Optimizing a Norman Window - What am I doing wrong?

**pseizure2000** · October 5th 2007, 06:37 PM

A Norman window has the shape of a semicircle atop a rectangle so that the diameter of the semicircle is equal to the width of the rectangle. What is the area of the largest possible Norman window with a perimeter of 45 feet?

$A(h) = 1/2 \pi r^2 + 2r ((45-\pi(r)-2r)/2r)$

$A'(h) = \pi(r) - \pi-2$

So solving for 0 i get $(\pi +2)/\pi$ as the critical point

pluging back in:

$1/2 \pi[(\pi +2)/\pi]^2 + 45-\pi[(\pi+2)/\pi)]-2[(\pi +2)/\pi)]<br />$

the hell is wrong here? i keep getting an incorrect answer. something simple i suppose

**ticbol** · October 5th 2007, 08:30 PM

Originally Posted by pseizure2000

A Norman window has the shape of a semicircle atop a rectangle so that the diameter of the semicircle is equal to the width of the rectangle. What is the area of the largest possible Norman window with a perimeter of 45 feet?

A(h) = 1/2 pi r^2 + 2r ((45-pi(r)-2r)/2r)

Why divide by 2r?

Divide by 2 only.

Perimeter, P = 45 = (1/2)(2pi*r) +2r +2y
2y = 45 -pi(r) -2r
y = (45 -pi(r) -2r) / 2 only.

**Soroban** · October 6th 2007, 02:22 AM

[size=13Hello, pseizure2000![/size]

A Norman window has the shape of a semicircle atop a rectangle
so that the diameter of the semicircle is equal to the width of the rectangle.
What is the area of the largest possible Norman window with a perimeter of 45 feet?

Code:

              * * *
          *           *
        *               *
       *                 *

      * - - - - + - - - - *
      |    r         r    |
      |                   |
    h |                   | h
      |                   |
      |                   |
      * - - - - - - - - - *
               2r

The radius of the semicircle is $r.$
Then the width of the rectangle is $2r.$
Let $h$ = height of the rectangle.

The perimeter is: . $\pi r + 2h + 2r \:=\:45\quad\Rightarrow\quad h \:=\:\frac{45 - 2r - 2\pi}{2}$ . [1]

The area of the window is: . $A \;=\;\frac{1}{2}\pi r^2 + 2rh$ . [2]

Substitute [1] into [2]: . $A \;=\;\frac{1}{2}\pi r^2 + 2r\left(\frac{45 - 2r - \pi r}{2}\right)$

. . which simplifies to: . $A \;=\;45r - \left(\frac{4+3\pi}{2}\right)r^2$ . [3]

Differentiate and equate to zero: . $A' \;=\;45 - (4+3\pi)r \;=\;0$

. . Hence: . $r \;=\;\frac{45}{4+3\pi}$

Substitute into [3]: . $A \;=\;45\left(\frac{45}{4+3\pi}\right) - \left(\frac{4+3\pi}{2}\right)\left(\frac{45}{4+3\p i}\right)^2<br />$

. . and we get: . $A \;=\;\frac{2025}{2(4+3\pi)} \;\approx\;\boxed{7542\text{ ft}^2}$

**fitswell** · March 26th 2011, 07:41 PM

sorry if this seems a bit elementary but how did you simplify to step #3?
where does the $3\pi$ in numerator come from?

I am not sure where I am making a mistake but I keep getting:
$A \;=\;45r - \left(\frac{\pi-4}{2}\right)r^2$

thanks!

Originally Posted by Soroban

. . which simplifies to: . $A \;=\;45r - \left(\frac{4+3\pi}{2}\right)r^2$ . [3]

**mr fantastic** · March 27th 2011, 02:18 AM

Originally Posted by fitswell

sorry if this seems a bit elementary but how did you simplify to step #3?
where does the $3\pi$ in numerator come from?

I am not sure where I am making a mistake but I keep getting:
$A \;=\;45r - \left(\frac{\pi-4}{2}\right)r^2$

thanks!

Yes there is a mistake in the post. An easy mistake to make.

I get $A \;=\;45r - \left(\frac{\pi+4}{2}\right)r^2$

Note the + ....

**fitswell** · March 27th 2011, 01:47 PM

yes, very easy to get lost in the numbers, many thanks for the rework!

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