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. See Exercise 52 on page 24.) If the perimeter of...

Section 1
PAGE
1
Solution: The basic shape of the window is given by the following diagram:
99
To maximize to greatest amount of light admitted, we must maximize the area of the window. Let x be the width of the rectangle and y be the width. Note that the radius r of the semicircle is half the width of the rectangle, that is,
2
x
r
=
. To find the objective area equation, we must find the area of the rectangle and the area of the semicircle. The area of the rectangle is given as follows:
xy
=
=
dth)
length)(wi
(
Rectangle

of

Area
Recall that the area of a circle is given by the formula
2

r
A
c
p
=
, where r is the circle’s radius. Since we have a semicircle, the area is half of this formula. Thus, keeping in mind that
2
x
r
=
, we have
2
2
2
2

8
1
4
2
1
2

2
1

2
1
Circle
-
Semi

of

Area
x
x
x
r
p
p
p
p
=
=
÷
ø
ö
ç
è
æ
=
=
Thus, the to get the total area, we add the area rectangle to the area of the semi-circle:
2

8
1
Circle)
-
Semi

of

(Area
Rectangle)

of

Area
(
Area

Total
x
xy
A
p
+
=
+
=
=
(Continued on next page)
Hence, the objective equation for this problem is
2

8
1
x
xy
A
p
+
=
To get the perimeter of the rectangle, we use the fact that the perimeter around the window is 30 ft. To get the perimeter, we must sum the length of sides around the rectangle and the semicircle. The...