document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=980840>Question 980840</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I'm completely failing to understand ratios. I'm trying to understand a 1 to 1.5 ratio of cylinders of differing diameters. V = pi x r squared x length. Ok got it. My question is, Why is a cylinder measuring 3\" in diameter 24\" long (678.58401 cu in) not the same as a cylinder measuring 1.5\" in diameter 48\" long (339.29201 cu in)? My simple minded logic say's to me that if I have one cylinder 3\" in diameter 24\" long (678.58401 cu in) and divide that by 2 (339.292005 cu in) and add that to the original (678.58401 cu in) for (1017.876015) that (1017.876015 cu in) should be the target 1 to 1.5 ratio. However I don't understand how to turn (1017.876015 cu in) into a length for a 4\" cylinder that is 1.5 to (678.58401 cu in). I suppose that's two questions but any feedback on this confusion would be greatly appreciated.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #601909 by </B><A HREF=/tutors/aboutme.mpl?userid=solver91311><B>solver91311(24713)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=solver91311><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=solver91311><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=601909\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font face=\"Times New Roman\" size=\"+2\">\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "The volume of a cylinder varies directly as the height but directly as the SQUARE of the radius.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "If you simply halve the radius, then the smaller cylinder is 1/4 of the volume of larger, but if you then double the length, the smaller radius cylinder is still only 1/2 the volume of the original cylinder.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "To begin with, all of your volume calculations are 4 times too large.  You are forgetting to divide the diameter by 2 to get the radius.  Be that as it may, I am having trouble deciphering which way you want to go with the 1:1.5 ratio.  So, I'm just going to assume that you want the larger diameter cylinder to have 1.5 times the volume of the smaller diameter cylinder and further, you want to halve the larger diameter and find the length that will result in the given ratio.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "The volume of your original cylinder is\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ V_1\ =\ \pi r^2\ h_1\">\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "And the new, smaller diameter cylinder\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ V_2\ =\ \pi\left(\frac{r}{2}\right)^2h_2\">\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "But we want the first volume to be time and a half of the second volume, so:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ V_1\ =\ 1.5V_2\">\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Substituting:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ \pi r^2\ h_1\ =\ 1.5\left(\pi\left(\frac{r}{2}\right)^2h_2\right)\">\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "A little simplification:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ h_1\ =\ \frac{1.5}{4}h_2\">\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Or, since you know the original cylinder's height:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ \frac{4h_1}{1.5}\ =\ h_2\">\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "So if you halve the radius and make the length of the new cylinder <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large 2\frac{2}{3}\"> times longer, the original cylinder will have a volume that is 1.5 times larger than the new cylinder.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "John<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE e^{i\pi}\ +\ 1\ =\ 0\"><BR>\n" );
document.write( "My calculator said it, I believe it, that settles it </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );