document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1205006>Question 1205006</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The volume of a frustum of a cone of height h is given by the formula<BR>\n" );
document.write( "  V = ⅓πh(R² + Rr + r²)<BR>\n" );
document.write( "Where R and r are the radii of its circular ends.<BR>\n" );
document.write( "  Hollow metal hemispheres of internal and external diameters 12.5cm and 17.5cm respectively, are to be cast from molten contents of a ladle in the shape of a frustum of a cone of depth 1.2m and radii of ends 0.6m and 0.635m. Calculate the number of hemispheres that can be cast.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #841592 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=841592\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "This can be done on a regular calculator, but I'll turn to <a  rel=nofollow HREF=\"https://www.geogebra.org/\">GeoGebra</a> instead.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Here is the particular GeoGebra workbook that I created for this problem. <BR>\n" );
document.write( "<a  rel=nofollow HREF=\"https://www.geogebra.org/calculator/gwhxa25b\">https://www.geogebra.org/calculator/gwhxa25b</a><BR>\n" );
document.write( "Normally I use GeoGebra to make some kind of graph. <BR>\n" );
document.write( "However, the graph part is blank. Focus on the column of items on the left side.<BR>\n" );
document.write( "I'll explain the inputs below. Each input is highlighted in <font color=blue>blue</font>\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<font color=blue>R = 0.635</font> and <font color=blue>r = 0.6</font> represent the radius values of the conical frustum. \r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<font color=blue>h = 1.2</font> is the height or depth of the ladle.<BR>\n" );
document.write( "So far everything is in meters, so I'll convert the diameters \"12.5cm and 17.5cm\" to meters<BR>\n" );
document.write( "12.5 cm = 0.125 m<BR>\n" );
document.write( "17.5 cm = 0.175 m<BR>\n" );
document.write( "Divide by 100 to convert from cm to meters.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "The formula <font color=blue>Vf = (1/3)*pi*h*(R^2+R*r+r^2)</font> represents the volume of the conical frustum. <BR>\n" );
document.write( "It is the formula your teacher gave you. <BR>\n" );
document.write( "GeoGebra will replace the letters h, R and r with their proper values. <BR>\n" );
document.write( "The result of the calculation is roughly Vf = 1.43788 when rounding to 5 decimal places.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "So far we've just taken care of the conical frustum only.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Recall that the volume of a full sphere is (4/3)*pi*(radius)^3\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "The larger sphere has volume (4/3)*pi*(external radius)^3<BR>\n" );
document.write( "The smaller sphere has volume (4/3)*pi*(internal radius)^3\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "The volume of a spherical outer shell, ignoring the hollow empty space, is:<BR>\n" );
document.write( "shell volume = (larger sphere volume) - (smaller sphere volume)<BR>\n" );
document.write( "shell volume = (4/3)*pi*(external radius)^3 - (4/3)*pi*(internal radius)^3<BR>\n" );
document.write( "shell volume = (4/3)*pi*( (external radius)^3 - (internal radius)^3 )<BR>\n" );
document.write( "shell volume = (4/3)*pi*( (eR)^3 - (iR)^3 )\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "where,<BR>\n" );
document.write( "eR = external radius<BR>\n" );
document.write( "iR = internal radius\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "In this case:<BR>\n" );
document.write( "eR = 0.175/2 = 0.0875<BR>\n" );
document.write( "iR = 0.125/2 = 0.0625<BR>\n" );
document.write( "both are in meters. <BR>\n" );
document.write( "Keep in mind the 0.175 m and 0.125 m represent diameters. <BR>\n" );
document.write( "Cut those in half to get their corresponding radius values.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "The formula (4/3)*pi*( (eR)^3 - (iR)^3 ) represents a full spherical shell. But we want a hemispherical shell instead. We'll cut that in half to get (2/3)*pi*( (eR)^3 - (iR)^3 )\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "So that's where the formula <font color=blue>Vh = (2/3)*pi*( (eR)^3 - (iR)^3 )</font> comes from.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "-------------------------------------\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Here's a quick recap of the inputs to type in (type them in the order mentioned)<BR>\n" );
document.write( "<font color=blue>R = 0.635</font><BR>\n" );
document.write( "<font color=blue>r = 0.6</font><BR>\n" );
document.write( "<font color=blue>h = 1.2</font><BR>\n" );
document.write( "<font color=blue>Vf = (1/3)*pi*h*(R^2+R*r+r^2)</font><BR>\n" );
document.write( "That takes care of the frustum<BR>\n" );
document.write( "Then for the hemispheres we have these inputs<BR>\n" );
document.write( "<font color=blue>eR = 0.0875</font><BR>\n" );
document.write( "<font color=blue>iR = 0.0625</font><BR>\n" );
document.write( "<font color=blue>Vh = (2/3)*pi*( (eR)^3 - (iR)^3 )</font>\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "GeoGebra should display these approximate values<BR>\n" );
document.write( "Vf = 1.43788<BR>\n" );
document.write( "Vh = 0.00089<BR>\n" );
document.write( "which represent the frustum volume and hemisphere volume respectively.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "The last calculation to make is to type in <font color=blue>ratio = Vf/Vh</font> and the result should be roughly 1612.41248<BR>\n" );
document.write( "Round this down to the nearest whole number and it shows we can make <font color=red size=4>1612</font> hemispheres.<BR>\n" );
document.write( "This assumes that there is no waste. \r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Again this could have been done with any calculator, but I used GeoGebra so we could name the inputs and use them at a later time. Likely other calculators have this capability as well. \r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "--------------------------------------------------------------------------<BR>\n" );
document.write( "--------------------------------------------------------------------------\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Answer: <font color=red size=4>1612</font><BR>\n" );
document.write( "</font> </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );