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Question 188830: A parking lot has a rectangular area of 40,000 sq. yards. The lenght is 200 yards more than twice the width. what are the dimensions of the parking lot?
Answer by BillKearns2(1)   (Show Source):
You can put this solution on YOUR website! First I would draw a rectangle.
The length is 200 yards more than twice the length
So set length to 2x+200 and the width is x
The area is 40,000 so: (2x+200)(x)=40000 Since L(w)=Area
(2x+200)(x)=40000
2x^2+200x=40000
Subtract 40,000 from both sides and set to zero
2x^2+200x-40,000=0
Factor (may be easier to factor out a 2 first.
2(x^2+100x-200)=0
2(x-100)(x+200)=0
now solve
2=0 No
x-100=0: x=100 possibly
x+200=0: x-200 no, cant have a negative side so use x=100
so insert 100 into x of original length
2x+200
2(100)+200=400
So, length is 400 and width is 100
Check your answer by seeing if 400 is actually 2times plus 200 of 100 (It is)
Check the area Area=LW
A=400(100)
A=40,000
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