Optimization maximum area Problem

[Saterial]

1) The question
A rectangular pen is to be built with 1200 m of fencing. The pen is to be divided into three parts using two parallel partitions.
A) Find the maximum possible area of the pen. (45000 m^2)
B) explain how the maximum area would change if each side of the pen had to be at least 180 m long.

2) Relevant Equations
A=lw

3) Attempt at solution

A=lw
=(1200-2w)w
=1200w-2w^2
dA/dw=1200-4w
0=1200-4w
w=300
l=600
A=(600)(300)
=180000 m^2

This was the method I was taught to solve this question. I don't understand how the answer is 45000 m^2.

b) 0>_w>_180

Solve for w at both 0 and 180

0>_l>_180

Solve for l at both 0 and 180

Find area using a=lw

Once again incorrect answer by myself. What am I doing wrong? Am I missing something?

 

[eumyang]

1) The question
A rectangular pen is to be built with 1200 m of fencing. The pen is to be divided into three parts using two parallel partitions.
A) Find the maximum possible area of the pen. (45000 m^2)
B) explain how the maximum area would change if each side of the pen had to be at least 180 m long.

2) Relevant Equations
A=lw

3) Attempt at solution

A=lw
=(1200-2w)w

You didn't take into account that the pen is divided into 3 partitions. The total length of fencing is not the perimeter of the rectangular pen (2L + 2W), but the perimeter plus 2 parallel partitions. If you let one partition be the same as the width, then the total fencing would be
2L + 4W = 1200

Solve for L, plug into A=LW, and if you redo your steps you'll get the maximum area.