Question 183949
find the dimensions of a rectangle "a" with the greatest area whose perimetter is 30 feet
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Perimeter = 2(L + W)
30 + 2(L+W)
L+W = 15
W = 15-L
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Area = LW
Substitute to get:
Area = L(15-L)
Area = -L^2 + 15L
That is a quadratic with a = -1,b = 15
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Maximum area occurs when L = -b/2a = -15/(-2) = 15/2
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Solve for W when L = (15/2)
W = 15 - (15/2) = 15/2
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The Width and the Length are both 15/2 ft.
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Cheers,
Stan H.