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Maximizing Window Area

Date: 02/24/97 at 20:37:16
From: pierre
Subject: Related rates

The business office wants to put in Norman windows, which are windows
in the shape of rectangles capped by a semicircle.  These windows will
have special trim that is purchased by length, so it is most cost 
efficient to get the most light for the least perimeter of window. 
What dimensions maximize the area of the windows?

Date: 02/25/97 at 19:32:55
From: Doctor Luis
Subject: Re: Related rates


I'll try to draw the window, just to visualize the problem: 

        *   *
     *         *      (circle w/ radius r)
   |             |
   |             |  height h        
   |             |

     length 2r

Obviously, the larger the dimensions of the window are, the larger
area you'll get. So the maximum area possible is obtained when both r 
and h (refer to diagram) are as large as possible. Of course, that
depends on how much money you have to spend, which fixes a value for 
the perimeter of the window. The problem, then, becomes:

"For a given perimeter p find the dimensions h and r which will give 
the greatest area for our window"
An expression for the area of the window is:

     a = area(window)
       = area(rectangle) + area(semi-circle)
       = 2rh + (pi*r^2)/2            [remember: r^2 means "r squared"]
An expression for the perimeter p of the window is:

    p = p(window)
      = p(rectangle) + p(semi-circle) - 2*(overlapping side)
      = (4r+2h) + (pi*r+2r) - 2*(2r)
      = 2h + (pi+2)r

Since a = a(r,h) is a function of two variables, we must look for a  
relationship between r and h in order to make the problem easier. 
Since p is a constant (it's given), r and h can't just have any 
values. They must satisfy the expression for p derived above.
Now, solving for h, we have:

   h = (p-(pi+2)r)/2  

Substituting this expression for h into a, we obtain:

   a = 2r(p-(pi+2)r)/2 + (pi*r^2)/2
   a =  r(p-(pi+2)r) + (pi*r^2)/2

Now, a is a function of a single variable (a = a(r)), so we can apply
the methods of maxima and minima learned in calculus to find out which 
value of r will maximize a(r):
i) set a'(r) = 0 and solve for r to find critical points
ii) figure out the sign of a''(r) at the critical points to find out
    if they are maxima or minima (or points of inflection).     

I'm sure you can work the rest of the problem now. :) (By the way, the 
critical point I found is r = p/(4 + pi).)

-Doctor Luis,  The Math Forum
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Associated Topics:
College Calculus
College Euclidean Geometry
High School Calculus
High School Euclidean/Plane Geometry

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