document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=450607>Question 450607</A>: <I><SPAN STYLE=\"word-break: break-all;\"> An isosceles triangle has perimeter 64m. Find the dimensions of this triangle that make its area a maximum </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #310193 by </B><A HREF=/tutors/aboutme.mpl?userid=richard1234><B>richard1234(7193)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=richard1234><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=richard1234><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=310193\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Suppose the side lengths are x, x, and 64-2x:\r<BR>\n" );
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document.write( "(64-2x won't show up so assume the base length is 64-2x)\r<BR>\n" );
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document.write( "By Heron's formula, we can find the area A of the triangle in terms of x (semiperimeter = 32)\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE A = \sqrt{32(32-x)(32-x)(32-(64-2x))} = \sqrt{32(32-x)^2(2x-32)} = (32-x)\sqrt{64x-1024}\">\r<BR>\n" );
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document.write( "Maximizing the area requires a little calculus. We can say that A is a function in terms of x, and find dA/dx. However, we can also note that the value of x that maximizes A will also be the value of x that maximizes A^2. To make things simpler, we can find A^2:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE A^2 = (32-x)^2(64x-1024)\">\r<BR>\n" );
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document.write( "Here, we take the derivative with respect to x:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \frac{d(A^2)}{dx} = -2(32-x)(64x-1024) + (32-x)^2(64)\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \frac{d(A^2)}{dx} = (32-x)(-128x + 2048 + 2048 - 64x) = (32-x)(-192x + 4096)\">\r<BR>\n" );
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document.write( "Here, the derivative is equal to zero when x = 32 or x = 64/3. Clearly, x = 64/3 maximizes the area of the triangle. This would also imply that the maximal area occurs when the triangle is equilateral.\r<BR>\n" );
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