# A farmer has 120 feet of fencing available to enclose 2 adjacent rectangular pens alongside his barn. No fencing is required alongside the barn. He needs to have a total of 930 sq ft....

A farmer has 120 feet of fencing available to enclose 2 adjacent rectangular pens alongside his barn. No fencing is required alongside the barn. He needs to have a total of 930 sq ft. Determine the width and the combined lengths of the pens.

There are two solutions. One is 29.49 ft, 31.53 ft.

Determine the other solution rounded to two decimals.

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To calculate the field sizes we must create equations based on the given information.We know that the area to be enclosed is 930 sq ft and we know that `l times b=A`

`therefore l times b=930` as the fields are adjacent to each other.

We also know that the perimeter fencing available is 120ft and in this case we need 3 sides (the two ends and one to divide the field) and one length (the other side is the barn which does not need fencing). Perimeter is based on sides and length so the usual 2l x 2b = P would not work but `l +3b =P` would work based on our number of sides

`therefore l+3b=120`

In both these equations `l = l and b = b` so we can solve them simultaneously.

`l timesb=930 and l+3b=120` so substitution will work best:

`l +3b=120` becomes `l=120-3b` and substituting:

`l times b=930` becomes `(120 -3b) times b = 930`

Simplified:

`therefore 120b - 3b^2 = 930`

Create a trinomial:

`3b^2 -120b +930 = 0` Divide by 3 to simplify

`b^2 - 40b+310 = 0`

Use the quadratic formula to find b but do not confuse b(breadth) with b in the equation. a=1;b=-40 and c=310 from the equation

`therefore b(breadth)= (-b +-sqrt(b^2-4ac))/(2a)`

`therefore b(breadth) =(-(-40) +-sqrt((-40)^2 - 4(1)(310)))/(2(1))`

`therefore b=(40+-sqrt(1600-1240))/2`

`therefore b = 29.49 or b = 10.51`

Now use the 10.51 (as we are given the 29.49 above) and substitute to find `l`

`l=120-3b` becomes `l = 120 - 3(10.51)`

`therefore l=88.46`

**Ans: therefore the second set of measurements are 10.51ft and 88.46ft**