SOLUTION: A Norman window consists of rectangle surmounted by a semicircle. If the perimeter of a Norman window is 20 ft, what should be the radius of the semicircle and the height of the

Algebra ->  Test -> SOLUTION: A Norman window consists of rectangle surmounted by a semicircle. If the perimeter of a Norman window is 20 ft, what should be the radius of the semicircle and the height of the       Log On


   

Question 1193560: A Norman window consists of rectangle surmounted by a semicircle. If the
perimeter of a Norman window is 20 ft, what should be the radius of the
semicircle and the height of the rectangle such that the window will admit most
light?
Answer by ikleyn(52787) About Me  (Show Source):
You can put this solution on YOUR website!
.
A Norman window consists of rectangle surmounted by a semicircle.
If the perimeter of a Norman window is 20 ft, what should be the radius of the semicircle
and the height of the rectangle such that the window will admit most light?
~~~~~~~~~~~~~~ 


Let x = width and diameter;

    y = height of the rectangle part.


Then the perimeter   

P = x+%2B+2y+%2B+%28pi%2Ax%29%2F2%29  ====>  x+%2B+%28pi%2Ax%29%2F2 + 2y = 20  ====>  y = 10+-+x%2F2+-+%28pi%2Ax%29%2F4.


The area A = xy + %281%2F2%29%2Api%2A%28x%2F2%29%5E2 = x%2A%2810-x%2F2+-+%28pi%2Ax%29%2F4%29 + %28pi%2F2%29%2A%28x%2F2%29%5E2 = 

           = 10x - x%5E2%2F2 - %28pi%2F4%29%2Ax%5E2 + %28pi%2F8%29%2Ax%5E2 = -x%5E2%2F2 + 10x - %28pi%2F8%29%2Ax%5E2



Then  the condition for the maximum area  %28dA%29%2F%28dx%29 = 0  takes the form


-x+%2B+10+-+%28pi%2F4%29%2Ax = 0,   or   x%2A%281%2Bpi%2F4%29 = 10  ====> x = 10%2F%281%2Bpi%2F4%29 = 10%2F%281+%2B+%283.14%2F4%29%29 = 5.60 ft.


The maximum area is when the radius of the semicircle is  5.60/2 = 2.80 ft and the height of the rectangular part is 


    y = 10+-+x%2F2+-+%28pi%2Ax%29%2F4 = 10+-+5.60%2F2+-+%283.14%2A5.60%29%2F4 = 2.804 ft.

Solved.