# Math Help - Finding the function

1. ## Finding the function

I started but got stuck...

A Norman Window has the shape of a rectangle surmounted by a semicircle. It has a perimeter of 40 ft.
A) find the function that models the area of the window
I ended up with this function, but I'm not sure it's right.
x²(-3╥ -4)
----------- +20x
8

B) find the dimensions, to two decimal places, that admits the greatest amount of light.

2. ## Re: Finding the function

Hello, karlyjo!

Did you make a sketch?

A Norman window has the shape of a rectangle surmounted by a semicircle.
It has a perimeter of 40 ft.

A) Find the function that models the area of the window

Code:
              * * *
*           *
*               *
*                 *

*                   *
* - - - - * - - - - *
|    r    :    r    |
|                   |
h |                   | h
|                   |
|                   |
*-------------------*
: - - - - 2r  - - - :
The radius of the semicircle is $r.$
The height of the rectangle is $h.$
The width of the rectangle is $2r.$

The circumference of the circle is $2\pi r.$
The perimeter of the semicircle is $\pi r.$
The perimeter of the rectangle is $2r + 2h.$

The total perimeter is 40 ft.
. . $\pi r + 2r + 2h \:=\:40 \quad\Rightarrow\quad h \:=\:\frac{40 - \pi r - 2r}{2}$ .[1]

The area of the semicircle is: $\tfrac{1}{2}\pi r^2$
The area of the rectangle is: $(2r)(h) \:=\:2rh$

The total area is: . $A \;=\;\tfrac{1}{2}\pi r^2 + 2rh$ .[2]

Substitute [1] into [2]: . $A \;=\;\tfrac{1}{2}\pi r^2 + 2r\left(\frac{40 - \pi r - 2r}{2}\right)$

And we have: . $A \;=\;40r - \tfrac{1}{2}(\pi + 4)r^2$

B) Find the dimensions, to two decimal places, that admits the greatest amount of light.

We want, of course, maximum area of the window.

We have: . $A \;=\;40r - \tfrac{1}{2}(\pi + 4)r^2$

Then: . $A' \:=\:40 - (\pi + 4)r \;=\;0 \quad\Rightarro\quad r \:=\:\frac{40}{\pi+4} \:=\:5.600991535$

Substitute into [1]: . $h \;=\;\frac{40 - \pi\left(\frac{40}{\pi+4}\right) - 2\left(\frac{40}{\pi+4}\right)}{2} \;=\;\frac{40}{\pi + 4}$

Therefore: . $\begin{Bmatrix} r &\approx & 5.60\text{ ft} \\ \\[-3mm] h &\approx & 5.60\text{ ft} \end{Bmatrix}$