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August 28, 2014

August 28, 2014

Posted by **<3** on Thursday, June 4, 2009 at 8:34pm.

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 47 feet?

- Math (: -
**Reiny**, Thursday, June 4, 2009 at 8:40pmI have seen this question many times.

Your answer might be right, try sketching your diagram with the half-circle on the long side of the rectangle, if I recall that is what the shape actually looks like.

Does your answer make sense now?

- Math (: -
**<3**, Thursday, June 4, 2009 at 9:06pmNo. Sorry :[

- Math (: -
**Reiny**, Thursday, June 4, 2009 at 9:49pmlet the radius of the circle be x

then the base of the rectangle on which the circle sits is 2x

let its height be y

perimeter = 2x + 2y + 1/2(2pix)

2x+2y+pix=47

y = (47-2x-pix)/2

Area = 2xy + (1/2)pi(x^2)

= 2x(47-2x-pix)/2 + (1/2)pi(x^2)

= x(47-2x-pix) + (1/2)pi(x^2)

= 47x - 2x^2 - pix^2 + (1/2)pi(x^2)

d(Area)/dx = 47 - 4x - 2pix + pix

= 0 for a max/min of Area

solving this I got x = 6.58

so the base is 13.16

and after substituting back for y, I got

y = 6.58 as well

subbing that back in my Area equation , I got Area = 154.657

How does that match up with your answers?

- Math (: -
**<3**, Thursday, June 4, 2009 at 9:52pmMy answer was around the thousands. That's how I knew it was wrong. Thanks!

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