SOLUTION: A rancher plans to use 340 yards of fencing for a corral and divide it into two parts with the fence parallel to the shorter sides of the corral. find the dimensions of the corral

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Question 1154296: A rancher plans to use 340 yards of fencing for a corral and divide it into two parts with the fence parallel to the shorter sides of the corral. find the dimensions of the corral if its area is to be 4700 square yards. I used a previously answered questions and my answer still came up wrong

two answers i came up with 45.43967 & 101.8405
or 65.4858 & 71.7712
Found 2 solutions by josgarithmetic, MathLover1:
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
w width (usually the shorter dimension is "width").
L length (the longer dimensions is usually "length").

Total fencing 2w%2Bw%2B2L=340
or
3w%2B2L=340


The area only depends on the entire rectangular region's width and length, so 4700=wL.


System to solve, in the two variables w and L would start as:
system%283w%2B2L=340%2CwL=4700%29.
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.
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Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
A rancher plans to use 340 yards of fencing for a corral and divide it into two parts with the fence parallel to the shorter sides of the corral.
=> it is a rectangle divided into two parts; so, if L is the longer side and W is the shorter side, you will have 2L and W whose sum is 340

find the dimensions of the corral if its area is to be 4700 square yards.

perimeter in this case would be:

2L+%2B+3W+=+340....solve for L
2L+=+340+-+3W
L+=+340%2F2+-+%283%2F2%29W
L+=+170+-+1.5W.........eq.1

The area:

A=L%2AW......plug in desired area and substitute L from eq.1
4700=%28170-1.5W%29%2AW.....solve for W
4700=170W-1.5W%5E2
1.5W%5E2-170W%2B4700=0......use quadratic formula
W=%28-%28-170%29%2B-sqrt%28%28-170%29%5E2-4%2A1.5%2A4700%29%29%2F%282%2A1.5%29
W=%28170%2B-sqrt%2828900-28200%29%29%2F3
W=%28170%2B-sqrt%28700%29%29%2F3
W=%28170%2B-10sqrt%287%29%29%2F3

exact solutions:
W=%28170%2B10sqrt%287%29%29%2F3 or
W=%28170-10sqrt%287%29%29%2F3
approximate solutions:
W=65.49 or
W=47.85

find L
L+=+170+-+1.5%28170%2B10sqrt%287%29%29%2F3.........eq.1
L+=+170+-+%28170%2B10sqrt%287%29%29%2F2 or
L+=+170+-+%28170-10sqrt%287%29%29%2F2

approximate solutions:
L+=+71.77 or
L+=+98.23
so, you have:
L+=+highlight%2898.23%29 and W=highlight%2847.85%29
L+=+highlight%2871.77%29 and W=highlight%2865.49%29

check the perimeter:
2L+%2B+3W+=+340
2%2A98.23+%2B+3%2A47.85+=+340
340.01=+340 round it
340=+340
or
2L+%2B+3W+=+340
2%2A71.77+%2B+3%2A65.49+=+340
340.01=+340 round it
340=+340
check the area:
4700=98.23%2A47.85
4700=4700.3055round it
4700=4700
4700=71.77%2A65.49
4700=4700.2173round it
4700=4700