document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=199523>Question 199523</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Canadian Postal Service regulations require that the sum of three dimensions of a rectangular package not exceed 3 m.  What are the dimensions of the largest rectangular box with square ends that can be mailed? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #149918 by </B><A HREF=/tutors/aboutme.mpl?userid=solver91311><B>solver91311(24713)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=solver91311><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=solver91311><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=149918\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font face=\"Garamond\" size=\"+2\">\r<BR>\n" );
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document.write( "I have to presume that you are a calculus student and are familiar with the process of finding the first and second derivatives of a polynomial function because that is the only way I know to solve this problem.\r<BR>\n" );
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document.write( "Another assumption that I have to make is that by \"largest rectangular box\" you mean the \"rectangular box with the largest volume\".\r<BR>\n" );
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document.write( "Given all of that:\r<BR>\n" );
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document.write( "The restriction is that the sum of the dimensions be not exceed 3 m and that at least one pair of opposite faces of the rectangular box are squares.\r<BR>\n" );
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document.write( "Let <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large x\"> represent the measure of the side of one of the square faces.  Then we can say that the remaining dimension is <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large 3 - 2x\">.\r<BR>\n" );
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document.write( "The volume is the product of the three dimensions so the volume function with respect to the dimension of the square end is:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ V(x) = x^2(3 - 2x) = -2x^3 + 3x^2\">\r<BR>\n" );
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document.write( "The feasible domain for <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large x\"> in this situation is <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large (0,1.5)\"> because if <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large x = 0\"> then you obviously have a zero volume box and if <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large x = 1.5\"> then the length must be zero because 2 times 1.5 is 3, and again you have a zero volume box.\r<BR>\n" );
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document.write( "We are interested in finding a local maximum of the volume function on the interval (0,1.5).\r<BR>\n" );
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document.write( "Take the first derivative:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ V'(x) = -6x^2 + 6x\">\r<BR>\n" );
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document.write( "Set the first derivative equal to zero and solve (Fermat's Theorem)\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ -6x^2 + 6x = 0\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ -6x(x - 1) = 0\">\r<BR>\n" );
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document.write( "Hence\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ x = 0\">\r<BR>\n" );
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document.write( "which must be excluded because it is not in the feasible interval, or\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ x = 1\">\r<BR>\n" );
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document.write( "Now that we know that there is a critical point at <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large x = 1\"> we need to apply the second derivative test to determine if it is a maximum, minimum, or a possible inflection point.\r<BR>\n" );
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document.write( "Take the second derivative:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ V''(x) = -12x + 6\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ V''(1) = -12(1) + 6 < 0\">\r<BR>\n" );
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document.write( "Therefore V(1) is a local maximum.  That means that the measure of each side of the square end of the box must be 1, and since 3 - 2 = 1, the measure of the length must be 1 as well.  The maximum volume rectangular box for a given sum of dimensions is a cube.\r<BR>\n" );
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document.write( "John<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE e^{i\pi} + 1 = 0\"><BR>\n" );
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