SOLUTION: Find the length and width of a rectangle whose perimeter is 36 feet and whose area is 80 square feet.

Question 438648: Find the length and width of a rectangle whose perimeter is 36 feet and whose area is 80 square feet.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Find the length and width of a rectangle whose perimeter
is 36 feet and whose area is 80 square feet.
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Perimeter Eq: 36 = 2(L+W)
Area Eq; LW = 80
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Modify:
L + W = 18
----
L(18-L) = 80
-L^2+18L-80 = 0
----
Use The Quadratic Formula to solve for "L":
L = [-18 +- sqrt(18^2-4*-1*-80)]/(-2)
---
L = [-18 +- sqrt(4)]/-2
---
Positive solution:
L = [-18-2]/-2 = 10 (length)
---
Since L+W = 18, W = 8 (width)
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Cheers,
Stan H.
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