Area problems involving quadratics A recntagular play enclosure for some dogs is to be made with $60$m of the fencing using the kennel as one side of the enclosure. The quadratic function that models the area of the enclosure is represented by the function $A(x)=-2x^2+60x$ where $A(x)$ represents the area enclosed and $x$ represents the width in meters.

so I havem multiple questions I am looking for some help with:
$1)$ why is the length of the play enclosure $60-2x$?
$2)$ How would I find the maximum area?
$3)$ what would the domain and range be?   
I am looking for some help with this as  I have many examples in my notes of similar problems but I am not sure how to do them, but once I understand this I can finish the rest, thanks!!!
 A: With optimization problems, it's important to go over what the question is asking, with the help of the diagram given, so why not do that first?
"A rectangular play enclosure for some dogs is to be made with $60$ m of the fencing using the kennel as one side of the enclosure."
You're only using the fencing to form three sides of the fencing. That means that the length of all three side lengths has to be $60$. If the other two sides have a variable length $x$, then the side parallel to the kernel has a length $60-2x$ because $x+x+(60-2x) = 60$. The diagram helps show this too.
There are two ways to find the maximum area (with and without calculus):


*

*Remember that any quadratic $f(x) = ax^2+bx+c$ may be written in vertex-form: $f(x) = a(x-h)^2+k$, where $h = -\dfrac{b}{2a}$ and $k = c-\dfrac{b^2}{4a}$ (this is really just completing the square). When $a < 0$, the vertex corresponds to a maximum. Since $A(x) = -2x^2+60x$ is a quadratic with a negative leading coefficient, you can use this in order to find where $A$ is maximized and what its value is.

*You can also use calculus (which is probably what the question "wants"). The extrema of a function occur at critical points or at endpoints (which isn't the case here). Since the derivative of a quadratic is defined at all points, all you need is to find $A'(x)$ and use $A'(x) = 0$ to find where the maximum occurs. Then, just plug that value of $x$ in the original function $A(x)$.
For the domain and range, you want to find where the values make sense. Obviously, none of the sides can be negative, so $x \geq 0$ and $60-2x \geq 0 \iff x \leq 30$. This gives $x \in [0, 30]$. However, if $x = 0$ or $x = 30$, you'll have a degenerate rectangle, and a side length of $0$ doesn't make sense either, so you get $x \in (0, 30)$.
A: The answers are
1) The length is $60 - 2x$ because the other $2$ sides are each $x$ and the total length must be $60$, i.e., $60 - 2x = 60 - x - x$.
2) At the boundary points for $x$, i.e., $x = 0$ and $x = 30$ (as the latter gives $60 - 2x = 0$), the area is $0$. Thus, you can find the maximum area by taking the derivative of the area function and setting it to $0$, or alternatively completing the square. I'll leave that for you to do.
3) By "domain and range", I assume you mean domain of $x$ and range is of the area. Well, with the domain, you need to have the sides be non-zero lengths (I'm assuming you allow a degenerate case of basically a "line" of a zero-area enclosure). This gives $x \ge 0$ and $60 - 2x \ge 0 \implies x \le 30$. Thus, the domain is $[0,30]$. The range will be from $0$ to the maximum value you found in $2$.
