document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=182631>Question 182631</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A wire 12 inches long is cut into two pieces. One piece is bent to form a square. The other is bent to form a rectangle which is 1 inch longer than it is wide. How long is each piece if the sum or the area is a minimum?\r<BR>\n" );
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document.write( "All I know is that it may help to use the equation <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=V=-b%2F2a ALIGN=MIDDLE  ALT=\"V=-b%2F2a\"   DISABLED_style= \"max-width:90%; height:auto;\" > </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #137146 by </B><A HREF=/tutors/aboutme.mpl?userid=kev82><B>kev82(151)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=kev82><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=137146\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> So the wire is cut into to pieces, some of it is used to make a rectangle, the rest a square. Lets's say the amount used for the rectangle is x. This means the amount used for the square is 12-x.\r<BR>\n" );
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document.write( "Lets's consider the square, it has perimeter 12-x, a square has 4 sides, each the same length. So the length of one side of this square will be 3-x/4. So the area of the square will be (3-x/4)^2.\r<BR>\n" );
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document.write( "The rectangle has perimeter x, the formula for perimeter is 2*(w+l), but we know from the question that l=w+1. So 2*(w+w+1) = x, 4w+2 = x, w =(x-2)/4. The area of a rectangle is w*l, so in this case it is. (x-2)(x+2)/16.\r<BR>\n" );
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document.write( "So the formula for the total area is\r<BR>\n" );
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document.write( "(3-x/4)^2 + (x-2)(x+2)/16\r<BR>\n" );
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document.write( "I don't like fractions very much, so I am going to simplify this by letting y=x/4\r<BR>\n" );
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document.write( "(3-y)^2 + y^2 - 1/4\r<BR>\n" );
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document.write( "Looking at this, you can see the dominating term is a positive y^2 (expand the first brackets if you are not convinced). This means the function looks approximately like y^2, which has a minimum at it's turning point. If we were looking for a maximum, then you wouldn't differentiate as the maximum of a y^2 graph is at its endpoints. Of course if it was dominated by negative y^2 the the max would be at the turning point and the min at the end points. Anyway, we have to differentiate to find this turning point.\r<BR>\n" );
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document.write( "Remember I'm differentiating wrt x, and dy/dx=1/4. (Use chain rule)\r<BR>\n" );
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document.write( "-2*(3-y)/4 + 2y/4 = 0\r<BR>\n" );
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document.write( "4y=6\r<BR>\n" );
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document.write( "y = 3/2\r<BR>\n" );
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document.write( "but remember y=x/4, so x=6. That means the minimum area is when you give exactly half of the string to each.  </SPAN>\n" );
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