SOLUTION: The perimeter of a rectangle is 50m. If the width was doubled and the length was increased by 20m, the perimeter would be 100m. What are the length and width of the rectangle?

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Question 380474: The perimeter of a rectangle is 50m. If the width was doubled and the length was increased by 20m, the perimeter would be 100m. What are the length and width of the rectangle?
Answer by mananth(16946) (Show Source):
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The perimeter of a rectangle is 50m. If the width was doubled and the length was increased by 20m, the perimeter would be 100m. What are the length and width of the rectangle?
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let width be w
length = l
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Area A = l*w=50 m^2
l= 50/w
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second rectangle
width = 2w
length = l+20
Perimeter of second rectangle =2*(2w+(l+20))
P=2(2w+l+20)
P=4w+2l+40
plug the value of l above
P=4w+2(50/w)+40
100=4w+100/w+40
multiply by w
100w=4w^2+100+40w
re write
4w^2-100w+40w+100=0
4w^2-60w+100=0
/4
w^2-15w+25=0
find the roots w1,w2 using quadratic formula
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discriminant = 125
w1=((15+sqrt(125))/2)
w1=13.1 meters ... the width
w2 will be negative so ignore
length = 50/13.1
length = 3.82 meters
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CHECK perimeter of second rectangle
plug w & l
2*(2*13.1+((3.82+20))=100
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m.ananth@hotmail.ca