document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=253062>Question 253062</A>: <I><SPAN STYLE=\"word-break: break-all;\"> At which vertex is the objective function C = 3x - 4y maximized? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #185517 by </B><A HREF=/tutors/aboutme.mpl?userid=drk><B>drk(1908)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=drk><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=185517\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> First, we solve for y as<BR>\n" );
document.write( "y = (3/4)x - C/4<BR>\n" );
document.write( "If x = 0, then y = C/4<BR>\n" );
document.write( "If y = 0, then x = C/3.<BR>\n" );
document.write( "--<BR>\n" );
document.write( "C = 3x - 4y.<BR>\n" );
document.write( "C(0,C/4) = 3*0 - 4(C/4) = -4<BR>\n" );
document.write( "C(C/3,0) = 3*(C/3) - 4*0 = C<BR>\n" );
document.write( "--<BR>\n" );
document.write( "Now it gets tricky.<BR>\n" );
document.write( "If C > -4 then (C/3) is a max.<BR>\n" );
document.write( "Otherwise, C/4 is the max. </SPAN>\n" );
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