SOLUTION: find the dimensions of a rectangle whose perimeter is 26 meters and are is 42 square meters

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Question 342111: find the dimensions of a rectangle whose perimeter is 26 meters and are is 42 square meters
Found 2 solutions by checkley77, stanbon:
Answer by checkley77(12844) (Show Source):
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P=2L+2W
26=2L+2W
2L=26-2W
L=(26-2W)/2
L=13-W
A=LW
42=(13-W)W
42=13W-W^2
W^2-13W+42=0
(W-7)(W-6)=0
W-7=0
W=7 ANS
L=13-7=6 ANS.
W-6=0
W=6 ANS
L=13-6=7 ANS.
PROOF
26=2*6+2*7
26=12+14
26=26
42=6*7
42=42

Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
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.