document.write( "Question 194404: A norman window is one that consists of a rectangle with a curved top to it. the curved top is actually a semicircle that sits on top of the rectangle. if the perimeter around the norman window is 20 feet, what radius of the semicircle will provide the greatest window area? \n" ); document.write( "
Algebra.Com's Answer #145874 by ankor@dixie-net.com(22740)\"\" \"About 
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A norman window is one that consists of a rectangle with a curved top to it.
\n" ); document.write( "the curved top is actually a semicircle that sits on top of the rectangle.
\n" ); document.write( " if the perimeter around the norman window is 20 feet, what radius of the
\n" ); document.write( " semicircle will provide the greatest window area?
\n" ); document.write( ":
\n" ); document.write( "Let H = the height of the rectangular portion of the window
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\n" ); document.write( "Let x = the radius of the semi circle on top
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\n" ); document.write( "Half the circumference = pi*x
\n" ); document.write( ":
\n" ); document.write( "Width of the window = 2x
\n" ); document.write( ":
\n" ); document.write( "The perimeter:
\n" ); document.write( "2H + 2x + pix = 20
\n" ); document.write( "Solve for H
\n" ); document.write( "2H = 20 - 2x - pix
\n" ); document.write( "H = \"%2820-2x-pix%29%2F2\"
\n" ); document.write( ":
\n" ); document.write( "Area formula: A = (H*2x) + .5pi*x^2
\n" ); document.write( "Replace H with \"%2820-2x-pix%29%2F2\"
\n" ); document.write( "A = 2x(\"%2820-2x-pi%2Ax%29%2F2\")+ \".5pi%2Ax%5E2\"
\n" ); document.write( "Cancel the 2
\n" ); document.write( "A = \"x%2820-2x-pi%2Ax%29\"+ \".5pi%2Ax%5E2\"
\n" ); document.write( "A = \"20x+-+2x%5E2+-+%28pix%5E2%29+%2B+%28.5pix%5E2%29\"
\n" ); document.write( "Leaving us with
\n" ); document.write( "A = \"-2x%5E2+%2B+20x+-+.5pix%5E2\"
\n" ); document.write( ".5*pi = 1.57
\n" ); document.write( "A = \"-2x%5E2+-+1.57x%5E2+%2B+20x\"
\n" ); document.write( "A = \"-3.57x%5E2+%2B+20x\"
\n" ); document.write( "A quadratic equation, find the vertex for max area; a=-3.57, b=20
\n" ); document.write( "x = \"%28-20%29%2F%282%2A-3.57%29\"
\n" ); document.write( "x = \"%28-20%29%2F%28-7.14%29\"
\n" ); document.write( "x = 2.8 ft radius for max area
\n" ); document.write( ":
\n" ); document.write( "Check solution by finding the perimeter with this radius
\n" ); document.write( "Find 2H
\n" ); document.write( "2H = 20 - 2(2.8) - 2.8pi
\n" ); document.write( "2H = 5.6 ft
\n" ); document.write( "Find perimeter:
\n" ); document.write( "2H + 2x + pix = 20
\n" ); document.write( "5.6 + 2(2.8) + 2.8*pi =
\n" ); document.write( "5.6 + 5.6 + 8.8 = 20
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\n" ); document.write( "note that max area is when the rectangular portion, is a square, 5.6 by 5.6\r
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