SOLUTION: A rancher plans to use 200 yards of fencing to enclose a rectangular corral and to divide it into two parts with a fence parallel to the shorter sides of the corral. Find the dime

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: A rancher plans to use 200 yards of fencing to enclose a rectangular corral and to divide it into two parts with a fence parallel to the shorter sides of the corral. Find the dime      Log On


   



Question 1099343: A rancher plans to use 200 yards of fencing to enclose a rectangular corral and to divide it into two parts with a fence parallel to the shorter sides of the corral. Find the dimensions of the corral if its area is to be 1600 square yards.
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
the corral will essentially have 3 widths, 2 ends and the one in the middle dividing it. There are two lengths.
the width=x, and there are 3 of them
what is left over is 200-3x, and that is 2 lengths, so that one length is half that or 100-(3/2)x
The area is their product, x(100-1.5x)=-1.5x^2+100x, and that equals 1600
-1.5x^2+100x-1600=0
1.5x^2-100x+1600=0
3x^2-200x+3200=0
x=(1/6)(200+/- sqrt (200^2-12(3200)), and sqrt term is 40000-38400=1600, and that is 40
the roots are (1/6)(240)=40 yds width, and the length is 100-60=40 yds as well. The width is 40 yds, length is 40 yds and area is 1600 yds^2,
also (1/6)160=26 2/3 yds width. The length is 100-40=60 yds, so the corral is 60*26 (2/3)=1600 yds^2.
40 x 40
60 x 26 2/3
graph%28300%2C300%2C20%2C50%2C-100%2C2000%2C-1.5x%5E2%2B100x%2C1600%29
graph%28300%2C300%2C-10%2C50%2C-100%2C2000%2C-1.5x%5E2%2B100x%2C1600%29