document.write( "Question 890510: The question is related to Maxima and Minima chapter of Differential calculus. The question is as under :\r
\n" ); document.write( "\n" ); document.write( "\" Show that the cone of the greatest volume which can be inscribed in a given sphere has an altitude equal to 2/3 of the diameter of the sphere.\"\r
\n" ); document.write( "\n" ); document.write( "Please send step by step solution of the above question.\r
\n" ); document.write( "\n" ); document.write( "Regards,\r
\n" ); document.write( "\n" ); document.write( "Khoka123 \n" ); document.write( "
Algebra.Com's Answer #539052 by Edwin McCravy(20055)\"\" \"About 
You can put this solution on YOUR website!

\n" ); document.write( "
\r\n" );
document.write( "The above is a mid-cross section.  \r\n" );
document.write( "\r\n" );
document.write( "The radius of the sphere = OB = OC = R\r\n" );
document.write( "\r\n" );
document.write( "The height of the cone = CD = h\r\n" );
document.write( "\r\n" );
document.write( "OD = h-R\r\n" );
document.write( "\r\n" );
document.write( "The radius of the cone is \r\n" );
document.write( "DB = \"sqrt%28OB%5E2-OD%5E2%29\" = \"sqrt%28R%5E2-%28h-R%29%5E2%29\" = \"sqrt%28R%5E2-%28h%5E2-2hR%2BR%5E2%29%29\" = \r\n" );
document.write( "\"sqrt%28R%5E2-h%5E2%2B2hR%2BR%5E2%29\" = \"sqrt%28-h%5E2%2B2hR%29\"\r\n" );
document.write( "\r\n" );
document.write( "Volume of cone = \"V+=+expr%281%2F3%29pi%2A%28radius%29%5E2%2Ah%7D=expr%28pi%2F3%29r%5E2%2Ah\"\r\n" );
document.write( "\r\n" );
document.write( "\"V+=expr%28pi%2F3%29%28sqrt%28-h%5E2%2B2hR%29%29%5E2%2Ah\"\r\n" );
document.write( "\r\n" );
document.write( "\"V+=expr%28pi%2F3%29%28-h%5E2%2B2hR%29%2Ah\"\r\n" );
document.write( "\r\n" );
document.write( "\"V+=expr%28pi%2F3%29%28-h%5E3%2B2Rh%5E2%29\"\r\n" );
document.write( "\r\n" );
document.write( "\"%28dV%29%2F%28dh%29=expr%28pi%2F3%29%28-3h%5E2%2B4Rh%29\"\r\n" );
document.write( "\r\n" );
document.write( "Put that = 0\r\n" );
document.write( "\r\n" );
document.write( "\"expr%28pi%2F3%29%28-3h%5E2%2B4Rh%29=0\"\r\n" );
document.write( "\r\n" );
document.write( "Divide through by \"pi%2F3\"\r\n" );
document.write( "\r\n" );
document.write( "   \"-3h%5E2%2B4Rh=0\"\r\n" );
document.write( "\r\n" );
document.write( "Divide through by h.  We aren't interested in when h=0\r\n" );
document.write( "\r\n" );
document.write( "   \"-3h%2B4R=0\"\r\n" );
document.write( "\r\n" );
document.write( "   \"4R=3h\"\r\n" );
document.write( "\r\n" );
document.write( "   \"4R%2F3=h\"\r\n" );
document.write( "\r\n" );
document.write( "Since the radius = \"expr%281%2F2%29D\", where D is the diameter, \r\n" );
document.write( "\r\n" );
document.write( "   \"4%28expr%281%2F2%29D%29%2F3+=+h\"\r\n" );
document.write( "\r\n" );
document.write( "   \"expr%282%2F3%29D+=+h\"\r\n" );
document.write( "\r\n" );
document.write( "So we have proved that the height is \"2%2F3\" the diameter of the sphere.\r\n" );
document.write( "\r\n" );
document.write( "Edwin
\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "
\n" );