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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 Norm,an window with a perimeter of 45 feet

Answer 1

Perimeter of window = perimeter of semicircle + 2l + w

=>2pir/2 + 2l + 2r (since w = 2r)

=>pi*r + 2l + 2r

r(pi + 2) + 2l = 45

2l + 5.141r = 45

l = (45 - 5.141r)/2

area of window = pir^2/2 + l*w

A = pi r^2/2 + 2r(45 -5.141r)/2

A = pir^2/2 + 45r - 5.141r^2

Area will be maximum when dA/dr = 0

dA/dr = (1/2)2pir + 45 - 2(5.141)r

dA/dr = pi r + 45 - 10.282r

dA/dr = -7.14r + 45

-7.14r + 45 = 0

7.14r = 45

r = 45/7.14 = 6.3 ft

so maximum area = pi(6.3)^2/2 + 45(6.3) - 5.141(6.3)^2

Amax = 141.8 sq.ft

Answer 2

So, we have a semicircle on top of a rectangle, where d = w

Build a perimeter equation :
2h + w + pi*w / 2 = 45
(The top of the rectangle is not used as the semi-circle is on top)

Now build an area equation:
A = hw + (pi* (w/2) ^2)

Go back to the perimeter equation and solve for h
2h = (45 -w - piw/2)
h = 22.5 - 2w/4 - piw/4)
h = 22.5 - (2 + pi) w/4

Sub into the Area equation

A= (22.5 - (2 + pi) w/4)w + (pi* (w/2) ^2)
A = 22.5 w - (2-pi) w^2 /4 + pi w^2/4
A = 22.5 w - 2w^2/4 - pi w^2/4 + pi w^2/4
A = 22.5 w - 2w^2/4
A = w( 22.5 - w/2)
The roots (where A = 0) are:
w = 0, w = 45
The max appears halfway between, or at w = 22.5
Back substitute to solve for h.

Answer 3

Area=w(2w)+piw^2/2=w^2 (2+pi/2)
Perimeter=2piw+4w=45
max area=?

w=45/(2pi+4)

Area={45/(2pi+4)}^2 (2+pi/2)

if pi=3 then
Area=3.5 * 20.25=70.875

Answer 4

rad = 6.30ffeet
length = 12.6feet
breath = 12.6feet

area = 408.14feet