My answer doesn't make sense. It's too big.
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 8:40pm
I 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 9:06pm
No. Sorry :[
Math (: - Reiny, Thursday, June 4, 2009 at 9:49pm
let 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)
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 9:52pm
My answer was around the thousands. That's how I knew it was wrong. Thanks!