Question 1171703
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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 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}}} = 32  ====>  y = {{{16 - x/2 - (pi*x)/4}}}.


The area A = {{{xy}}} + {{{(1/2)*pi*(x/2)^2}}} = {{{x*(16-x/2 - (pi*x)/4)}}} + {{{(pi/2)*(x/2)^2}}} = {{{16x}}} - {{{x^2/2}}} - {{{(pi/4)*x^2}}} + {{{(pi/8)*x^2}}} = {{{-x^2/2}}} + {{{16x}}} - {{{(pi/8)*x^2}}}



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


{{{-x + 16 - (pi/4)*x}}} = 0,   or   {{{x*(1+pi/4)}}} = 16  ====> x = {{{16/(1+pi/4)}}} = {{{16/(1 + (3.14/4))}}} = 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 = {{{16 - x/2 - (pi*x)/4}}} = {{{16 - 8.96/2 - (3.14*8.96)/4}}} = 4.49 ft.
</pre>

Solved.


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This problem was solved in the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/Finding-the-maximum-area-of-the-window-of-a-special-form.lesson>Finding the maximum area of the window of a special form</A> 

in this site (Problem 2).