It is required to enclose a rectangular filed by a fence and then divide it . . .?

Question

into two lots by a fence parallel to the short side. If the area of the field is 25,000 ft^2, find the lengths of the sides so that the total length of the fence will be a minimum.


This is an optimization quesiton. Any ideas?

Answer 1

Let x = length
then width = 25,000/x
So 2x + 3*25000/x = amount of fence requred.
dy/dx = 2 -75000/x^2
Set the above = 0 getting:
2-75000/x^2 = 0
2x^2 = 75000
x^2 = 37500
x = 50sqrt(15) approximately = 193.65 feet = length of fence
25000/193.65 approximately = 129.1 feet = width

Answer 2

We can't blindly assume that the exact shape that fits the given area and minimizes the amount of fence is going to be a 2x1 rectangle. In fact, you can find a solution that uses less fence.

Let L be the length of the total area and W be the width, both in feet. Then LW = 25,000.

The total amount of fence is going to be twice the length (to cover those two sides), plus twice the width (to cover the other two sides) plus one more width (which is the length of the partition in the middle). This means 2L + 3W = the total fence length. Using the substitution L=25000/W, we get
2(25000/W) + 3W. We want to find the value of W that minimizes this function. So take the derivative and set it equal to zero.

-50,000(w^-2) + 3 = 0
3 = 50,000(w^-2)
(w^2)3 = 50,000
w^2 = 50,000/3
w = √(50,000/3), which is about 129.099 ft. This gives L as approximately 193.649 ft. Note that this gives a smaller length of fence (645.496 ft) than if we were to assume the exact shape is a 2x1 rectangle (670.82 ft).

Answer 3

Yes. This an optimization question.

25,000 / w........Area = (Length)(width)
________ .......25,000 = Lw
|.......|........| .......25,000 / w = L
|w....|w.....|w
|.......|........| ........Let f(w) = total amount of fencing
--------------
25,000 / w

f(w) = w + w + w + 25,000 / w + 25,000 / w
f(w) = 3w + 50,000 / w

Find the critical values of w at which f(x) is at a minimum by setting the first derivative of f(w) equal to zero.

f(w) = 3w +50,000w^(-1)
f'(w) = 3 - 50,0000x^(-2)
f'(w) = 3 - 50,000 / w^2

0 = 3 - 50,000 / w^2
0 = 3w^2 - 50,000
50,000 = 3w^2
50,000 / 3 = w^2
+or- sqrt( 50,000 / 3) = w
+or- 100sqrt(15) / 3 = w

w is positive, so w = 100sqrt(15) / 3 = 129.1 feet, approximately. The three short sides are 129.1 feet.

The long side is 25,000 / 129.1 = 193.6 feet, approximately.

Given the conditions of this problem, the minimum amount of fencing is f( 100sqrt(15) / 3 ) = 774.6 feet, approximately.

Answer: The three short parallel sides are 129.1 feet each and the long side is 193.6 feet.

Answer 4

Smallest fence will be cutting the rectangle into two squares.

Therefore the sides of the rectangle are x and 2x

Area = 2x^2

x = 111.80ft

2x = 223.60ft