norman window

Question

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?

Answer 1

draw a diagram and call the sides "h" and the bottom "2r"

P=(1/2)C+2h+2r

P=∏r+2h+2r because C=2∏r and we want 1/2 of C

P=2h+r(∏+2)

47=2h+r(∏+2)

2h=47-r(∏+2)

h=(47-r[∏+2])/2

area=area of 1/2 of the circle plus area of the rectangle

A=(1/2)∏r²+2rh (2r=bottom of rectangle and h=side of rectangle)

A=(1/2)∏r²+2r(47-r[∏+2])/2

A=(1/2)∏r²+r(47-r[∏+2])

A=(1/2)∏r²+47r-r²(∏+2)

A=(1/2)∏r²+47r-r²∏-2r²

A=r²([∏/2-∏-2)+47r

A=r²(0.5∏-2)+47r

use (p,q) where p=radius and q=maximum area

p=-b/2a

p=-47/2(-0.5∏-2)

p=47/(∏+4)

2p=diameter because p=radius

2p=94/(∏+4)

2p=94/(7.14159)

2p=13.162 which is the bottom of the rectangle

now find half of C

(1/2)C=(1/2)∏(13.162)

(1/2)C=20.675

now find the length of the sides of the rectangle

you used up 20.675 feet for half the circumference and 13.162 feet for the bottom length

you used up 33.837 feet out of 47 feet

this leaves 13.163 feet for both equal sides

13.163/2=6.582 for each side

now find the area of the semicircle and the area of the rectangle

A=lw

A=(13.162)(6.582)

A=86.632 square feet for the area of the rectangle

A=(1/2)∏r² for the area of half of the circle

A=(1/2)∏(6.582²)

A=(1/2)∏(43.323)

A=68 square feet for the area of the semicircle

add the areas

A=68+86.632

A=154.632 square feet for the area of the window