SOLUTION: A landscaper wants to put a cement walk with uniform width around a garden that measures 20x40 feet, She has enough cement to cover 660 ft squared. How wide should it be to us al

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Question 250431: A landscaper wants to put a cement walk with uniform width around a garden that measures 20x40 feet, She has enough cement to cover 660 ft squared. How wide should it be to us all the cement?
Found 3 solutions by drk, checkley77, Greenfinch:
Answer by drk(1908)   (Show Source):
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The original walkway: 20x40. New walkway with cement is (20 + 2x) x (40 + 2x). The total walkway area - garden area is what she will cement. She has enough cement for 660 ft ^2.

By quadratic we get
X~4.75 feet.

Answer by checkley77(12844)   (Show Source):
You can put this solution on YOUR website!
20*40+660=(20+x)(40+x)
800+660=800+60x+x^2
x^2+60x-660=0

x=(-60+_sqrt[60^2-4*1*-660])/2*1
x=(-60+-sqrt[3600+2640])/2
x=(-60+-sqrt6240)/2
x=(-60+-78.99)/2
x=(-60+78.99)/2
x=19.99/2
x=9.497 ans. for the width of the cement walk.
Proof:
20*40+660=(20+9.497)(40+9.497)
800+660=29.497*49.497
1460~1460

Answer by Greenfinch(383)   (Show Source):
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Assume width to be p feet. Then amount of path is Outside (40 x 20 feet) less the inside (40 - 2p)(20 - 2p) feet
or 800 - (800 - 80p - 40p +4p^2) = + 120p -4p^2
This looks right because it is the outside distance (120 feet) times the width of p less the four corners which are p squared in area
So 120 p - 4 p^2 = 660
rearranging 4p^2 - 120 p + 660 = 0
or p^2 - 40 p + 165 = 0
using formula p = {40 -/+ sqrt{40^2 - (4 x 1 x 165)} }/2
which I make 20 -/+ sqrt 235
The - option gives 4.67 feet
The + option gives 35.33 feet
The 4.67 is the right option, The other is 40 - 4.67 which is greater than one side of the rectangle.