Question 68077
(30-2X)(20-2X)=400
600-40X-60X+4X^2=400
4X^2-100X+600-400=0
4X^2-100X+200=0
X^2-25X+50=0
using the quadratic equation we get
x=(-b+-sqrt[b^2-4ac])/2a
x=(25+-sqrt[-25^2-4*1*50])/2*1
x=(25+-sqrt625-200])/2
x=(25+-sqrt425)/2
x=(25+-20.6155)/2
x=(25+20.6155)/2
x=45.6155
x=(25-20.6155)/2
x=4.3845/2
x=2.192 answer for the width of the path.
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IF YOU DRAW A DIAGRAM OF THE PROBLEM (WHICH I ALWAYS RECOMMEND) YOU'LL SEE THAT THE RECTANGLE INSIDE ANOTHER RECTANGLE WITH THE OUTSIDE DIMENTIONS OF 20 & 30 FT
YOU'LL SEE THAT THE INNER RECTANGLE LENGTH IS 30 & 20 -2 TIMES THE BORDER (X) THUS THE TERMS 30-2X & 20-2X TO DESCRIBE THE INNER RECTANCLE.