document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=985275>Question 985275</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The Equation of a Parabolic Curve is given by y = -x^2 + 12. What is the Maximum Area of a Rectangle inscribed in the Parabola?. I am asked to provide a Geometrical solution or Algebraic Solution. I am working for hours to find a Solution using coordinate geometry but would appreciate it if I can get help.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #606104 by </B><A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><B>jim_thompson5910(35256)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=606104\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> I'm assuming this rectangle has one edge on the x axis. Let P be one point of the rectangle. Also, let P be on the positive x axis to the left of the root of the given parabola.\r<BR>\n" );
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document.write( "The distance from (0,0) to P is some unknown quantity x. Due to symmetry, the edge of the rectangle that rests on the x axis is going to be 2*x units long.\r<BR>\n" );
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document.write( "The height of this rectangle will be -x^2+12 because the upper corner point is on the parabola (draw a picture and this should be clear hopefully).\r<BR>\n" );
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document.write( "So we have length = 2x and width = -x^2 + 12\r<BR>\n" );
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document.write( "The area of this rectangle is...\r<BR>\n" );
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document.write( "A = L*W<BR>\n" );
document.write( "A = (2x)*(-x^2 + 12)<BR>\n" );
document.write( "A = -2x^3 + 24x\r<BR>\n" );
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document.write( "There are two ways to find the max area\r<BR>\n" );
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document.write( "1) Use a graphing calculator to find the peak max point on y = -2x^3 + 24x. This is like the vertex of a parabola, but not entirely. I like to think of it like it though. If you have a TI calculator, you can use the min/max feature to find the peak point.\r<BR>\n" );
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document.write( "2) Use calculus to find the max without using a graphing calculator. This is only if you're in a calculus class.\r<BR>\n" );
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document.write( "Using either method (I recommend method #1 since it sounds like you're in a high school algebra class), you should find that the max occurs at the point (2,32). \r<BR>\n" );
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document.write( "So if x = 2, the length is 2*x = 2*2 = 4 and the width is -x^2 + 12 = -(2)^2 + 12 = 8. Notice how L*W = 4*8 = 32\r<BR>\n" );
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document.write( "The max area possible is 32 square units (length = 4, width = 8) </SPAN>\n" );
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