Question 1075503: Daisy and Johnny are mowing a rectanglar lawn that is 50m by 80m. Johnny mows a strip of uniform width inside the perimeter of the lawn and Daisy mows the inner rectangle. How wide is the strip if Johnny mows 4 times as much as Melinda?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! It helps to draw this.
x is the width of the perimeter that Johnny mows, and it affects the inner rectangle on both widths and both lengths.
The inner rectangle is (50-2x)(80-2x)=4000-260x+4x^2
The inner rectangle and the perimeter must add up to 4000, so the perimeter is 260x-4x^2
The perimeter is 4 times the inner rectangle.
Therefore, (260x-4x^2)=4(4000-260x+4x^2)
260-4x^2=16000-1040x+16x^2
0=16000-1300x+20x^2
factor out a 20
x^2-65x+800=0
x=(1/2)(65+/- sqrt (4225-3200); sqrt 1025=32.02
x=48.51m and 16.49m
The first is impossible, since 2x would give a negative number.
The perimeter is 16.49 m ANSWER
The inner rectangle is 17.02m x 47.02m =800.28 m^2
The strip is 4000-800.28=3199.72 m^2 , 4 times that within rounding error.
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You could come at this problem by realizing that 4 times hers makes the two mowing 3200 m^2 and 800 m^2.
If the inner rectangle is 800 m^2, it has to be 50-2x and 80-2x=800
4000-260x+4x^2=800
4x^2-260x+3200=0
x^2-65x+800=0, and solve.
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