Let's find the perimeter of the window.
The bottom side is

. The left and right sides make

.
The perimeter of a circle is

, so the perimeter of a semicircle must be

, The radius is

, so that gives

for the curve. All of that is equal to 12.

We only want to use one variable to create the area formula, so let's solve for

.


Now that we have a value for

in terms of

, we can find the area in terms of

.
The area of the rectangle is going to be

, which then becomes


The area of the semicircle is going to be

.
Since

,

.


Now let's add the areas of the rectangle and semicircle.



If you wanted to factor out

like you did, this would become

Now what we want to do is find what

is when

is at its highest point, Once we have the value for

we can also find the value for

, of course.
Let's put our equation in the general form of a quadratic.

Now we can use the vertex formula

.
(

and

refer to

.)




Now let's plug that in for

.
Since our final answers are in decimal form and not exact form, we can make our lives a little easier here and just use

.

<span>

</span><span>

Let's take our answers for

and

and round to 2 decimal places.


</span>