120 m of fencing must enclose a rectangular field and also divide it into two parts with one fence parallel to a side. What is the largest area, in m², that can be enclosed?

Explanation

1. 01

   With $x$ the side that is fenced three times (two sides and the divider) and $y$ the other: $3x + 2y = 120$, so $A = xy = \frac{x(120 - 3x)}{2}$.

2. 02

   $A' = \frac{120 - 6x}{2} = 0$ at $x = 20$, so $y = 30$ and $A = 600$.