document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=40752>Question 40752</A>This question is from textbook <A HREF=http://www.amazon.com/exec/obidos/ASIN/0321227573/algebrahomeworkh>college algebra </A><BR>\n" );
document.write( ": <I><SPAN STYLE=\"word-break: break-all;\"> I got part a and b of this question , need help with d & C <BR>\n" );
document.write( "(of course a word problem) A farmer wishes to enclose a rectangular region bordering a river with fencing, as shown in the diagram.  Suppose that x represents the length of each of the three parallel pieces of fencin.  She has 600 ft of fencing available.<BR>\n" );
document.write( "( the diagram is a rectangle, the top part is water and the rest is fencing, there is a vertical dotted line down the center, it and both the left and right side are labeled x the bottom is just fencing)<BR>\n" );
document.write( "a) What is the length of the remaining piece of fencing in terms of x?<BR>\n" );
document.write( "---My answer, 600-3x<BR>\n" );
document.write( "b) Determine a function(A) that represents the total area of the enclosed region give any restrictions for x.<BR>\n" );
document.write( "  Since Length and Width cannot be negative i put 600-3x is greater than or equal to zero and solved that, 200 is greater than or equal to x, and x is greater than or equal to x so my solution is <BR>\n" );
document.write( "0is less than or equal to x is less than or equal to 200.<BR>\n" );
document.write( "NOW FOR THE ONES IM LOST ON>>>>><BR>\n" );
document.write( "c) What dimensions for the total enclosed region would give an area of 22,500 sqaure feet?<BR>\n" );
document.write( "and<BR>\n" );
document.write( "d) What is the maximum area that can be enclosed?<BR>\n" );
document.write( "ELisa </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #26145 by </B><A HREF=/tutors/aboutme.mpl?userid=Fermat><B>Fermat(136)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Fermat><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/cgi-bin/show-face.mpl?user=Fermat><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=26145\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Your answer to part a) is correct, but not part b) I'm afraid.<BR>\n" );
document.write( "Let x = vertical dimension<BR>\n" );
document.write( "Let y = horizontal dimension<BR>\n" );
document.write( "a) total length of fencing is 600 = 3x + y<BR>\n" );
document.write( ".: y = 600 - 3x<BR>\n" );
document.write( "===============<BR>\n" );
document.write( "b) Enclosed area, A = length times width<BR>\n" );
document.write( "A = xy<BR>\n" );
document.write( "A = x(600-3x) , x <= 200<BR>\n" );
document.write( "=============<BR>\n" );
document.write( "c) you are given a value for the area required. So set A = 22,500<BR>\n" );
document.write( "Now, A = x(600-3x)<BR>\n" );
document.write( "So,<BR>\n" );
document.write( "22,500 = x(600-3x)<BR>\n" );
document.write( "22500 = 600x - 3x²<BR>\n" );
document.write( "x² - 200x + 7500 = 0<BR>\n" );
document.write( "(x-50)(x-150) = 0<BR>\n" );
document.write( "x = 50, x = 150<BR>\n" );
document.write( "===============<BR>\n" );
document.write( "using y = 600-3x,<BR>\n" );
document.write( "y = 450, y = 150<BR>\n" );
document.write( "================<BR>\n" );
document.write( "So there are two solutions. Both solutions are valid. They both give an enclosed area of 22,500 square feet<BR>\n" );
document.write( "d) A = 600x - 3x²<BR>\n" );
document.write( "Now differentiate the expression for A and set to zero to get a turning point.<BR>\n" );
document.write( "dA/dx = 600 - 6x<BR>\n" );
document.write( "setting dA/dx = 0 gives,<BR>\n" );
document.write( "600 - 6x = 0<BR>\n" );
document.write( "x = 100<BR>\n" );
document.write( "=======<BR>\n" );
document.write( "Check to see that this is a maximum, by differentiating again. If d²A/dx² is negative at x = 100, then the turning point is a maximum. (If d²A/dx² is positive at x = 100, then the turning point is a minimum)<BR>\n" );
document.write( "dA/dx = 600 - 6x<BR>\n" );
document.write( "d²A/dx² = -6, which is < 0 therefore a maximum.<BR>\n" );
document.write( "Since x = 100, y = 600 - 3x = 600 - 300 = 300<BR>\n" );
document.write( "y = 300<BR>\n" );
document.write( "=======<BR>\n" );
document.write( "area = xy<BR>\n" );
document.write( "Max area = 90,000 square feet<BR>\n" );
document.write( "============================= </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );