Question

Farmer Joe wants to fence an area of 1.4 million square feet in a rectangular field and then divide it in half with a fence parallel to one of the sides of the rectangular. What is the least amount of fencing required if the partitian is :
a) Horizontal
b) Vertical

Answer 1

Let w be the width of the field, and l be the length

w * l = 1, 400, 000

So he has 2w + 2l feet of fence to fence that field

Then, if he halves that field with a horizontal fence, he uses another
l feet of fence. So he uses 2w + 3l feet of fence. Let F be the fencing
required:

F = 2w + 3l

But l = 1,400,000/w, so:

F = 2w + 4,200,000/w

dF/dw = 2 - 4,200,000/w^2

We find the minimum value of F when dF/dw = 0:

2 - 4,200,000/w^2 = 0

Hence 2w^2 = 4,200,000

Hence w^2 = 2,100,000

Hence w = 1.45 * 10^3 gives the minimum, and the value of F at that
minimum is 5.8 * 10^3 feet

Thus for a horizontal partitian, the least amount of fencing is roughly
5,800 feet.

For a vertical partitian:

F = 3w + 2l

Hence F = 3w + 2,800,000/w

So dF/dw = 3 - 2,800,000/w^2

This is equal to 0 when 3w^2 = 2,800,000 so when w = 966.0918

The value of F when w = 966.0918 is given by 5.8 * 10^3 feet

So the least amount of fencing required is the same for both vertical
and horizontal partitians, so it doesn't matter how the farmer should
divide the field in half.