SOLUTION: A rectangular garden is to be surrounded by a walkway of constant width. The garden's dimensions are 30 ft by 40 ft. The total area, garden plus walkway, is to be 1800 ft2. What

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Question 71023: A rectangular garden is to be surrounded by a walkway of constant width. The garden's dimensions are 30 ft by 40 ft. The total area, garden plus walkway, is to be 1800 ft2. What must be the width of the walkway to the nearest thousandth?
Found 2 solutions by stanbon, checkley75:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A rectangular garden is to be surrounded by a walkway of constant width. The garden's dimensions are 30 ft by 40 ft. The total area, garden plus walkway, is to be 1800 ft2. What must be the width of the walkway to the nearest
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Draw the picture.
Let the width of the walkway be "w" ft.
Then the width of the garden plus walkway is "30+2x"
and the length of the garden plus walkway is "40+2x"
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EQUATION:
(30+2x)(40+2x) = 1800 sq ft.
4(15+x)(20+x) = 1800
x^2+35x+300=450
x^2+35x-150=0
Use quadratic formula to get:
x=[-35+sqrt(35^2-4*-150)]/2
x=[-35+-sqrt1825]/2
x=[-35+-42.72]/2
x=7.72/2
x=3.86 ft
Cheers,
Stan H.

Answer by checkley75(3666) About Me  (Show Source):
You can put this solution on YOUR website!
garden=30*40=1200 ft^2
garden+walkway=1800 ft^2
let walkway width be x ft. then
(30+2x)*(40+2x)=1800
1200+80x+60x+4x^2=1800
4x^2+140x+1200-1800=0
4x^2+140x-600=0
using the quadratic equation we get
x=(-b+-sqrt[b^2-4ac])/2a
x=(-140+-sqrt[140^2-4*4*-600])/2*4
x=(-140+-sqrt[19600+9600)/8
x=(-140+-sqrt29200])/8
x=(-140+-170.88)/8
x=(-140+170.88)/8
x=30.88/8
x=3.86 feet for the walkway