SOLUTION: A garden area is 30 ft long and 20 ft wide. A path of uniform width is set around the edge. If the remaining garden area is 400 ft2, what is the width of the path?

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Question 89369: A garden area is 30 ft long and 20 ft wide. A path of uniform width
is set around the edge. If the remaining garden area is 400 ft2, what is the width of the path?
Answer by ptaylor(2198) About Me  (Show Source):
You can put this solution on YOUR website!
SEE PROBLEM #89194
Original garden area is 30*20 or 600 sq ft
Let x=width of path
Now we'll divide the path up into 4 rectangles:
Two of the rectangles are 30 ft long and x ft wide and they have a combined area of: 2*30x or 60x sq ft
The other two rectangles are (20-2x) ft long and x ft wide and they have a combined area of 2*(20-2x)*x or 40x-4x^2 sq ft
Now we are basically told that the area of the path is (600-400)or 200 sq ft. So:
60x+40x-4x^2=200 subtract 200 from each side
60x+40x-4x^2-200=200-200 collect like terms
100x-4x^2-200=0 divide each term by -4
x^2-25x+50=0 quadratic in standard form. Solve using the quadratic formula:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x+=+%2825+%2B-+sqrt%28+%28-25%29%5E2-4%2A1%2A50+%29%29%2F%282%2A1%29+
x+=+%2825+%2B-+sqrt%28+625-200+%29%29%2F%282%29+
x+=+%2825+%2B-+sqrt%28+425+%29%29%2F%282%29+
x+=+%2825+%2B-+sqrt%28+425+%29%29%2F%282%29+
x+=+%28%2825+%2B-+20.62%29%29%2F%282%29%29
Lets look at:
x+=+%28%2825+%2B+20.62%29%29%2F%282%29%29;
x+=+%28%2845.62%29%29%2F%282%29%29;
x+=+22.81 ft--------------------NO!
Lets look at the other value for x:
x+=+%28%2825+-+20.62%29%29%2F%282%29%29;
x+=+%28%284.38%29%2F%282%29%29;
x=2.19 ft---------------------------Answer
CK
We'll see if the area of the path is 200 sq ft
2*30*(2.19)+2*(20-4.38)*2.19=
131.4+68.42=199.82 ~~~200 sq ft
We obviously have some round-off error that you can work on
Hope this helps----ptaylor