large as possible.

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  1. 31 March, 15:39
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    The dimensions would be 16 because 16+16+16+16=64 so if you wanted to fin the are it would be 16*16=256 squared inches
  2. 31 March, 15:43
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    I have found that when the length=width for any rectangle, the area is maximized

    note: all squares are rectangles but not all rectangles are squares, so don't tell me that a rectangle can't be a square because rectangle is defined as having 4 right angles and a square is a special rectangle with 4 right angles and 4 equilateral sides

    so

    64=p=2 (L+W)

    64=2 (L+W)

    we assume L=W for max area

    64=2 (L+L)

    64=2 (2L)

    64=4L

    divide both sides by 4

    16=L

    area=LW or L^2 for a square

    area=16*16=256 m²

    if you wanted calculus way

    64=P=2 (L+W)

    divide by 2

    32=L+W

    32-L=W

    area=LW

    subsitute 32-L for W

    area=L (32-L)

    area=32L-L²

    take derivitive of both sides

    area dy/dL=32-2L

    find where it equals 0

    0=32-2L

    2L=32

    L=16

    is it a max or min?

    dy/dL is negative when L=17 and positive when L=15

    so it goes up then down

    it is a max

    L=16

    32-L=W

    32-16=W

    16=W

    area=16*16=256

    area=256 m²
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