document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=957901>Question 957901</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Please help me solve this question:<BR>\n" );
document.write( "You will be constructing a rectangular deck against your bungalow, using 32 feet of railing and will leave a 4 foot gap in the railing for access to the stairs. Determine the dimensions that will maximize the area of the deck, and the maximum area. \r<BR>\n" );
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document.write( "What work I've done so far:<BR>\n" );
document.write( "- The deck only has 3 sides (32 ft railing and a 4 ft gap)<BR>\n" );
document.write( "36/3 = 3S/3<BR>\n" );
document.write( "= 12 for Width of each side (therefore 12ft squared is the max area of the deck)\r<BR>\n" );
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document.write( "For max area I did<BR>\n" );
document.write( "A= l x W<BR>\n" );
document.write( "A= (12)X (12)<BR>\n" );
document.write( "A= 144, Therefore 144 ft squared is the max area)\r<BR>\n" );
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document.write( "Gap for railing = 4ft.<BR>\n" );
document.write( "A= max area - gap<BR>\n" );
document.write( "A= 144-4<BR>\n" );
document.write( "A= 140 ft squared.\r<BR>\n" );
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document.write( "Any help will be greatly appreciated!!<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #585421 by </B><A HREF=/tutors/aboutme.mpl?userid=addingup><B>addingup(3677)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=addingup><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=addingup><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=585421\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Total perimeter (3 sides only): 32+4= 36 (32 of railing + 4ft opening for stairs)<BR>\n" );
document.write( "I’ll call the Length X and the Width Y:<BR>\n" );
document.write( "Maximize A = area of the rectangle = (Length)(Width) = X*Y<BR>\n" );
document.write( "given a perimeter of X+2Y= 36 (only 1 X, the other is against the bungalow-see my illustration)<BR>\n" );
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document.write( "Now we need to solve for one of the variables.  Since  X+2Y= 36, <BR>\n" );
document.write( "2Y= 36-X; Y= (36-X)/2;  Y= 18- X/2<BR>\n" );
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document.write( "Now that we solved for Y, we have that:<BR>\n" );
document.write( "A= X*Y= X*(18- X/2)= 18X-X^2/2<BR>\n" );
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document.write( "We want to maximize A:<BR>\n" );
document.write( "A’= 18-X. The only time A’=0 is when X= 18. <BR>\n" );
document.write( "So, 0<=X<=36. Why? Because we make it such that the only critical points of A are when X=0, 18, and 36, and the maximum area will be given by one of them.<BR>\n" );
document.write( "Here we go:<BR>\n" );
document.write( "At the critical number X= 0: A= 18(0) - (0)^2/2= 0 square feet – toss this answer out.<BR>\n" );
document.write( "At the critical number X= 18: A= 18(18) – (18)^2/2 = 162<BR>\n" );
document.write( "At the critical number X= 36: A= 18(36) – (36)^2/2 = 0 – toss this one out, too.\r<BR>\n" );
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document.write( "So our largest possible area is 162 square feet and the dimensions are: length (X)= 18 feet  and width (Y)= 18 - X/2= 18 - 9= 9 feet.<BR>\n" );
document.write( "Check: 18 + 2(9)= 36 feet, 32 of railing+4 feet opening on the railing for the stairs.<BR>\n" );
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