Hi,
This is the same answer I gave on your question the other time you entered it!!
If the garden is 30 ft long and a path of x ft is added onto each end, then the total length is now 2x + 30. Likewise. if the garden is 20 ft wide and a path of x ft is added onto each side, then the total width is now 2x + 20. so the new area is found by making
Y1 = (2x + 30)(2x + 20). This will make Y1 the area of the rectangle. Since the garden was 30 x 20 ft its area was 600 sq ft and since the path has another 400 feet squared, the total area of the larger rectangle is 1000 sq ft.If you let Y2 = 1000 and set your window to appropriate max and mins, you can solve for the intersection of the 2 equations.
That intersection occurs when x = 3.5078 feet, so that is the width of the path on each end.
If you do not have a calculator to graph these you could also set (2x + 30)(2x + 20) = 1000 and this could be solved using the quadratic formula after simplifying the expression.
(2x + 30)(2x + 20) = 1000
4x² + 100x + 600 = 1000
4x² + 100x - 400 = 0
Dividing by 4:
x² + 25x - 100 = 0
x = (-25 ± √(25² - 4(1)(-100))/(2*1)
x = (-25 ± √(625 + 400))/(2)
x = (-25 ± √(1025))/(2)
x = (-25 ± 32.2))/(2)
The only positive answer for x is (-25 + 32.02))/(2) = 7.02/2 = 3.51
To double check if you take (2x + 30)(2x + 20) = 1000 and substitute 3.51 for x, it becomes:
(2x + 30)(2x + 20) = 1000
(2(3.51) + 30)(2(3.51) + 20) = 1000
(7.02 + 30)(7.02 + 20) = 1000
(37.02)(27.02) = 1000
1000.2804 = 1000
Since this rounded value of x nearly gives 1000 sq ft, this shows that 3.51 ft. is correct for the width of the path.
I hope that helps!! :-)
2007-07-06 14:50:41
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answer #1
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answered by Pi R Squared 7
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The area of the garden including the path
=30*20=600 ft^2
let the width of the road be w ft.
Therefore the area of the garden without the path
=(30-2w)(20-2w)
=600-60w-40w+4w^2
=4w^2-100w+600
By the problem,
4w^2-100w+600=400
or,4w^2-100w+200=0
or w^2-25w+50=0
w={25+-sqrt(625-200)}/2
=(25+-20.6)/2
Therefore,w=22.8 or2.2ft
rejecting 22.8 ft as it is more than the width of the garden itself,we accept w=2.2 ft
Hence the width of the road is 2.2 ft
[Please note that the question submitted by you was incomplete.It has been solved by me only on the assumption that you meant to ask the width of the road]
2007-07-06 21:01:49
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answer #2
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answered by alpha 7
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The path will have an area of 200 ft^2
So 30x+30x +2(20-2x) = 200 {x = width of path}
60x + 40 -4x = 200
56x = 160
x = 2.86 ft
2007-07-06 21:00:34
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answer #3
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answered by ironduke8159 7
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25 - (5 times the square root of 17) all divided by 2
2007-07-06 21:05:14
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answer #4
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answered by Anonymous
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Assume the question is: how wide is the edge?
(30-2x)(20-2x) = 400
600 - 100x + 4x^2 = 400
4x^2 - 100x +200 = 0
x^2 - 25x + 50 = 0
x^2 - 25x +156.5 = 106.5
(x-12.5)^2 = 106.5
x -12.5 = sqrt(106.5)
x = 12.5 +/- 10.32 = 22.82, 2.18
22.82 is wider than the narrow way, so the answer is the path is 2.18' wide or 2' 3.84"
2007-07-06 21:07:04
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answer #5
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answered by Steve A 7
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what's your question?
2007-07-06 20:51:12
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answer #6
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answered by Anonymous
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