<PRE><B>Question 342111</B>
find the dimensions of a rectangle whose perimeter is 26 meters and area is 42 square meters.
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Perimeter = 2(L + W)
26 = 2(L+W)
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L+W = 13
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LW = 42
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Solve for L and W:
L = 13-W
Substitute:
(13-W)W = 42
-W^2+13W-42 = 0
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Quadratic with a = -1, and b = 13
Maximum area occurs when W = -b/(2a) = -13/(-2) = 6.5
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Dimensions of the rectangle:
width = W = 6.5 meters
length = 13-6.5 = 6.5 meters
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Cheers,
Stan H.</PRE>