Precalc question

Question

A Norman window is a rectangle with a semicircle on top. Suppose that the perimeter of a particular Norman window is to be 24 feet. What should its dimensions be in order to maximize the area of the window and, therefore, allow in as much light as possible?

Answer 1

draw a diagram

x=width

y=left length and right length

x/2=radius of the semicircle

first find the perimeter

P=x+y+y+(1/2)C

P=x+2y+(1/2)(2∏r) but r=x/2

P=x+2y+(1/2)(2∏[x/2])

P=x+2y+(∏/2)x

P=2y+x(1+[∏/2])

you have 24 feet

24=2y+x(1+∏/2)

solve for y

2y=24-x(1+∏/2)

y=12-(x/2)(1+∏/2)

y=12-x/2--∏x/4

y=12-x([1/2]+[∏/4])

y=12-x([2+∏]/4)

now find the area (of the rectangle and half of the circle)

A=xy+(1/2)(∏r^2)

remember we just found the value of y and r=x/2

A=x(12-x[(2+∏)/4])+(1/2)∏(x/2)^2

A=x(12-x[(2+∏)/4])+(1/2)∏(x^2/4)

A=12x-x^2[(2+∏)/4]+(∏/8)x^2

A=12x-x^2([[2+∏]/4]-[∏/8])

A=12x-x^2([[4+2∏]/8]-∏/8)

A=12x-x^2([4+∏]/8)

use -b/2a for x, the "maximizing" point

-12/(-[4+∏]/4)=

48/(4+∏)=x

you now have a value for x and a value for y

lastly find the value of half of the circumference, C

the radius r=x/2

x/2=(48/[4+∏])/2

r=24/(4+∏)

(1/2)C=(1/2)(2)(∏)(24/[4+∏])

(1/2)C=24∏/(4+∏)

now you have all the dimensions: x, y, and (1/2)C

if you add these four values together, x+y+y+(1/2)C, you will see that you get 24 !

Answer 2

L+2h+(d*pi/360)α=24'

α=angle/radius

Not enough imfo.