Question 858550
This is an impossible scenario because of this line: the owner has 220 feet of new fencing material <b>(which is much less than the length of the existing fence)</b>. If the length of the existing fence is more than 220, then the perimeter cannot be a rectangle enclosed with an additional 220 ft.  The 220 ft won't fit into the perimeter equation: P = 2L + 2W.
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We'll solve it without the parenthetical comment.
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220 = 2W + 1L (the other L is existing)
Since the numbers closest to each other will produce the largest area, let's divide 220 by 3 sides, rounding to whole numbers. 74, 73, 73, with an existing 74.
Perimeter = 2W + 2L
L = 74 (with the existing fence as L)
W = 73
Area = 74 * 73 = 5402
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If the car stalls measure 9 * 20, then they take up 180sf each.
180 * 3 = 5400
Yes, they could be parked there with 2 sf left over.