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.

 

[quicknote]

I am unable to provide a definitive answer to your question as I do not have access to the specific details of the problem, such as the dimensions of the back of the house and the specific fencing materials being used. However, I can provide some general guidance on how to approach this type of problem.

Firstly, it is important to understand the question and what is being asked. In this case, Mrs. Star wants to maximize the area of the dog pen given a fixed amount of fencing (600 feet) and using the back of the house as one side of the pen. This means that the total perimeter of the pen will be 600 feet, with one side being the back of the house and the remaining three sides being made of fencing.

To find the maximum area, we need to use the formula for the area of a rectangle, which is length x width. We know that the width will be equal to the length of the back of the house (let's call this dimension "a") minus twice the width of one side of the pen (let's call this dimension "b"). Therefore, the width of the pen will be a - 2b.

Next, we need to express the length and width of the pen in terms of one variable. Since we are trying to maximize the area, it makes sense to express the length in terms of the width. This can be done using the fact that the total perimeter of the pen is 600 feet. So we can set up an equation:

Perimeter = 2(length + width) = 600
Substituting in our expressions for the length and width, we get:
2[(a - 2b) + b] = 600
Simplifying:
2(a - b) = 600
a - b = 300
a = 300 + b

Now, we can substitute this expression for "a" into our formula for the area of the pen:
Area = length x width
Area = (300 + b) x (a - 2b)
Area = (300 + b)(300 + b - 2b)
Area = (300 + b)(300 - b)
Area = 90000 - b^2

To find the maximum area, we can use the vertex form of a quadratic equation, which is:
y = a(x - h)^2 + k
where (h, k) is the vertex of the parab