Question 123616: Local building codes require that a rectangular factory building must be surrounded by a parking lot of uniform width. The total area occupied by the factory and the parking lot must be double the factory area. If the factory is to be 80m wide and 120m long. How wide must the parking lot be? HELP:)
Answer by checkley71(8403)   (Show Source):
You can put this solution on YOUR website! 80*120=9,600 m^2 is the area of the factory.
therefore the total area of the factory & the parking lot must be: 2*9,600=19,200.
the formula for this problem is:
(80+2x)(120+2x)=19,200
9,600+240x+160x+4x^2=19,200
4x^2+400x+9,600-19,200=0
4x^2+400x-9,600=0
4(x^2+100-2,400)=0
4(x-20)(x+120)=0
x-20=0
x=20 answer for the width of the parking lot.
proof
(80+40)(120+40)=19,200
120*160=19,200
19,200=19,200
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