# Finding the function

• Dec 10th 2012, 10:58 AM
karlyjo
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.
• Dec 10th 2012, 02:57 PM
Soroban
Re: Finding the function
Hello, karlyjo!

Did you make a sketch?

Quote:

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 $\displaystyle r.$
The height of the rectangle is $\displaystyle h.$
The width of the rectangle is $\displaystyle 2r.$

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

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

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

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

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

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

Quote:

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

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

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

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