Question 1193560
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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?
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<pre>
Let x = width and diameter;

    y = height of the rectangle part.


Then the perimeter   

P = {{{x + 2y + (pi*x)/2)}}}  ====>  {{{x + (pi*x)/2}}} + {{{2y}}} = 20  ====>  y = {{{10 - x/2 - (pi*x)/4}}}.


The area A = {{{xy}}} + {{{(1/2)*pi*(x/2)^2}}} = {{{x*(10-x/2 - (pi*x)/4)}}} + {{{(pi/2)*(x/2)^2}}} = 

           = {{{10x}}} - {{{x^2/2}}} - {{{(pi/4)*x^2}}} + {{{(pi/8)*x^2}}} = {{{-x^2/2}}} + {{{10x}}} - {{{(pi/8)*x^2}}}



Then  the condition for the maximum area  {{{(dA)/(dx)}}} = 0  takes the form


{{{-x + 10 - (pi/4)*x}}} = 0,   or   {{{x*(1+pi/4)}}} = 10  ====> x = {{{10/(1+pi/4)}}} = {{{10/(1 + (3.14/4))}}} = 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/2 - (pi*x)/4}}} = {{{10 - 5.60/2 - (3.14*5.60)/4}}} = 2.804 ft.
</pre>

Solved.