1. ## 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.

2. 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.
What does x stand for?

3. Hello, Evan.Kimia!

Your answer is "close", so you're doing most of it Right.
You probably made a very simpler error.

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 $\displaystyle x$
so that the greatest possible amount of light is admitted.

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

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

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

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

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

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

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

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

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

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

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

Substitute into [1]: .$\displaystyle y \:=\:10 - \left(\frac{\pi+2}{2}\right)\left(\frac{20}{\pi+4} \right) \quad\Rightarrow\quad\boxed{y \;=\;\frac{20}{\pi+4}}$
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://mathhelpforum.com/calculus/134512-optimization.html", "fetch_time": 1526888000000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-22/segments/1526794863967.46/warc/CC-MAIN-20180521063331-20180521083331-00147.warc.gz", "warc_record_offset": 183012614, "warc_record_length": 10271, "token_count": 808, "char_count": 2550, "score": 4.125, "int_score": 4, "crawl": "CC-MAIN-2018-22", "snapshot_type": "latest", "language": "en", "language_score": 0.8004401922225952, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00019-of-00064.parquet"}