Get the latest news and information from us. Join our newsletter!
What should maximum dimensions be to enclose the maximum area using all available fencing?
You have 640m of fencing available. You need to fence in a rectangular area of fields that is then subdivided into three parts with fencing parallel to one of the sides of the rectangle. What should the dimensions be to enclose the maximum area using all of the available fencing?
Answer: Ok...this should be good. Here is a diagram:
http://farm4.static.flickr.com/3269/3098140145_13c3fd4e8e_o.jpg
Here:
x = length
y = width (also the side parallel to where it is all divided)
Given:
Amount of fence = 640
Need max area.
In this example the total amount of fence used is:
fence_used = perimeter of field + fence used inside enclosure
640 = (2*x + 2*y) + ( 2y )
640 = 2x + 4y
640 - 2x = 4 y
y = 160 - (1/2)*x
Area becomes:
area = x*y
area(x) = x*(160 - (1/2)*x)
area(x) = 160x -(1/2)*x^2
To find max and min of a function find area'(x), make it equal to zero and solve for x:
area'(x) = 160 - x
0 = 160 - x
x = 160 --> this should be your max value
Now since:
y = 160 - (1/2)*x
Plug in x:
y = 160-80 = 80
For max area the dimensions should be:
x = 160
y = 80
