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.
List the dimensions of each individual area.
Let's assume the rectangular region has length L and width W.
Since the total length of the fencing used is 120 meters, and each region is divided using an equal amount of fencing, the length of each side is 120/3 = 40 meters.
So, the dimensions of the rectangular region are 40 meters x L meters.
To find the maximum area possible, we need to maximize the length L while keeping the total fencing length the same. The perimeter of the rectangular region is 2L + 2W.
Since we know the lengths of two sides (40 meters each), the remaining fencing length is (120 - 2*40) = 40 meters.
This remaining 40 meters of fencing should be used to maximize the width W since the longer the width, the larger the area.
So, W should be 40/2 = 20 meters.
Therefore, the dimensions of the individual areas are as follows:
1) One of the equal regions: 40 meters x 20 meters
2) Another equal region: 40 meters x 20 meters
3) The remaining region: 40 meters x L meters
To find the value of L, we know the total length of the fencing used is 2L + 2W, which is 120 meters.
So, 2L + 2(20) = 120
2L + 40 = 120
2L = 120 - 40
2L = 80
L = 80/2
L = 40 meters
Therefore, the dimensions of each individual area are:
1) 40 meters x 20 meters
2) 40 meters x 20 meters
3) 40 meters x 40 meters
The maximum area for the entire region is the sum of the areas of the three individual areas:
Area 1 = (40 meters x 20 meters) = 800 square meters
Area 2 = (40 meters x 20 meters) = 800 square meters
Area 3 = (40 meters x 40 meters) = 1600 square meters
Maximum area = Area 1 + Area 2 + Area 3
Maximum area = 800 square meters + 800 square meters + 1600 square meters
Maximum area = 3200 square meters
Therefore, the maximum area possible for the entire region is 3200 square meters.