Question 276319
Jim has designed a rectangle with an area of 100 sq feet and a perimeter of 401 ft. What are the dimentions of Jims rectangle?

let the length be x
and the width be y.

2x+2y = 401
and xy = 100

y= 100/x

2x+ 2*100/x=401

2x^2 +200 = 401x

2x^2-401x+200=0

2x^2-400x -x +200=0

2x(x-200)-1(x-200)=0

(x-200)(2x-1)=0

x= 200 or 1/2


The rectangle will have dimensions of 200 by 1/2