document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=565891>Question 565891</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A 24-inch-wide sheet of metal is to be bent into a rectangular trough with the cross section shown in the illustration. Find the dimensions that will maximize the amount of water the trough can hold. That is, find the dimensions that will maximize the cross-sectional area.(depth and width = 24 inches) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #366003 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=366003\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font face=\"Times New Roman\" size=\"+2\">\r<BR>\n" );
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document.write( "Before I even start, I want you to go back and read the question as you posted it.  \"...shown in the illustration.\"  Right.  This is Algebra.com -- NOT the Psychic Hot Line.  First lesson:  Mathematics requires using your head for something besides a hat rack.\r<BR>\n" );
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document.write( "Be that as it may, I'm going to presume that you want to fold up two sides making an unknown depth (equal on both sides) on a trough with no top and a bottom that perforce measures 24 inches minus 2 times the depth.  To the extent that my assumptions about the true nature of your problem are correct, begin with the concept that the area of a rectangle is given by length (or depth in this case) times width. \r<BR>\n" );
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document.write( "For this problem we let <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large x\"> represent the amount folded up or the depth of the trough, and then the width of the rectangle (the bottom of the trough must be <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large 24\ -\ 2x\">.  Multiplying length times width gives area as a function of the depth:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ A(x)\ =\ 24x\ -\ 2x^2\">\r<BR>\n" );
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document.write( "Written in standard <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large \left(\rho(x)\ =\ ax^2\ +\ bx\ + c\right)\"> form:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ A(x)\ =\ -2x^2\ +\ 24x\">\r<BR>\n" );
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document.write( "<i><b>Algebra Solution</b></i>  Recognize that this is a quadratic polynomial function that will graph to a parabola.  Since the lead coefficient is negative, the parabola must open downward, hence the vertex is a maximum.  Using <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large x_v\ =\ \frac{-b}{2a}\ =\ \frac{-24}{2(-2)}\ =\ 6\">.  Hence, fold up 6 inches to get the maximum area, giving you a 6 by 12 (24 - 12) rectangle.\r<BR>\n" );
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document.write( "<i><b>Calculus Solution</b></i>  Take the first derivative:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ \frac{dA}{dx}\ =\ -4x\ +\ 24\">\r<BR>\n" );
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document.write( "Set the first derivative equal to zero and solve:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ -4x\ +\ 24\ =\ 0\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 6\">\r<BR>\n" );
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document.write( "Hence extreme point at <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large x\ =\ 6\">\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 \ \ \ \ \ \ \ \ \ \ \frac{d^2A}{dx^2}\ =\ -4\ <\ 0\ \forall\ x\"> in the domain of A.\r<BR>\n" );
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document.write( "Hence the extreme point is a maximum.\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" );
document.write( "My calculator said it, I believe it, that settles it<BR>\n" );
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