A rectangular region, placed against a wall of a house, is divided into three equal regions using a total of 120 meters of fencing. Calculate the maximum area possible for the entire region.
The dimensions of each individual area.
Let the length of the rectangular region be x meters and the width be y meters.
Since the entire region is divided into three equal regions, each region has dimensions of x/3 meters by y meters.
The total length of fencing used is 120 meters, so we have:
2x + 4y = 120
x + 2y = 60
Solving for x in terms of y:
x = 60 - 2y
The area of the entire region is given by:
A = 3xy
Substitute x = 60 - 2y:
A = 3(60 - 2y)y
A = 180y - 6y^2
To find the maximum area, we take the derivative of A with respect to y and set it equal to zero:
dA/dy = 180 - 12y = 0
12y = 180
y = 15
Substitute y = 15 back into x = 60 - 2y:
x = 60 - 2(15) = 30
Therefore, the dimensions of each individual area are 30 meters by 15 meters.
The maximum area for the entire region is:
A = 3(30)(15) = 1350 square meters.