Problema Solution
A rancher has 600 feet of fencing to put around a rectangular field and then subdivide the field into 2 identical smaller rectangular plots by placing a fence parallel to one of the field's shorter sides. Find the dimensions that maximize the enclosed area. Write your answers as fractions reduced to lowest terms.
Answer provided by our tutors
First, let's write an equation for the amount of fencing needed for the end result (two identical co-joined fields) where 'l' stands for the long side and 's' stands for the short side.
fencing=3s+4l
600=3s+4l
The enclosed area is:
A=2s(l+l)=4sl
To maximize the enclosed area, we need to make 'A' dependent on only one variable. Then we can take the derivative of 'A' with respect to that variable and set the derivative equal to 0 to find the max point.
A=4s((600-3s)/4)=4s (150-3/4s)=600s-6(s^2)
A'=600-12s
0=600-12s
s=50
Now we can find l from 600=3s+4l:
600=3(50)+4l
l=125.5
Therefore, the large rectangle has dimensions 251x50 ft and each smaller one has dimensions 125.5x50 ft (or 252/2 x 50 ft)