document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1005984>Question 1005984</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A window is being built and the bottom is a rectangle and the top is a semicircle. If there is 12 m of framing materials what must the dimension of the window be to let in the most light? \r<BR>\n" );
document.write( "\n" );
document.write( "Note: This is a calculus optimization problem. I need to solve it algebraically, not with calculus (ex: first derivative test...). Thanks. I appreciate your help.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #622173 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=622173\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Imagine that amount of light is according to window area.<BR>\n" );
document.write( "Bottom of rectangle is x, height of rectangle is y, radius of the semicircle is x/2.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Length of the framing materials is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=pi%28x%2F2%29%2Bx%2B2y=12 ALIGN=MIDDLE  ALT=\"pi%28x%2F2%29%2Bx%2B2y=12\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "and the area of the window is  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=xy%2B%281%2F2%29pi%2A%28x%2F2%29%5E2 ALIGN=MIDDLE  ALT=\"A=xy%2B%281%2F2%29pi%2A%28x%2F2%29%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" >.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "That should get you started.  Either solve the framing equation for x in terms of y and substitute into the A function; or solve the framing equation for y in terms of x and substitute into the A function; and then simplify the function.  Is it quadratic, or does it have a quadratic numerator?\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "--SUMMARY OF PART OF THE METHOD, NOT TO COMPLETION:<BR>\n" );
document.write( "Solve the framing equation for y,<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y=6-pi%2Ax%2F4-x%2F2 ALIGN=MIDDLE  ALT=\"y=6-pi%2Ax%2F4-x%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" >;<BR>\n" );
document.write( "Substitute into the area function A and simplify, very detailed work steps, to get <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=highlight%28A=%286-%28%28pi%2B8%29%2F8%29x%29x%29 ALIGN=MIDDLE  ALT=\"highlight%28A=%286-%28%28pi%2B8%29%2F8%29x%29x%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >.<BR>\n" );
document.write( "This is parabola with vertex at a maximum, and it occurs exactly in the middle of the two zeros of A, which is why A is shown in its factored form, so you can more easily identify the zeros. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );