# Optimizing Problem of a Window

• November 27th 2009, 04:08 PM
RangerKimmy
Optimizing Problem of a Window
"A Norman window has the shape of a rectangle surmounted by a semicircle. (Thus, the diameter of the semicircle is equal to the width of the rectangle.) If the perimeter of the window is 22 ft, find the dimensions of the window so that the greatest possible amount of light is admitted."

(I've seen some forum posts about optimizing Norman Windows, but I've still gotten stuck. I was hoping to get some help with my problem.)

EDIT: Okay, I'm still having trouble, even though I've reworked it to fit the problem properly and solve for optimized dimensions.

Perimeter: $\pi r + 2h + 2r \:=\:22\quad\Rightarrow\quad h \:=\:\frac{22 - 2r - \pi r}{2}$

Substituting: $P \;=\;\pi r + 2\left(\frac{22 - 2r - \pi r}{2}\right) + 2r$

And now I have trouble. I know I needed to plug in my solved h to eliminate a variable, then simplify, and then take the derivative.
But simplifying it brings it back to P = 22, which... isn't what I need.

Simplifying: $P \;=\;\pi r + 22 - 2r - \pi r + 2r$

$P \;=\; 22$

Clearly this isn't the right method! I don't know what I'm doing wrong. Usually I'm supposed to work with Area too, but that shouldn't be involved this time, should it?
• November 27th 2009, 04:41 PM
qmech
You're right
The calculation looks good. Don't be scared of
$
\pi + 4
$

It's just the exact value of a number equal to about 7.14159...
Your answer will be as precise as the approximation you use. For a rough estimate, just use
$
\pi + 4 \approx 7
$
• November 27th 2009, 04:47 PM
RangerKimmy
Quote:

Originally Posted by qmech
The calculation looks good. Don't be scared of
$
\pi + 4
$

It's just the exact value of a number equal to about 7.14159...
Your answer will be as precise as the approximation you use. For a rough estimate, just use
$
\pi + 4 \approx 7
$

Thanks, but it turns out I read the question incorrectly.. I was looking for dimensions, not area. And now I have entirely NO idea how to do this problem because I am used to working with area, not dimensions.
• November 27th 2009, 04:49 PM
mr fantastic
Quote:

Originally Posted by RangerKimmy
Thanks, but it turns out I read the question incorrectly.. I was looking for dimensions, not area. And now I have entirely NO idea how to do this problem because I am used to working with area, not dimensions.

Go back to your calculations. Have you found the height and width of the rectangle? These give you the dimensions.
• November 27th 2009, 07:02 PM
RangerKimmy
Quote:

Originally Posted by mr fantastic
Go back to your calculations. Have you found the height and width of the rectangle? These give you the dimensions.

I'm still doing this incorrectly..