View Full Version : Math Help Pt.II
Jul 18, 2004, 02:37 PM
yea yea..i need help again...my mentors arnt teaching really explaining this clearly so can i get help from u guys?
fencing in goats
Wajibu is planning his retirement. although he loves teaching he was told his retirement plan would pay him what he is presently earning.. He plans to teach for only one more year. in planning his retirement he purchased some land in sacramento and plans to move there and raise goats. the land he bough is one acre. the only building on the land is a barn 150' by 40'. he bought 50 baby goats to raise. he used all his savings on the land and the goats that he didnt realise that he didnt buy a fence. looking troughout the barn her founf 110' of fencing and two fence posts. using all the stuff he found what is the the largest size area he can build for the goats///
ok so can u help?
Jul 18, 2004, 02:53 PM
I think you will want to build a circualar pen in order to maximize your fencing area. As far as I can tell, the fact that the barn is 150' x 40' is irrelevant.
Since pi times diamteter equals circumfrance, 110 (circumfrance) divided by pi will equal the diamter, or around 35. The formula to find the surface area is pi times radius squared, so 17.5 squared time pi equals 962.9, which should be the maximum area. I'm not exatly sure where the two fence posts come in, though.
Jul 18, 2004, 03:13 PM
I think that the fencing is straight. As you only have 2 fence posts you cannot build curves. You also cannot use the fence on it's own. What you can do is put 2 posts in the ground and use the wall of your barn as 1 side of your pen. Then you have to split your 110 meters of fence out. So you want 33.3333333 meters of fence per side (and use 33.333333 meters of your barn wall). So your max air is 33.333333 squared which is 1111.1111111111 square meters.
I've just noticed your in feet so switch meters for feet. The correct answer to this is probably to form an equation that describes the area contained with regard to 2 variables (x and y) and solve for max area.
Jul 18, 2004, 03:51 PM
are you taking multivariable calculus? use the lagrangian. the constraint should be 110-x-2y and you're solving for an area xy. savvy? y is the short side.
Jul 18, 2004, 03:57 PM
in the hope that your professor requires you to show work, the answer should end up as the short side being 27.5 feet and the long side being 55 feet with an area of 1512.5
Jul 18, 2004, 03:59 PM
20 seconds of Calculator playing suggest that the numbers should be roughly 55' x 27.5', yielding an area slightly over 1512 square feet. Over 30 square feet per goat. Not great, not terrible.
Of course, 30 seconds of math would probably yield a better solution. But if the area is < 1512 square feet, you haven't found the max.
Edit: jasylonian beat me to it....
Jul 18, 2004, 04:54 PM
Solution: Just kill the damn goats and sell the meat on eBay.
Jul 19, 2004, 02:40 AM
Maybe the area of the land (1 acre - you could convert that to square feet) and/or the shape of the barn constrain the choices for how big an area you can fence.
Jul 20, 2004, 12:22 AM
meh i got this and my teacher was like ok