Maximizing Dog Pen Area: Help Mrs. Star Find the Best Solution!

[minase]

I took a test on my algerbra1 class and one of the question says the following.
Question
Mrs. Star wants to start a doggie day care. She plans to purchase 600feet of fencing to build a rectangular dog pen. Her plans also include using the back of the house as one side of pen. The area of the pen depends on how far away from the house the pen extends. Mrs. Star wants the Dogs to have the maximum area in which to play. Can you help her to find the maximum area?
And this was my answer
X represents the width of one side. Since in a rectangle two opposite sides are equal therefore the other side equal x. And Mrs. Star wants to purchase 600 feet of fencing and her plan was to include using the back of the house as one side of the pen. This means the width is always equal to 600-2x even the though the length may vary.
Area=length×width
A= (600-2x) (x)
A= 600x-2x^2
A= -2x^2+600x
A= -2(x^2 -300x)
-45000+A= -2(x^2-300x+2250)
-45000+A= -2(x-150) ^2
A= -2(x-150) ^ 2+45000
Therefore the maximum area Mrs. Star can build is 45,000 feet.
My friend here got a different answer I was wondering If I got it right. If I got it wrong can you please explain the steps how to do it.

 

[Atomos]

That is absolutely correct. Your friend is flawed hin his/her logic.