SOLUTION: A dog owner has 250 feet of fencing to enclose a rectangular run for his dogs. If he wants the maximum possible area, what should the length and width of the rectangle be?

Question 205701: A dog owner has 250 feet of fencing to enclose a rectangular run for his dogs. If he wants the maximum possible area, what should the length and width of the rectangle be?

Answer by RAY100(1637) (Show Source): You can put this solution on YOUR website!
Perimeter of rectangle = 2* Length + 2* Width
.
P =250 = 2L +2W,,,,,dividing by 2
.
125 = L+W,,,,,W=125-L
.
Area = L*W = L*(125-L) = 125L -L^2
.
making a table
.
L,,,,,,,Area
.
1,,,,,,,124
.
60,,,,,,3900
.
62,,,,,,3906
.
62.5,,,,3906.25,,,,,max,,,,L=W=62.5
.
63,,,,,,3906
.
The max area of the rectangle occurs when L=W=62.5 ft,,,,
.
This makes rectangle into a square
.

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