Question 1171703
<pre>
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:

{{{drawing(2800/9,500,-7,7,-7,11,locate(-.2,-5,2x),

green(line(-5,5,5,5)),line(5,5,5,-5),line(5,-5,-5,-5),line(-5,-5,-5,5),
locate(-2.4,5,x),circle(0,5,.2),locate(2.5,5,x),locate(5.2,.5,h), locate(-5.6,.5,h),
arc(0,5,10,-8,0,180) )}}}

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

{{{Area = pi*x^2}}}

so the area of the semicircle on top is {{{expr(1/2)pi*x^2}}}

The area of the rectangle is {{{base*height=(2x)(h)=2xh}}}

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

{{{y=expr(1/2)pi*x^2+2xh}}}

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

{{{Perimeter = "'circumference'"= 2pi*x}}}

so the perimeter ("circumference") of the semicircle on top is {{{expr(1/2)2pi*x}}} or {{{pi*x}}}

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

{{{perimeter = pi*x+h+h+2x}}}

And since the perimeter is 32 feet,

{{{perimeter = pi*x+2h+2x=32}}}

               {{{2h=32-pi*x-2x}}}

Substituting for 2h in the equation for the area:

{{{y=expr(1/2)pi*x^2+2xh}}}

{{{y=expr(1/2)pi*x^2+2h*x}}}

{{{y=expr(1/2)pi*x^2+(32-pi*x-2x)*x}}}

{{{y=expr(1/2)pi*x^2+32x-pi*x^2-2x^2}}}

{{{y=expr(1/2)pi*x^2-pi*x^2-2x^2+32x}}}

{{{y=(expr(1/2)pi-pi-2)x^2+32x}}}

{{{y=(-expr(1/2)pi-2)x^2+32x}}}

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 {{{-b/(2a)}}}, where 'a' is the
coefficient of x<sup>2</sup>, and 'b' is the coefficient of x.

{{{-b^""/(2a^"")=-32^""^""/(2(-expr(1/2)pi-2)^"")=-32^""/(-pi-4^"")=32^""/(pi+4^"")=4.480793228}}}

So the x coordinate when the area is a maximum is 

{{{x=32/(pi+4)}}}, which is the radius of the semicircle.

Finally we find the height of the rectangle, by substituting 

{{{32/(pi+4)}}} for x in:

{{{2h=32-pi*x-2x}}}
{{{2h=32-x(pi+2)}}}
{{{2h=32-(32^""/(pi+4^""))(pi+2)}}}
Divide every term by 2
{{{h=16-(16^""/(pi+4^""))(pi+2)}}}
{{{h=16-16(pi+2)/(pi+4)}}}
{{{h=16-(16pi+32)/(pi+4)}}}
{{{h=16(pi+4)/(pi+4)-(16pi+32)/(pi+4)}}}
{{{h=(16pi+64)/(pi+4)-(16pi+32)/(pi+4)}}}
{{{h=((16pi+64)-(16pi+32))/(pi+4)}}}
{{{h=(16pi+64-16pi-32)/(pi+4)}}}
{{{h=32/(pi+4)}}}

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

{{{drawing(400,400,-3,3,-3,3,green(line(-2,0,2,0)),
locate(0,-2,2x),locate(-1,0,x),locate(1,0,x),locate(2.1,-1,h),locate(-2.15,-1,h),circle(0,0,.04),
line(-2,0,-2,-2),line(-2,-2,2,-2),line(2,-2,2,0), arc(0,0,4,-4,0,180))}}}


 


Edwin</pre>