document.write( "Question 1019815: A Wire Length L Is To Cut Into Two Pieces, One Of Which Is To Be Bent Into The Shape Of A Circle And The Other Into The Shape Of An Equilateral Triangle. Find The Length Of Each Piece So That The Sum Of The Enclosed Areas Is A Minimum
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Algebra.Com's Answer #635948 by KMST(5328)\"\" \"About 
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\"R\"= radius of the circle
\n" ); document.write( "\"x\"= side of the triangle
\n" ); document.write( "\"L\" length of the wire before cutting it
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\n" ); document.write( "\"2pi%2AR\"= length of wire used for the circle
\n" ); document.write( "\"3x=L-2pi%2AR\"= length of wire used for the triangle\r
\n" ); document.write( "\n" ); document.write( "\"3x=L-2pi%2AR\"<-->\"x=%28L-2pi%2AR%29%2F3\"
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\n" ); document.write( "Area of the triangle =
\n" ); document.write( "Area of the circle =\"pi%2AR%5E2\"
\n" ); document.write( "Total area =\"y=pi%2AR%5E2%2Bsqrt%283%29%28L-2pi%2AR%29%5E2%2F36\"
\n" ); document.write( "\"y=pi%2AR%5E2%2Bsqrt%283%29%28L%5E2-4L%2Api%2AR%2B4pi%5E2%2AR%5E2%29%2F36\"
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\n" ); document.write( "That is a quadratic function.
\n" ); document.write( "Quadratic functions of the form \"y=ax%5E2%2Bbx%2Bc\" with \"a%3E0\"
\n" ); document.write( "have a minimum at \"x=%28-b%29%2F2a\" .
\n" ); document.write( "In the case of ,
\n" ); document.write( "\"x=R\" , \"a=%28pi%2Bsqrt%283%29pi%5E2%2F9%29%3E0\" , and \"b=-sqrt%283%29L%2Api%2F9\" ,
\n" ); document.write( "so the minimum is at
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