document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1198876>Question 1198876</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A molasses can is made with a cylindrical base having a radius of 6<BR>\n" );
document.write( "cm, that is topped by a smaller cylindrical pouring spout with radius<BR>\n" );
document.write( "2 cm. When upright, the top of the molasses is 14 cm above the<BR>\n" );
document.write( "base of the can, but when turned over, the top of the molasses is<BR>\n" );
document.write( "19⅓ cm above the base. Find the total height of the can in cm.\r<BR>\n" );
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document.write( "Diagram of Molasses Can:<BR>\n" );
document.write( "https://ibb.co/6RwWmPy\r<BR>\n" );
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document.write( "Ms.Ikelyn previously referred me to a solution by web2.0calc - however that solution is not valid as the poster did not provide a diagram as I have, hence, Cphil (answerer) had to make assumptions.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #832545 by </B><A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><B>math_tutor2020(3817)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><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=832545\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "For anyone curious, the web2.0calc post is here<BR>\n" );
document.write( "<a  rel=nofollow HREF= \"https://web2.0calc.com/questions/a-molasses\">https://web2.0calc.com/questions/a-molasses</a><BR>\n" );
document.write( "where the user CPhill has posted his solution.\r<BR>\n" );
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document.write( "This is the original diagram<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/display-illustration.mpl?tutor=math_tutor2020&illustration_name=Screenshot_166aa.png\"><BR>\n" );
document.write( "and I'll add these labels<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/display-illustration.mpl?tutor=math_tutor2020&illustration_name=Screenshot_165aa.png\">\r<BR>\n" );
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document.write( "x = height of the larger cylinder (figure 1) such that <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+0+%3C+x+%3C=+14 ALIGN=MIDDLE  ALT=\"+0+%3C+x+%3C=+14\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "14-x+y = height of the smaller cylinder\r<BR>\n" );
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document.write( "V1 = volume of the larger cylinder (figure 1)<BR>\n" );
document.write( "V2 = volume of the extra molasses (figure 1)\r<BR>\n" );
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document.write( "V3 = volume of the smaller cylinder (figure 2)<BR>\n" );
document.write( "V4 = volume of the extra molasses (figure 2)\r<BR>\n" );
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document.write( "Each volume (V1 through V4) represents a cylinder of some kind.<BR>\n" );
document.write( "V2 and V4 are part of the top cylinder for each respective figure.\r<BR>\n" );
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document.write( "Notice in figure 2, I show the calculation of:<BR>\n" );
document.write( "58/3 - (14-x+y) = x-y+16/3<BR>\n" );
document.write( "The 58/3 is the improper fraction form of the mixed number 19 & 1/3, aka 19 + 1/3<BR>\n" );
document.write( "Subtracting off the small cylinder's height, 14-x+y, will get us the remaining height of the molasses for figure 2. Then we have z to finish it off.\r<BR>\n" );
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document.write( "Volume of a cylinder = pi*(radius)^2*(height)<BR>\n" );
document.write( "V1 = pi*r^2*h<BR>\n" );
document.write( "V1 = pi*6^2*x<BR>\n" );
document.write( "V1 = 36pi*x\r<BR>\n" );
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document.write( "V2 = pi*r^2*h<BR>\n" );
document.write( "V2 = pi*2^2*(14-x)<BR>\n" );
document.write( "V2 = 4pi*(14-x)\r<BR>\n" );
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document.write( "V3 = pi*r^2*h<BR>\n" );
document.write( "V3 = pi*2^2*(14-x+y)<BR>\n" );
document.write( "V3 = 4pi*(14-x+y)\r<BR>\n" );
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document.write( "V4 = pi*r^2*h<BR>\n" );
document.write( "V4 = pi*6^2*(x-y+16/3)<BR>\n" );
document.write( "V4 = 36pi*(x-y+16/3)\r<BR>\n" );
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document.write( "Assuming there aren't any leaks in the container, we can say this:<BR>\n" );
document.write( "V1+V2 = V3+V4<BR>\n" );
document.write( "which basically says rearranging the contents of the container will not change the overall volume.\r<BR>\n" );
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document.write( "So,<BR>\n" );
document.write( "V1+V2 = V3+V4<BR>\n" );
document.write( "36pi*x+4pi*(14-x) = 4pi*(14-x+y)+36pi*(x-y+16/3)<BR>\n" );
document.write( "9x+(14-x) = (14-x+y)+9(x-y+16/3)\r<BR>\n" );
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document.write( "The jump from step 2 to step 3 involved dividing both sides by 4pi. <BR>\n" );
document.write( "Conveniently all of the pi terms cancel, and things get a bit more simple.\r<BR>\n" );
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document.write( "Now we don't need to find x, and we only need the value of y.<BR>\n" );
document.write( "This way we can compute 14+y later on (see figure 1).\r<BR>\n" );
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document.write( "As the steps below will indicate, the x values cancel out and we can solve for y as such:\r<BR>\n" );
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document.write( "9x+(14-x) = (14-x+y)+9(x-y+16/3)<BR>\n" );
document.write( "9x+14-x = 14-x+y+9x-9y+48<BR>\n" );
document.write( "8x+14 = 14+8x-8y+48<BR>\n" );
document.write( "8x+14-8x-14 = -8y+48<BR>\n" );
document.write( "0 = -8y+48<BR>\n" );
document.write( "8y = 48<BR>\n" );
document.write( "y = 48/8<BR>\n" );
document.write( "y = 6\r<BR>\n" );
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document.write( "Then,<BR>\n" );
document.write( "<font color=red>total height</font> = 14+y = 14+6 = <font color=red>20 cm</font><BR>\n" );
document.write( "when referencing figure 1\r<BR>\n" );
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document.write( "It's not a coincidence we get the same solution CPhill did on the link I posted above. <BR>\n" );
document.write( "This is because the x terms cancel out.<BR>\n" );
document.write( "So for the sake of simplicity, we could have x = 14.\r<BR>\n" );
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document.write( "Answer: <font color=red>20 cm</font><BR>\n" );
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