# optimization

• Mar 18th 2010, 06:17 PM
Evan.Kimia
optimization
I cant seem to get the right answer.

A Norman window has the shape of a rectangle surmounted by a semicircle. If the perimeter of the window is 20 ft, find the value of x so that the greatest possible amount of light is admitted.

for my critical number i get 80/(4+pi) for my critical number of x which isnt working. (Headbang)
• Mar 18th 2010, 07:57 PM
ione
Quote:

Originally Posted by Evan.Kimia
I cant seem to get the right answer.

A Norman window has the shape of a rectangle surmounted by a semicircle. If the perimeter of the window is 20 ft, find the value of x so that the greatest possible amount of light is admitted.

for my critical number i get 80/(4+pi) for my critical number of x which isnt working. (Headbang)

What does x stand for?
• Mar 18th 2010, 08:01 PM
Soroban
Hello, Evan.Kimia!

You probably made a very simpler error.

Quote:

A Norman window has the shape of a rectangle surmounted by a semicircle.
If the perimeter of the window is 20ft, find the value of $x$
so that the greatest possible amount of light is admitted.

For my critical number, i get 80/(pi + 4) . It shoud be: . ${\color{blue}\frac{20}{\pi + 4}}$

Code:

              * * *           *          *         *              *       *                *       * - - - - * - - - - *       |    r        r    |       |                  |     y|                  |y       |                  |       |                  |       * - - - - - - - - - *               2r

The radius of the semicircle is $r.$
Hence, the width of the rectangle is $2r.$
The height of the rectangle is $y.$

The perimeter of the semicircle is: . $\pi r$
The perimeter of the rectangle is: . $2r + 2y$

The total perimeter is 20:
. . $\pi r + 2r + 2y \:=\:20 \quad\Rightarrow\quad y \:=\:10 - \left(\frac{\pi+2}{2}\right)r$ .[1]

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

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

Substitute [1] into [2]: . $A \;=\;\tfrac{1}{2}\pi r^2 + 2r\bigg[10 - \frac{\pi+2}{2}\,r\bigg]$

. . And we have: . $A \;=\;\tfrac{1}{2}\pi r^2 + 20r - (\pi+2)r^2$

Differentiate and equate to zero: . $A' \;=\;\pi r + 20 - 2(\pi+2)r \;=\;0$

. . . $(\pi+4)r \:=\:20 \quad\Rightarrow\quad\boxed{ r \:=\:\frac{20}{\pi+4}}$

Substitute into [1]: . $y \:=\:10 - \left(\frac{\pi+2}{2}\right)\left(\frac{20}{\pi+4} \right) \quad\Rightarrow\quad\boxed{y \;=\;\frac{20}{\pi+4}}$