help solve this, quadratic applications

Question

Local park officials want to enclose a rectangular section of the park using 7,500 meters of fence. They further want to subdivide the enclosure into two (not necessarily equal) parts with fence parallel to one pair of sides. What is the largest area that can be enclosed? If necessary, approximate the area to the nearest square meter.

Answer 1

7500 = 2L + 3W = 2 equal sides of length + 3 other sides, to enclose and split the enclosure in two

2L = 7500 - 3W

L = 3750 - 3W/2

Area = LW = (3750-3W/2)W = 3750W - 3W^2/2

take the derivative and set it equal to zero to find the Width which maximizes Area

A' = 3750 - 3W = 0

3W = 3750

W = 1250 meters

L= 3750- 3W/2 = 3750 -3(1250)/2 = 3750 - 3(625) = 3750 - 1875 = 1875 meters

Largest area = LW = 1875 x 1250 = 2,343,750 square meters