You can
put this solution on YOUR website! Let the three sides be y, x and y
Sum of the three sides x + 2y = 80 , so y = (80-x)/2
Area of the rectangle A = xy = x(80-x)/2 = 40x -x^2/2
By complete square A = -1/2(x^2 - 80x + (80/2)^2) + 1600/2
= -1/2(x - 40)^2 + 800
Since (x - 40)^2 >=0, -1/2(x - 40)^2 <= 0,
A = -1/2(x - 40)^2 + 800 <= 800,
when x = 40, y = (80-40)/2 = 20.
Hence, when the dimensions of the field are 20,40 and 20 meters
the max area is 800 sq. meters.
Kenny
PS. Also, try to get rid of some redunant words which are not
related to math in this question. Like :
There is a river that runs through her property so she decides to increase the size of the garden by using the river as one side of the rectangle.
You can
put this solution on YOUR website!As per question x represent the length of the side of the rectangle
along the river. Let the width of garden be y.
The length of fence = x + 2y = 80
and the area of the garden = xy
We need to maximize the area of the garden.
Suppose x = 40+k where k is some constant either positive or negative
The equation for fencs is
On putting x = 40+k in the above equation:
The area of garden is xy.
On putting the value of x and y in terms of k in area:
To maximize the area, we need to take the value of k = 0.
Then, Area = 800 square meter
and x = 40 and y = 20.
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