document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=975755>Question 975755</A>: <I><SPAN STYLE=\"word-break: break-all;\"> You have two square plots formed with 100m of fencing. Determine the equation of the curve of best fit, and use the equation to determine the value of x that minimizes the area of each square.<BR>\n" );
document.write( "(Must use quadratics, no derivatives/calculus)<BR>\n" );
document.write( "Only thing I've seen to do is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A+=+X%5E2+%2B+Y%5E2 ALIGN=MIDDLE  ALT=\"A+=+X%5E2+%2B+Y%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" > then <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=100+=+4x+%2B+4y ALIGN=MIDDLE  ALT=\"100+=+4x+%2B+4y\"   DISABLED_style= \"max-width:90%; height:auto;\" >. <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=25=x%2By ALIGN=MIDDLE  ALT=\"25=x%2By\"   DISABLED_style= \"max-width:90%; height:auto;\" >, <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y=25-x ALIGN=MIDDLE  ALT=\"y=25-x\"   DISABLED_style= \"max-width:90%; height:auto;\" >.\r<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=X%5E2+%2B+%2825-x%29%5E2 ALIGN=MIDDLE  ALT=\"A=X%5E2+%2B+%2825-x%29%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=x%5E2+%2B+625+-+25x+-+25x+%2B+x%5E2 ALIGN=MIDDLE  ALT=\"A=x%5E2+%2B+625+-+25x+-+25x+%2B+x%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=x%5E2+-+50x+%2B+625 ALIGN=MIDDLE  ALT=\"A=x%5E2+-+50x+%2B+625\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "But then solutions for x are imaginary, what do I do now? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #597438 by </B><A HREF=/tutors/aboutme.mpl?userid=Alan3354><B>Alan3354(69443)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Alan3354><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=Alan3354><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=597438\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> You have two square plots formed with 100m of fencing. Determine the equation of the curve of best fit, and use the equation to determine the value of x that minimizes the area of each square.<BR>\n" );
document.write( "(Must use quadratics, no derivatives/calculus)<BR>\n" );
document.write( "Only thing I've seen to do is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A+=+X%5E2+%2B+Y%5E2 ALIGN=MIDDLE  ALT=\"A+=+X%5E2+%2B+Y%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" > then <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=100+=+4x+%2B+4y ALIGN=MIDDLE  ALT=\"100+=+4x+%2B+4y\"   DISABLED_style= \"max-width:90%; height:auto;\" >. <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=25=x%2By ALIGN=MIDDLE  ALT=\"25=x%2By\"   DISABLED_style= \"max-width:90%; height:auto;\" >, <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y=25-x ALIGN=MIDDLE  ALT=\"y=25-x\"   DISABLED_style= \"max-width:90%; height:auto;\" >.\r<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=X%5E2+%2B+%2825-x%29%5E2 ALIGN=MIDDLE  ALT=\"A=X%5E2+%2B+%2825-x%29%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=x%5E2+%2B+625+-+25x+-+25x+%2B+x%5E2 ALIGN=MIDDLE  ALT=\"A=x%5E2+%2B+625+-+25x+-+25x+%2B+x%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" > ***************<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=x%5E2+-+50x+%2B+625 ALIGN=MIDDLE  ALT=\"A=x%5E2+-+50x+%2B+625\"   DISABLED_style= \"max-width:90%; height:auto;\" > **************** <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A+=+2x%5E2+-+50x+%2B+625 ALIGN=MIDDLE  ALT=\"A+=+2x%5E2+-+50x+%2B+625\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "But then solutions for x are imaginary, what do I do now?<BR>\n" );
document.write( "--------------<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=2x%5E2+-+50x+%2B+625 ALIGN=MIDDLE  ALT=\"A=2x%5E2+-+50x+%2B+625\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "=======================<BR>\n" );
document.write( "There are no real zeroes for x, but that's not relevant.<BR>\n" );
document.write( "--------<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=2x%5E2+-+50x+%2B+625 ALIGN=MIDDLE  ALT=\"A=2x%5E2+-+50x+%2B+625\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "The minimum of the parabola is the vertex @ x = -b/2a<BR>\n" );
document.write( "x = -50/-4 = 12.5 --> minimum Area<BR>\n" );
document.write( "Minimum area = 312.5 sq m<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );