document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=972903>Question 972903</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A culvert designed with a semicircular cross-section of diameter d is to be<BR>\n" );
document.write( "redesigned to have an isosceles trapezoidal cross-section (angled sides are<BR>\n" );
document.write( "of equal length) by inscribing the trapezoid in the semi circle.\r<BR>\n" );
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document.write( "What is the length of the bottom base x of the trapezoid if its area is to be<BR>\n" );
document.write( "a maximum? What is the maximum area of the trapezoidal cross-section in<BR>\n" );
document.write( "terms of d?<BR>\n" );
document.write( "Hint: First find an expression for the area of cross section in terms of x<BR>\n" );
document.write( "and d.\r<BR>\n" );
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document.write( "Amy. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #595150 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=595150\">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( "First draw the picture.  Bottom half of a circle with center O.  Sketch in the trapezoid.  Points of intersection of the bottom base of the trapezoid and the semi-circle are A and B.  Drop a vertical radius from O that bisects the bottom base.  Point of intersection of this radius and the bottom base is C.  Construct radius OA.\r<BR>\n" );
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document.write( "Note the right triangle OAC.  The hypotenuse is the radius OA which measures <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large \frac{d}{2}\">.  Since OC bisects AB, the short leg of the triangle measures <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large \frac{x}{2}\">.  Then Mr. Pythagoras assures us that the other leg of the triangle, which is the height of the trapezoid, measures <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large \frac{\sqrt{d^2\ -\ x^2}}{2}\">.\r<BR>\n" );
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document.write( "Since the area of a trapezoid is the average of the two bases times the height, we can hold <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large d\"> constant and write the function for the area of the trapezoid in terms of <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large x\"> thus:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ A(x)\ =\ \left(\frac{d\ +\ x}{2}\right)\left(\frac{\sqrt{d^2\ -\ x^2}}{2}\right)\">\r<BR>\n" );
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document.write( "Taking the first derivative (use the product rule and a bunch of ugly algebra that I will leave as an exercise for the student)\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ A'(x)\ =\ \frac{2x^2\ +\ dx\ +\ d^2}{4\sqrt{d^2\ -\ x^2}}\">\r<BR>\n" );
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document.write( "Set the first derivative equal to zero and solve for <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large x\"> in terms of <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large d\">, we get (after multiplying through by the denominator)\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ 2x^2\ +\ dx\ -\ d^2\ =\ 0\">\r<BR>\n" );
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document.write( "Then a little quadratic formula work the details of which you can work out, and then discarding the absurd negative root, we end up with:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{d}{2}\">\r<BR>\n" );
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document.write( "Then plug this value for <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large x\"> back into the area function that we previously derived and simplify.  Again, I leave the details to you, but you should end up with:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ A_{max}\ =\ \frac{3d^2\sqrt{3}}{16}\">\r<BR>\n" );
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document.write( "Or not quite 32 and a half square feet if you started with a 10 foot diameter culvert.\r<BR>\n" );
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document.write( "Piece of cake; easy as pie.  Or is that piece of pie, easy as cake...I can never remember.\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\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large \ \\r\"> </SPAN>\n" );
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