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?

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Question 151799: 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 Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Let's start with the formulas for the area and the perimeter of a rectangle.
A+=+L%2AW
P+=+2%28L%2BW%29
The perimeter is given as 250 feet, so we can write:
2%28L%2BW%29+=+250 Dividing both sides by 2, we get:
L%2BW+=+125 Rewrite this as:
L+=+125-W and substitute it into the formula for the area (A = L*W),
A+=+%28125-W%29%2AW Simplifying this, we get:
A+-+125W-W%5E2 Now this is a quadratic equation and if you were to graph this, you see a parabola that opens downward, thus there would be a maximum value of A (also known as the vertex).
You can find the value of W at this vertex by:
W+=+%28-b%29%2F2a This comes from the standard form for a quadratic equation:y+=+ax%5E2%2Bbx%2Bc
In this problem, a = -1 and b = 125, so, making the appropriate substitutions, we get:
W+=+%28-125%29%2F2%28-1%29
W+=+62.5 Now this is the value of W (the width) that would make the area (A) a maximum.
The length (L) is:
L+=+125-W
L+=+125-62.5
L+=+62.5
So the length and the width of the rectangle would be L = 62.5 feet and W = 62.5 feet. In other words, the rectangle is a square.