A farmer has 3000 m of fencing and wishes to enclose a rectangular area and then divide it into 3 congruent rectangular areas with 2 dividing fences parallel to one side. Find the largest total area that can be enclosed by this fencing.

Answer 1

let the length of the large rectangle be x m
let the width of the large rectangle by y m
So I see:
2y + 4x = 3000
y = 1500 - 2x

area = xy = x(1500 - 2x) = -2x^2+ 1500x
d(area)/dx = -4x + 1500 = 0 for max of area
-4x = -1500
x = 375

largest area = -2(375)^2 + 1500(375) m^2 = 281,250 m^2