Calculus Homework: Length of Fence

Question

A rancher wants to fence in an area of 3000000 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?

Length of fence = ______ feet.

Answer 1

Let l = length

w = width of rectangle

area = 3000000 = iw or l = 3000000/w

fencing = f = 3w + 2l

f = 3w + 2*3000000/w

differentiate wrt w

f' = 3 - 6000000w^-2

set this equal to zero to find minimum

0 = 3 - 6000000w^-2

w = √2000000 = 1000√2

l = 3000/√2 = 1500√2

f = 3w + 2l = 6000√2 feet

 

Answer 2

I'll use the variable A for the 3000000 sq ft area enclosed by the fence (just so everything looks neater when I type it out here).

 

We know the area is a rectangle, so A must be equal to length*width. If we mark the shorter side of the fence as x, the length of the other side has to be A/x. The total perimeter of the rectangle (and thus the required length of fence to enclose A) is P=2x+2A/x.

 

______________A/x________________

|                                                        |

|x                        A                            | x

|                                                        |

|________________________________|

                       A/x

 

We are not told how the rancher splits A into two, but there are only two possibilities:

______________A/x________________                           _______________A/x______________
|                          |                            |                           |                                                      |
|                          |                            | x                        |_______________________________| x

|                          |                            |                           |                                                      |

|_______________|_______________  |                          |_______________________________|

If the rancher wants to use the least amount of fencing, A should be split parallel to the shorter side, so that the length of the dividing fence is x.

 

This means the total fencing used equals the perimeter plus the divider: F= (2x+2A/x)+x = 3x+2A/x. We can now use optimization to find the length x that will use up the least amount of fencing F.

 

Taking the derivative:

F' = 3-2A/x²

Set equal to zero and solve for the critical points:

0=3-2A/x²

3=2A/x²

x²=2A/3

x=±√[2A/3].

 

We can reject the x value corresponding to a nonphysical answer, and substitute the usable value into our function for the total fencing.