To solve the problem of finding the maximum area of the enclosed rectangle, follow these steps:

Step 1: Define Variables

Let the sides of the rectangle be (x) and (y). Assume the dividing fence is parallel to (y), so the total fence length is:
(2x + 3y = 600) (2 outer sides of (x), 3 sides of (y) including the divider).

Step 2: Express Area in Terms of One Variable

From the fence length equation:
(2x = 600 - 3y \implies x = \frac{600 - 3y}{2}).

The area (A = x \cdot y):
(A = \left(\frac{600 - 3y}{2}\right)y = 300y - \frac{3}{2}y^2).

Step 3: Find Maximum Area

This is a quadratic function (A(y) = -\frac{3}{2}y^2 + 300y) (opening downward, so maximum at vertex).

The vertex of (ay^2 + by + c) is at (y = -\frac{b}{2a}):
(y = -\frac{300}{2 \cdot (-\frac{3}{2})} = 100).

Substitute (y = 100) into (x):
(x = \frac{600 - 3(100)}{2} = 150).

Step 4: Calculate Maximum Area

(A = x \cdot y = 150 \times 100 = 15000).

Answer: (\boxed{15000})

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