document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=70941>Question 70941</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Suppose there is a semicircle below the rectangle as well as above it. Find the dimensions of the window which admit most sunlight for a given perimeter 8m </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #50698 by </B><A HREF=/tutors/aboutme.mpl?userid=ankor@dixie-net.com><B>ankor@dixie-net.com(22740)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ankor@dixie-net.com><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=ankor@dixie-net.com><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=50698\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Suppose there is a semicircle below the rectangle as well as above it. Find the dimensions of the window which admit most sunlight for a given perimeter 8m.<BR>\n" );
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document.write( "The most sunlight would be admitted at the max area and I seem to remember that<BR>\n" );
document.write( "a circle is the shape to have max area for a given perimeter.<BR>\n" );
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document.write( "Let x = the radius of the two semi-circles<BR>\n" );
document.write( "Then 2x = the length of the rectangle (horizontal measurement)<BR>\n" );
document.write( "Width (W) will be the vertical measurement  <BR>\n" );
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document.write( "Perimeter:<BR>\n" );
document.write( " Twice the rectangle width + Circumference = 8<BR>\n" );
document.write( "2W + 2pi*x = 8<BR>\n" );
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document.write( "2W = 8 - 2pi*x<BR>\n" );
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document.write( "W = (4 - pi*x); divided equation by 2<BR>\n" );
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document.write( "Area = Rectangle area + area of the circle<BR>\n" );
document.write( "A = W(2x) + pi*x^2<BR>\n" );
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document.write( "A = 2x(4-pi*x) + pi*x^2; Substituted (4-pi*x) for W<BR>\n" );
document.write( "A = 8x - 2pi*x^2 + pi*x^2<BR>\n" );
document.write( "A = 8x - pi*x^2<BR>\n" );
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document.write( "Write it as a quadratic equation; y = area<BR>\n" );
document.write( "y = -(pi*x^2) + 8x<BR>\n" );
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document.write( "Find the axis of symmetry to find the max area: a = -pi, b = 8<BR>\n" );
document.write( "x = -8/2(-pi)<BR>\n" );
document.write( "x = 1.27 meters is the radius of the semi circles at max area<BR>\n" );
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document.write( "Width of the rectangle = 4 - pi*x<BR>\n" );
document.write( "W = 4 - pi*1.27<BR>\n" );
document.write( "W = 4 - 4<BR>\n" );
document.write( "W = 0: <BR>\n" );
document.write( "Just as we suspected, the max area will be when there is no rectangle,<BR>\n" );
document.write( "and the window is a circle.<BR>\n" );
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document.write( "Area if it is a circle: pi*1.27^2 = 5.067 sq meters<BR>\n" );
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document.write( "Let's say the width is 1 meter. Then the total circumference of semi-circles<BR>\n" );
document.write( " will be 6 meters\"<BR>\n" );
document.write( "Find the radius if this is the case:<BR>\n" );
document.write( "2*pi*r = 6<BR>\n" );
document.write( "r = 6/(2pi)<BR>\n" );
document.write( "r = .955 meters<BR>\n" );
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document.write( "The length of the rectangle will be twice that: 1.91 meters<BR>\n" );
document.write( "Area of the rectangle 1*1.91 = 1.91 sq meters<BR>\n" );
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document.write( "The area of the circle = pi*.955^2 = 2.865 sq meters <BR>\n" );
document.write( "Total area 1.91 + 2.865 = 4.77 sq meters, less than a circle will be.<BR>\n" );
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document.write( "Note to student, let me know if I am wrong about this or any comment you have<BR>\n" );
document.write( "about this, I would glad to hear. <BR>\n" );
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