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.
Let the length of the region be x meters and the width be y meters.
Since the region is divided into three equal regions, each region will have dimensions of x/3 meters by y meters.
The amount of fencing used for the length of the region will be 2x meters, and the amount of fencing used for the width of the region will be 2y meters. Since there are three equal regions, the total fencing used will be:
2x + 2y + 2x + 2y + 2x + 2y = 120
6x + 6y = 120
x + y = 20
y = 20 - x
To find the maximum area possible, we need to maximize the area of the entire region, which is given by A = 3xy.
Substitute y = 20 - x into the area equation:
A = 3x(20 - x)
A = 60x - 3x^2
To find the maximum area, we need to find the maximum value of the function A. The maximum value of a quadratic function ax^2 + bx + c occurs at x = -b/2a.
In our case, a = -3, b = 60, and c = 0. Therefore:
x = -60/(2*-3) = 10
Therefore, the maximum area of the entire region is:
A = 60(10) - 3(10)^2 = 600 - 300 = 300 square meters.