document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=629201>Question 629201</A>: <I><SPAN STYLE=\"word-break: break-all;\"> bob owns a watch repair shop. he has found that the cost of operating his shop is given by c(x)=3x^2-168x+52, where x is the number of watches repaired. how many watches must he repair to have the lowest cost? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #396091 by </B><A HREF=/tutors/aboutme.mpl?userid=303795><B>303795(602)</B></A><img src=\"/images/stars/stars-11.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=303795><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=396091\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The minimum cost will have a first derivative with a value of zero.<BR>\n" );
document.write( "ie <BR>\n" );
document.write( "c(x)=3x^2 - 168x + 52<BR>\n" );
document.write( "c'(x) = 6x - 168<BR>\n" );
document.write( "ie 6x - 168 = 0<BR>\n" );
document.write( "6x = 168<BR>\n" );
document.write( "x = 168/6<BR>\n" );
document.write( "= 28 </SPAN>\n" );
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