Please HELP with this College Algebra problem! Its due tomorrow.... HELP!!!!?

Question

I'm trying to find the total feets of fencing that i will need for the following problem.?
I will build a fence that encloses two sides of my barn, and to save costs i will use the creek, which runs parallel to the short side of my barn, as a border. If the dimensions of my barn are 150 feet by 225 and the fence in area is 84000 square yards, not including the area of the barn. That means that i will need a total of how many ??? feet of fencing?
Please show the formula!!!!!
Thanks

Answer 1

Here's what you do. Not knowing whether it is a square or a rectangle shaped fenced area, you would:
base * width = area
(knowing that you are using the short side of the barn and have the area, however first you need to get all the numbers into equal terms-feet & feet or yards & yards, not feet & yards. Knowing that there are 3 feet in a yard, divide the short side of the barn by 3. So 150/3 gives you 50. So-
50 * x = 84000
Take and divide 84000 by 50 and you get 1680. Now that we know that you would 1680-50(for the barn) - 50(for the creek) = 1580. That is still in yards so, take 1580 and multiply by 3 to get feet. 4740 feet of fencing.
I would double check your units of measure in the question, to make sure that you have them correct. It seems awfully big for an answer. (I also recommend not using the creek as a source for fencing. It doesn't work, the animals still get out.)

Answer 2

Does the question state that the fence (and the creek) have to be the same distance from the barn, all the way around on all sides? You mention two sides, but minus the creek, there would be three sides fenced in. So this is confusing.

But if it is the same distance between the barn and the fence on all four sides (including the creek) then given the barn is 75 yds by 50 yds, and given the distance to the fence/creek is a constant X added on all four sides:

Then (50*X)(2) [front and back of the barn/short side]
+ (75 + 2X)(X)(2) [both sides along the barn/long side]
= 84,000
100X + 150X + 4X^2 = 84,000
4X^2 + 250X - 84,000 = 0
2X^2 + 125X - 42,000 = 0

I would use the Quadratic Formula to solve for X.
Then to get the length on all four sides
50 + 2X is the short side (counted one time only)
75 + 2X is the long side (counted twice)
Add those together (remember not to count one of the
short sides that the creek is used for) to get the length of fencing.

That is IF the fence has to be the same distance from the barn on all four sides, and IF you are using fencing on THREE sides not TWO.

However, if this problem is about MINIMIZING the amount of fencing to ensure 84,000 of fenced in area, and the distance from the barn can vary as long as the area is 84,000 and the fencing is MINIMUM then that is a different problem altogether!

The problem as given does not seem to come out even, so I wonder if the question is not worded correctly? Add details?

Answer 3

Rearrange this as x² - 5 = 0. Now this is of the variety ax² + bx + c = 0, the area a=a million, b=0, and c=-5. in case you're required to place those coefficients in the time of the quadratic formula besides, write x = ( -0 ± ?(0² - 4(a million)(-5)) ) / 2(a million) = ( 0 ± ?(0 + 20) ) / 2 = ±(?20)/2 = ±(2?5)/2 = ±?5. answer: x = ?5 or x = -?5.