Without further information it is impossible to say what the dimensions are; only an equation can be given relating the width to length which will have infinitely many solutions as there are 2 unknowns:
Let one dimension of the playground be x and the other be y; let the extra fence be constructed parallel to the side of length x. Then:
total fence used = perimeter of rectangle + fence across the middle
→ 648 ft = 2x + 2y + x
→ 2y + 3x = 648 ft
As we only have 1 equation but 2 unknowns, there are infinitely many possible solutions, eg:
2 ft wide by 321 ft long, split with a fence parallel to the 2 ft side;
4 ft wide by 318 ft long, split with a fence parallel to the 4 ft side;
129 ft wide by 130 ft long, split with a fence parallel to the 130 ft side.
If the problem included other information, for example: the area of the playground is a maximum, then:
From above:
2y + 3x = 648 → y = 324 - 3x/2
and area = xy
→ area = x(324 - 3x/2)
→ area = 324x - 3x²/2
To find the maximum area, completing the square gives:
area = 3/2 (216x - x²)
→ area = -3/2 (x² - 216x)
→ area = -3/2 ((x - 108)² - 108²)
→ area = 3/2 (108² - (x - 108)²)
As x varies (x - 108)² is greater than or equal to 0, and the greater it is the smaller the area is; the maximum area is when (x
(x - 108)² = 0
→ x - 108 = 0
→ x = 108
→ y = 324 - 3 × 108 / 2 = 162
Thus the maximum area is when the playground is 162 ft long by 108 ft wide with the dividing fence parallel to the 108 ft side.
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