document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1191566>Question 1191566</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find the dimensions of the right circular cylinder of greatest volume that can be inscribed in a right-circular cone with a radius of 5 cm and a height of 12 cm. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #823446 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13200)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=greenestamps><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=greenestamps><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=823446\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "Here is a 2-dimensional sketch of a right circular cylinder inscribed in a right circular cone.<br><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=drawing%28400%2C400%2C-8%2C8%2C-2%2C14%0D%0A%2Cline%28-5%2C0%2C5%2C0%29%2Cline%28-5%2C0%2C0%2C12%29%2Cline%285%2C0%2C0%2C12%29%2Cline%280%2C0%2C0%2C12%29%2Cline%28-10%2F3%2C4%2C10%2F3%2C4%29%0D%0A%2Cline%28-10%2F3%2C0%2C-10%2F3%2C4%29%2Cline%2810%2F3%2C0%2C10%2F3%2C4%29%0D%0A%2Clocate%280%2C0%2CA%29%2Clocate%28.2%2C3.8%2CB%29%2Clocate%280%2C13%2CC%29%2Clocate%285%2C0%2CD%29%2Clocate%283.6%2C4.2%2CE%29%0D%0A%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "We are given AD = 5 and CD = 13, so we can conclude AC = 12.<br><BR>\n" );
document.write( "Let x be the radius of the cylinder, BE.  Similar triangles ACD and BCE give us BC = (12/5)x = 2.4x.  The height of the cylinder is then 12-2.4x.<br><BR>\n" );
document.write( "Express the volume of the cylinder in terms of x and maximize the volume by finding where the derivative of the volume function is 0.<br><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=V=%28pi%29%28r%5E2%29%28h%29=%28pi%29%28x%5E2%29%2812-2.4x%29=pi%2812x%5E2-2.4x%5E3%29 ALIGN=MIDDLE  ALT=\"V=%28pi%29%28r%5E2%29%28h%29=%28pi%29%28x%5E2%29%2812-2.4x%29=pi%2812x%5E2-2.4x%5E3%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=dV%2Fdx+=+pi%2824x-7.2x%5E2%29+=+%28pi%29%2824x%281-.3x%29%29 ALIGN=MIDDLE  ALT=\"dV%2Fdx+=+pi%2824x-7.2x%5E2%29+=+%28pi%29%2824x%281-.3x%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=.3x+=+1 ALIGN=MIDDLE  ALT=\".3x+=+1\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x+=+1%2F.3+=+10%2F3 ALIGN=MIDDLE  ALT=\"x+=+1%2F.3+=+10%2F3\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "The volume is maximum when the radius of the cylinder is 10/3.<br><BR>\n" );
document.write( "That makes the height <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=12-2.4%2810%2F3%29=12-8=4 ALIGN=MIDDLE  ALT=\"12-2.4%2810%2F3%29=12-8=4\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "ANSWER: radius 10/3; height 4<br><BR>\n" );
document.write( " </SPAN>\n" );
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