Let's find the perimeter of the window.
The bottom side is [tex]x[/tex]. The left and right sides make [tex]2y[/tex].
The perimeter of a circle is [tex]2\pi r[/tex], so the perimeter of a semicircle must be [tex]\pi r[/tex], The radius is [tex]\frac{1}2x[/tex], so that gives [tex]\frac{1}2\pi x[/tex] for the curve. All of that is equal to 12.
[tex]x+2y+\frac{1}2\pi x=12[/tex]
We only want to use one variable to create the area formula, so let's solve for [tex]y[/tex].
[tex]2y=12-x-\frac{1}2\pi x[/tex]
[tex]y=6-\frac{1}2x-\frac{1}4\pi x[/tex]
Now that we have a value for [tex]y[/tex] in terms of [tex]x[/tex], we can find the area in terms of [tex]x[/tex].
The area of the rectangle is going to be [tex]xy[/tex], which then becomes
[tex]A_r=x(6-\frac{1}2x-\frac{1}4\pi x[/tex]
[tex]A_r=6x-\frac{1}2x^2-\frac{1}4\pi x^2[/tex]
The area of the semicircle is going to be [tex]\frac{1}2\pi r^2[/tex].
Since [tex]r=\frac{1}2x[/tex], [tex]A_{sc}=\frac{1}2\pi (\frac{1}2x)^2[/tex].
[tex]A_{sc}=\frac{1}2\pi \frac{1}4x^2[/tex]
[tex]A_{sc}=\frac{1}8\pi x^2[/tex]
Now let's add the areas of the rectangle and semicircle.
[tex]A=A_r+A_{sc}[/tex]
[tex]A=6x-\frac{1}2x^2-\frac{1}4\pi x^2+\frac{1}8\pi x^2[/tex]
[tex]A=6x-\frac{1}2x^2-\frac{1}8\pi x^2[/tex]
If you wanted to factor out [tex]\frac{1}8[/tex] like you did, this would become
[tex]\boxed{A(x)=\frac{1}8(48x=4x^2-\pi x^2)}[/tex]
Now what we want to do is find what [tex]x[/tex] is when [tex]A[/tex] is at its highest point, Once we have the value for [tex]x[/tex] we can also find the value for [tex]y[/tex], of course.
Let's put our equation in the general form of a quadratic.
[tex]A(x)=(-\frac{1}2-\frac{1}8\pi )x^2+6x[/tex]
Now we can use the vertex formula [tex]x=\frac{-b}{2a}[/tex].
([tex]a[/tex] and [tex]b[/tex] refer to [tex]ax^2+bx+c[/tex].)
[tex]x=\frac{-6}{2(-\frac{1}2-\frac{1}8\pi)}[/tex]
[tex]x=\frac{-6}{-\frac{1}4\pi -1}[/tex]
[tex]x=\frac{-24}{-\pi -4}[/tex]
[tex]\boxed{x=\frac{24}{\pi +4}}[/tex]
Now let's plug that in for [tex]y=6-\frac{1}2x-\frac{1}4\pi x[/tex].
Since our final answers are in decimal form and not exact form, we can make our lives a little easier here and just use [tex]x\approx3.36059492[/tex].
[tex]y\approx6-\frac{1}2(3.36059492)-\frac{1}4\pi(3.36059492)[/tex]
[tex]y\approx6-(1.68029746+2.63940507809)[/tex]
[tex]\boxed{y\approx1.68029746191}[/tex]
Let's take our answers for [tex]x[/tex] and [tex]y[/tex] and round to 2 decimal places.
[tex]\boxed{x\approx 3.36\ ft}[/tex]
[tex]\boxed{y\approx 1.68\ ft}[/tex]
