document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1107733>Question 1107733</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A sheet of metal, 60cm wide and of indefinite length is bent to form an open rectangular gutter. What height of the gutter will give the maximum cross-sectional area? What height will give a cross-sectional area of 300cm^2? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #722802 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=722802\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A sheet of metal, 60cm wide and of indefinite length is bent to form an open rectangular gutter.<BR>\n" );
document.write( " What height of the gutter will give the maximum cross-sectional area?<BR>\n" );
document.write( " What height will give a cross-sectional area of 300cm^2?<BR>\n" );
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document.write( "A rough picture of the end of the gutter<BR>\n" );
document.write( " h|_w_|h<BR>\n" );
document.write( "h = the height of the gutter,<BR>\n" );
document.write( "w = the width<BR>\n" );
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document.write( "2h + w = 60<BR>\n" );
document.write( "w = (60-2h), we can use this form for substitution<BR>\n" );
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document.write( "Cross sectional area = width * height<BR>\n" );
document.write( "f(h) = the cross sectional area, replace w with (60-2h)<BR>\n" );
document.write( "f(h) = h(60-2h)<BR>\n" );
document.write( "f(h) = 60h - 2h^2<BR>\n" );
document.write( "max h occurs at the axis of symmetry a = -b/(2a), where a = -2, b = 60<BR>\n" );
document.write( "h = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-60%29%2F%282%2A-2%29 ALIGN=MIDDLE  ALT=\"%28-60%29%2F%282%2A-2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "h = +15 cm is the height for max area<BR>\n" );
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document.write( "\"What height will give a cross-sectional area of 300cm^2?\"<BR>\n" );
document.write( "60h - 2h^2 = 300<BR>\n" );
document.write( "Arrange as a quadratic equation<BR>\n" );
document.write( "-2h^2 + 60h - 300 = 0<BR>\n" );
document.write( "simplify, divide by -2<BR>\n" );
document.write( " h^2 - 30h + 150 = 0<BR>\n" );
document.write( "use the quadratic formula; a=1, b=-30, c=150<BR>\n" );
document.write( "I got two solutions<BR>\n" );
document.write( "h = 6.34 cm, then w = 60 - 2(6.34) = 47.32 cm is the width<BR>\n" );
document.write( "and<BR>\n" );
document.write( "h = 23.66 cm, then w = 60 - 2(23.66) = 12.68 cm is the width<BR>\n" );
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document.write( "Check this, find the cross section area with these value<BR>\n" );
document.write( "6.34 * 47.32 = 300 sq cm<BR>\n" );
document.write( "and<BR>\n" );
document.write( "23.66 * 12.68 = 330 sq cm<BR>\n" );
document.write( "   </SPAN>\n" );
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