# Question #2041c

##### 1 Answer
Feb 6, 2017

The maximum area is $\frac{128}{4 + \pi} \left(\approx 17.92\right)$. This occurs when:

radius of the semicircle is $\frac{16}{4 + \pi} \left(\approx 2.24\right)$
rectangular window is $\frac{16}{4 + \pi} \times \frac{32}{4 + \pi} \left(\approx 2.24 \times 4.48\right)$

#### Explanation:

Let us set up the following variables:

$\left\{\begin{matrix}r & \text{Radius of the semicircle" & "(feet)" \\ h & "Height of the rectangular window" & "(feet)" \\ A & "Total area enclosed by the window" & "(sq feet)}\end{matrix}\right.$

Our aim is to find $A \left(h , r\right)$, as a function of a single variable and to maximize the total area, $A$, wrt that variable (It won't matter which variable we do this with as we will get the same result). ie we want a critical point of $A$ wrt the variable.

The total perimeter is that of $3$ sides of the rectangle and the semicircle; we are told that this perimeter is $16$ feet

$16 = \left(h + 2 r + h\right) + \left(\frac{1}{2}\right) \left(2 \pi r\right)$
$\setminus \setminus \setminus = 2 r + 2 h + \pi r$

$\therefore 2 h = 16 - 2 r - \pi r$
$\therefore \setminus \setminus h = \frac{1}{2} \left(16 - 2 r - \pi r\right)$

And the total Area is that of a rectangle and a semicircle:

$A = \left(h\right) \left(2 r\right) + \left(\frac{1}{2}\right) \left(\pi {r}^{2}\right)$
$\setminus \setminus \setminus = 2 h r + \frac{1}{2} \pi {r}^{2}$
$\setminus \setminus \setminus = 2 \left(\frac{1}{2} \left(16 - 2 r - \pi r\right)\right) r + \frac{1}{2} \pi {r}^{2}$
$\setminus \setminus \setminus = 16 r - 2 {r}^{2} - \pi {r}^{2} + \frac{1}{2} \pi {r}^{2}$
$\setminus \setminus \setminus = 16 r - 2 {r}^{2} - \frac{1}{2} \pi {r}^{2}$

We now have the Area, $A$, as a function of a single variable $r$, so differentiating wrt $x$ we get:

$\frac{\mathrm{dA}}{\mathrm{dr}} = 16 - 4 r - \pi r$

At a critical point we have $\frac{\mathrm{dA}}{\mathrm{dr}} = 0 \implies$

$16 - 4 r - \pi r = 0$
$\therefore \setminus \setminus \setminus \setminus 4 r + \pi r = 16$
$\therefore \setminus \setminus \setminus r \left(4 + \pi\right) = 16$
$\therefore \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus r = \frac{16}{4 + \pi} \left(\approx 2.24\right)$

With this value of $r$ we have:

$A = 16 \left(\frac{16}{4 + \pi}\right) - 2 {\left(\frac{16}{4 + \pi}\right)}^{2} - \frac{1}{2} \pi {\left(\frac{16}{4 + \pi}\right)}^{2}$
$\setminus \setminus \setminus = \frac{128}{4 + \pi} \left(\approx 17.92\right)$

And:

$h = \frac{1}{2} \left(16 - 2 \left(\frac{16}{4 + \pi}\right) - \pi \left(\frac{16}{4 + \pi}\right)\right)$
$\setminus \setminus = \frac{16}{4 + \pi} \left(\approx 2.24\right)$

We can visually verify that this corresponds to a maximum by looking at the graph of $y = A \left(r\right)$:

graph{16x - 2x^2 - 1/2 pi x^2 [-5, 10, -5, 22]}

And also check that the perimeter is correct:

$P = 2 r + 2 h + \pi r$
$\setminus \setminus = 2 \left(\frac{16}{4 + \pi}\right) + 2 \left(\frac{16}{4 + \pi}\right) + \pi \left(\frac{16}{4 + \pi}\right)$
$\setminus \setminus = 16$