Cost-effective design for fencing and partitioning a rectangular ranch field?

Click For Summary
The discussion focuses on two homework assignments related to fencing a rectangular ranch field. The first assignment involves minimizing the total length of fence needed for an area of 1,900,000 square feet, with calculations leading to a potential shortest side of approximately 1125.46 feet. The second assignment requires determining the dimensions of a 460 square foot area to minimize costs, with a calculated width of about 21.306 feet and a corresponding length of 21.590 feet. Clarifications are provided regarding the relationships between dimensions and costs in both problems. The thread emphasizes the importance of understanding how to derive and minimize functions related to fencing design.
mayo2kett
Messages
23
Reaction score
0
hi i have two homework assignment I'm kinda stuck on they are very similar i was hoping someone could help me...

1) A rancher wants to fence in an area of 1,900,000 square feet in a rectangular field and then divide it in half with a fence down the middle parallel to one side. What is the shortest length of fence that the rancher can use?

2) A rancher wants to fence in an area of 460 square feet in a rectangular field using fencing material costing 1.5 dollars per foot, and then divide it in half down the middle with a partition, parallel to one side, constructed from material costing 0.4 dollars per foot.
Assuming that the partition is parallel to the side which gives the width of the field, find the dimensions of the field of the cheapest design.
Length=
Width=
What is the (total) cost of the cheapest design?
 
Physics news on Phys.org
1) Let A be one side of the rectangular fence and B be the other side. Since the area of the fence must be 1,900,000 square feet, we can relate A and B like this:

B = 1,900,000/A

Now write down the function for the total length of the fence:

f(A) = 3A + 2B = 3A + 3,800,000/A

Find the derivative of that function, see where it has a minimum and find the value of the function in that minimum point.

2) Very similar problem. Again we will use W for width and L for length and relate them like so:

L = 360/W

The function for the total cost of the fence is:

f(W) = 2*1.5*W + 2*1.5*L + 0.4*W = 3.4W + 1080/W

Again find the derivative of that function, see where it has a minimum and find the value of the function in that minimum point.
 
for question one i did:
f(A)'=3-(3800000/A^2)
A=1125.46
is this shortest side? i wasn't sure where to go from here

for the second question i have:
f(w)'=3.04-(1380/w^2)
and w=21.306
so then i put the 21.306 into the original equation l=460/w=460/21.306=21.590
at this point I'm not sure what the l and the w stand for, are they the length or the sides of the fence or are they the cost of each side since we added in the cost of the different types of fencing??
 
1) You found A, how does it relate to B? Find B and see which of them is smaller.

2) L and W represent the length of the width and length of our fence. The cost was added in the f(W) function (the $1.5, $1.5 and $0.4 coefficient).
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
Replies
1
Views
2K
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
6K
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K