SOLUTION: A Norman window consists of rectangle surmounted by a semicircle. If the perimeter of a Norman window is 32 ft, what should be the radius of the semicircle and the height of the

Let x = width and diameter;

    y = height of the rectangle part.


Then the perimeter   

P =   ====>   +  = 32  ====>  y = .


The area A =  +  =  +  =  -  -  +  =  +  - 



Then  the condition for the maximum area   = 0  takes the form


 = 0,   or    = 16  ====> x =  =  = 8.96 ft.


The maximum area is when the radius of the semicircle is  8.96/2 = 4.48 ft and the height of the rectangular part is 


    y =  =  = 4.49 ft.

Let x = the radius of the semicircle, which is half the base of the rectangle.
Let h = the height of the rectangle. 

This drawing is not to scale:



The area of the semicircle is half the area of a whole circle.
The area of a whole circle circle is



so the area of the semicircle on top is 

The area of the rectangle is 

So what we want to maximize is the total area y:



The perimeter of the entire window is given as 32 feet.

The perimeter ("circumference") of the semicircle is half the perimeter ("circumference") of a whole circle.
The perimeter "circumference" of a whole circle is



so the perimeter ("circumference") of the semicircle on top is  or 

The total perimeter is the perimeter of the semicircle plus the left, right,
and bottom sides of the rectangle.



And since the perimeter is 32 feet,



               

Substituting for 2h in the equation for the area:















From this point you can get the maximum value by just algebra,
using the vertex formula, or by using calculus.  Ikleyn used
calculus above so I'll just use algebra.

The x-coordinate of the vertex of a parabola is , where 'a' is the
coefficient of x2, and 'b' is the coefficient of x.



So the x coordinate when the area is a maximum is 

, which is the radius of the semicircle.

Finally we find the height of the rectangle, by substituting 

 for x in:




Divide every term by 2









The radius of the semicircle equals the height. So the rectangle is twice as
wide as it is high.  So it looks like this:




 


Edwin