SOLUTION: Hi guys, I am really stuck with this problem, maybe somebody could help me out. A rancher has 2100 feet of fencing with which to construct adjacent, equally sized rectangular

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: Hi guys, I am really stuck with this problem, maybe somebody could help me out. A rancher has 2100 feet of fencing with which to construct adjacent, equally sized rectangular      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 1057201: Hi guys,
I am really stuck with this problem, maybe somebody could help me out.
A rancher has 2100 feet of fencing with which to construct adjacent, equally sized rectangular pens as shown in the figure above. What dimensions should these pens have to maximize the enclosed area?
This is a picture that belonged to it as well.
http://i.imgur.com/qh13W07.jpg
x=
y=
Maximum area=
Thanks for your help!
Frank
Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!
So there are two things going on here. You have 2100ft of fence in which to make the adjacent pens and you want to do that such that the area is maximized.
The relationships are as follows:
(1) Perimeter = 2x+y+2x+y+y = 2100 (3 y's b/c there's a divider in the middle)
(2) Area = 2xy
Eqn (1) can be simplified to +4x%2B3y=2100+
So the plan here is to solve for y then substitute into the Area formula. I will then apply a small amount of calculus to find the maximum area:
+y=%281%2F3%29%282100-4x%29+ (1a)
Now substitute this for 'y' in (2):
+Area+=+2%28x%29+%28%281%2F3%29%282100-4x%29%29+
+Area++=+2x%28700-%284%2F3%29%28x%29%29+
+Area+=+1400x+-+%288%2F3%29%28x%5E2%29+
Now take derivative of Area with respect to x:
++d%28Area%29%2Fdx+=+1400-%2816%2F3%29x+
Maximum (or minimum) will be where derivative is zero, so set this to zero:
++1400-%2816%2F3%29x++=+0+
++x+=+262.5+ ft
From eqn(1a), this gives
+y+=+350+ ft
+Max+Area+=+%282%29%28262.5%29%28350%29+=+183750+ ft%5E2
As an exercise, you can adjust x to some other values on either side of 262.5 (e.g. 260, 270) to prove to yourself this is the maximum area (use eqn (1a) to find y for each value of x that you pick then use (2) to calculate the area).