as always in problems like this, the maximum area is achieved when the fencing is divided equally among lengths and widths.
Since you have 5 lengths and 2 widths, the field is 200x80
To go through the math, you have
5x+2y = 800
a = xy = x(800-5x)/2 = 400x - 5/2 x^2
The vertex of this parabola is at x = -b/2a = 400/5 = 80
A total of 800 m of fencing material is used to enclose a rectangular field and to divide the field into four portions by fences parallel to one of the sides of the field. Find the maximum area which can be enclosed in this rectangle and it’s dimensions
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