document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1098711>Question 1098711</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A wire 10 meters long is to be cut into two pieces. One piece will be shaped as a square, and the other piece will be shaped as a circle. The figure shows 10m as the full length up and down. On the other side of the line it shows 4x, and 10-4x. It shows a square with an x, and a circle with nothing next to it.\r<BR>\n" );
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document.write( "(a) Express the total area A enclosed by the pieces of wire as a function of the length x of the side of the square.\r<BR>\n" );
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document.write( "(b) What is the domain A?\r<BR>\n" );
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document.write( "(c) Graph A=A(x). for what value of x is A the smallest?\r<BR>\n" );
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document.write( "My book doesn't show clearly how to set up this problem. I hope I was clear enough with the dimensions. I didn't know which category to post as it is in functions and their graphs, but didn't want to put it in the function category that said NOT graphing. Sorry if I have misplaced or not been clear. I will add on if needed. I don't just post because I want the answer, I truly want to know how to set the problem up so that I can successfully do the problem right on homework, and on a test. Thank you so much for taking the time to read my question and help show me how to do my homework problem! it's very appreciated!  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #713106 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=713106\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Using the notation there, the square has area x^2 and the circle's area is f(radius), and the radius can be found by taking 10-4x, the circumference, and dividing by 2pi, so the area of the circle is pi^2, after reducing the radius to (5-2x)/pi<BR>\n" );
document.write( "The sum of these areas is x^2+[pi*(5-2x)^2/pi^2]<BR>\n" );
document.write( "Take the derivative and set equal to 0: 2x+(1/pi)(2(5-2x)(-2))=0\r<BR>\n" );
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document.write( "This is 2x + (1/pi)(4x-20)=0<BR>\n" );
document.write( "multiply by pi<BR>\n" );
document.write( "2x*pi+4x-20=0=x*pi+2x-10<BR>\n" );
document.write( "x(pi+2)=10<BR>\n" );
document.write( "x=10/(pi+2)<BR>\n" );
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document.write( "Minimum is when A=3.5 m^2 and x=1.4 m </SPAN>\n" );
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