SOLUTION: The first problem on this page illustrates an important relationship between perimeter and area. Let’s change the problem slightly so that it applies to many cases at once. Inst

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Question 751009: The first problem on this page illustrates an important relationship between perimeter and area. Let’s change the problem slightly so that it applies to many cases at once. Instead of 60 feet, let’s say you have 4x feet of fencing, where x is any number. What, in terms of x, should the dimensions of the largest rectangular enclosure be, whose perimeter is 4x feet? What would the area be?
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
The first problem on this page illustrates an important relationship between perimeter and area. Let’s change the problem slightly so that it applies to many cases at once. Instead of 60 feet, let’s say you have 4x feet of fencing, where x is any number. What, in terms of x, should the dimensions of the largest rectangular enclosure be, whose perimeter is 4x feet? What would the area be?
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perimeter = 2(L+W) = 4x
L+W = 2x
L = (2x-W)
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Area = L*W
A = (2x-W)W
----
Area = -W^2 + 2xW
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Cheers,
Stan H.
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