document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=445838>Question 445838</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A right-circular cylinder has a radius r cm and a height h cm.  Given that h + r = 30, find the maximum volume of the cylinder.\r<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 #307002 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=307002\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> We know that, from the constraint, h = 30 - r. Hence, the volume V is\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE V = (\pi r^2)(30 - r) = -\pi r^3 + 30\pi r^2\"> If we suppose V is a function of r, we can take the derivative of V 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{dV}{dr} = -3\pi r^2 + 60 \pi r\">\r<BR>\n" );
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document.write( "The derivative is zero when r = 0 or r = 20. Clearly, r = 0 would imply V = 0. It can be checked that r = 20, h = 10 maximizes the volume, which is\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE V = (20^2\pi)(10) = 4000\pi\"> cm^3. </SPAN>\n" );
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