document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=950318>Question 950318</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Theorem: For any given area A, the rectangle that has the least perimeter is a square.\r<BR>\n" );
document.write( "\n" );
document.write( "Part I:\r<BR>\n" );
document.write( "\n" );
document.write( "Complete the proof that is outlined below. Submit each step as part of your final answer.\r<BR>\n" );
document.write( "\n" );
document.write( "Proof:\r<BR>\n" );
document.write( "\n" );
document.write( "In terms of the side length, x, the formula for the perimeter of a square: P= 2x+((2A)/x)\r<BR>\n" );
document.write( "\n" );
document.write( "Define a new variable: v=sqrt(x)-(sqrt(A))/(sqrt(x))\r<BR>\n" );
document.write( "\n" );
document.write( "    In terms of the new variable, v, compute and simplify: 2v^2 = _____.<BR>\n" );
document.write( "    Rewrite this equation to get the formula for P alone on one side. Replace the formula by the variable P. Write that result by completing the equation below:\r<BR>\n" );
document.write( "\n" );
document.write( "P = 2v^2 + __________\r<BR>\n" );
document.write( "\n" );
document.write( "Part II:\r<BR>\n" );
document.write( "\n" );
document.write( "    In the equation above, P is a function of v, and A is a constant. The minimum value of this function occurs when v = _____.<BR>\n" );
document.write( "    Explain why this is true.<BR>\n" );
document.write( "    Substitute this value of v into v=sqrt(x)-(sqrt(A))/(sqrt(x))  and solve for x.<BR>\n" );
document.write( "    This is the value of the length x that yields the minimum value of P. For this length, what is the value of the width?<BR>\n" );
document.write( "    How do the length and width compare?<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #580474 by </B><A HREF=/tutors/aboutme.mpl?userid=josgarithmetic><B>josgarithmetic(39626)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=josgarithmetic><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=580474\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The plan you hope to follow is obscure.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "More directly you have known area A, unknown dimensions x and y, and function p for perimeter.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=xy=A ALIGN=MIDDLE  ALT=\"xy=A\"   DISABLED_style= \"max-width:90%; height:auto;\" > and p is a function, <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=p=2x%2B2y ALIGN=MIDDLE  ALT=\"p=2x%2B2y\"   DISABLED_style= \"max-width:90%; height:auto;\" >.<BR>\n" );
document.write( "Substitute for y, <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=p=2x%2B2%28A%2Fx%29 ALIGN=MIDDLE  ALT=\"p=2x%2B2%28A%2Fx%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > just as you found.  This p is a rational function.  Use of derivative would be best.  You could really go low-level and try the difference quotient and look for a limit, but too complicated and timeconsuming.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=dp%2Fdx=2%2B%28d%2Fdx%29%282A%2Fx%29 ALIGN=MIDDLE  ALT=\"dp%2Fdx=2%2B%28d%2Fdx%29%282A%2Fx%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=dp%2Fdx=2%2B%28-1%292A%2Ax%5E%28-2%29 ALIGN=MIDDLE  ALT=\"dp%2Fdx=2%2B%28-1%292A%2Ax%5E%28-2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=dp%2Fdx=2-2A%2Fx%5E2 ALIGN=MIDDLE  ALT=\"dp%2Fdx=2-2A%2Fx%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" >, and you want to know for what x value is this rate equal to 0, which will be the minimum for p.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2-2A%2Fx%5E2=0 ALIGN=MIDDLE  ALT=\"2-2A%2Fx%5E2=0\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1-A%2Fx%5E2=0 ALIGN=MIDDLE  ALT=\"1-A%2Fx%5E2=0\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1=A%2Fx%5E2 ALIGN=MIDDLE  ALT=\"1=A%2Fx%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E2=A ALIGN=MIDDLE  ALT=\"x%5E2=A\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x=sqrt%28A%29 ALIGN=MIDDLE  ALT=\"x=sqrt%28A%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "You can do the same process but substituting for x instead of for y, and you will find <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y=sqrt%28A%29 ALIGN=MIDDLE  ALT=\"y=sqrt%28A%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x=y=sqrt%28A%29 ALIGN=MIDDLE  ALT=\"x=y=sqrt%28A%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >  will be the dimensions for the smallest perimeter, p.  This is a square. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );