document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=564691>Question 564691</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A cylinder has a volume of 1200cm3 and the least surface area possible. What are the dimensions of the cylinder?<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 #365445 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=365445\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Assume the cylinder has a circular base. We want to minimize\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE A = 2\pi r^2 + 2\pi rh\">, given <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE V = \pi r^2h = 1200\">. Couple ways to do this:\r<BR>\n" );
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document.write( "Solution 1: Let <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE h = \frac{1200}{\pi r^2}\">. Then <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE A = 2\pi r^2 + 2\pi r(\frac{1200}{\pi r^2}) = 2\pi r^2 + \frac{2400}{r}\">. Taking derivative with respect to r,\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \frac{dA}{dr} = 4\pi r - \frac{2400}{r^2}\">. This equals zero when <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE 4\pi r = \frac{2400}{r^2}\">, or when <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE r = \sqrt[3]{\frac{600}{\pi}}\">. &#8718;\r<BR>\n" );
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document.write( "Solution 2: Using the method of Lagrange multipliers, there exists a constant <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \lambda\"> with A(r,h) and V(r,h) such that\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \bigtriangledown A = \lambda \bigtriangledown V\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \langle 4r\pi + 2\pi h, 2\pi r \rangle = \lambda \langle 2rh, r^2 \rangle\">, therefore\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE 4r\pi + 2\pi h = 2rh\lambda \Rightarrow \lambda = \frac{2\pi}{h} + \frac{\pi}{r}\"><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE 2\pi r = r^2\lambda \Rightarrow \lambda = \frac{2\pi}{r}\">\r<BR>\n" );
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document.write( "Equating these two,\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \frac{2\pi}{h} + \frac{\pi}{r} = \frac{2\pi}{r} \Rightarrow \frac{2}{h} = \frac{1}{r} \Rightarrow h = 2r\">, finding the actual dimensions follows. &#8718;\r<BR>\n" );
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document.write( "There are probably other solutions involving AM-GM inequality or other techniques. The preferred solution to use is probably dependent on the course you are taking. </SPAN>\n" );
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