document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1116692>Question 1116692</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A Norman window has the shape of a rectangle surmounted by a semicircle. (Thus the diameter of the semicircle is equal to the width of the rectangle. See the figure below.) If the perimeter of the window is 32 ft, find the value of x so that the greatest possible amount of light is admitted. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #731595 by </B><A HREF=/tutors/aboutme.mpl?userid=josgarithmetic><B>josgarithmetic(39617)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=josgarithmetic><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=731595\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2pi%2A%28x%2F2%29%281%2F2%29 ALIGN=MIDDLE  ALT=\"2pi%2A%28x%2F2%29%281%2F2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >, to account for the round part of the perimeter.<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2pi%2Ax%281%2F4%29 ALIGN=MIDDLE  ALT=\"2pi%2Ax%281%2F4%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28pi%2F2%29x ALIGN=MIDDLE  ALT=\"%28pi%2F2%29x\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "If the other rectangular dimension is y, then accounting for perimeter 32,<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28pi%2F2%29x%2Bx%2B2y=32 ALIGN=MIDDLE  ALT=\"%28pi%2F2%29x%2Bx%2B2y=32\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Solving that for y:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2y=32-x-%28pi%2F2%29x ALIGN=MIDDLE  ALT=\"2y=32-x-%28pi%2F2%29x\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y=16-x%2F2-%28pi%2F4%29x ALIGN=MIDDLE  ALT=\"y=16-x%2F2-%28pi%2F4%29x\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Area <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=xy ALIGN=MIDDLE  ALT=\"A=xy\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=x%2816-x%2F2-%28pi%2F4%29x%29 ALIGN=MIDDLE  ALT=\"A=x%2816-x%2F2-%28pi%2F4%29x%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=16x-%281%2F2%29x%5E2-%28pi%2F4%29x%5E2 ALIGN=MIDDLE  ALT=\"A=16x-%281%2F2%29x%5E2-%28pi%2F4%29x%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=16x-%281%2F2%2Bpi%2F4%29x%5E2 ALIGN=MIDDLE  ALT=\"A=16x-%281%2F2%2Bpi%2F4%29x%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( ".<BR>\n" );
document.write( "to continue, finding and setting derivative to 0 would be the best way...\r<BR>\n" );
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document.write( ".\r<BR>\n" );
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document.write( ". </SPAN>\n" );
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