document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1001517>Question 1001517</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Q: Find the area A of the largest rectangle with base on the x-asix that can be inscribed in the region R bounded above the by the growth of y = 9 -x^2 and below by the x-axis.\r<BR>\n" );
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document.write( "A:I know that y = -x^2 + 9 is an inverted parabola that is shifted upwards 9 units because + 9 and hase x points on -3 and +3.\r<BR>\n" );
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document.write( "I know that the area of a rectangle is A=L*W or in this case I made it A=x*y for the x and y axis. So the x base stretches from both the [-3,3] and the y goes from [0,9] what I don't get is why is the Area set up like A=2x*y because I know we have the top and bottom part of the rectangle for X so 2X and the left and right side of the rectangle 2Y so shouldn't we take into account that there are two sides of Y representing the left and right side of the rectangle?\r<BR>\n" );
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document.write( "But it's not for some reason basic reason I am sure that I forgot. I think I might be confusing permiter or something. I get that confused a lot.\r<BR>\n" );
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document.write( "I think the answer goes:<BR>\n" );
document.write( "A = 2x*y<BR>\n" );
document.write( "A = 2x*(-x^2+9)<BR>\n" );
document.write( "A = -2x^2 + 18x<BR>\n" );
document.write( "then optimizing it<BR>\n" );
document.write( "A' = -4x + 18<BR>\n" );
document.write( "find the critical value of x<BR>\n" );
document.write( "x=-/+&#8730;(9/2) but since we are talking about distance it's always the pos end.<BR>\n" );
document.write( "so,<BR>\n" );
document.write( "x=+&#8730;(9/2)<BR>\n" );
document.write( "then plugging this back into my base form before the derivative:<BR>\n" );
document.write( "A =  - 2(&#8730;(9/2)^2 + 18(&#8730;(9/2)\r<BR>\n" );
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document.write( "Is this correct? what is my issue with the Area formula?\r<BR>\n" );
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document.write( "Thank you </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #618656 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><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=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=618656\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> your steps are:\r<BR>\n" );
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document.write( "I think the answer goes:<BR>\n" );
document.write( "A = 2x*y<BR>\n" );
document.write( "A = 2x*(-x^2+9)<BR>\n" );
document.write( "A = -2x^2 + 18x ****************************************\r<BR>\n" );
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document.write( "A = -2x^2 + 18x is where you went wrong.<BR>\n" );
document.write( "2x * (-x^2 + 9) is equal to -2x^3 + 18x.<BR>\n" );
document.write( "find the derivative of that and you get -6x^2 + 18.<BR>\n" );
document.write( "set that equal to 0 and you get -6x^2 + 18 = 0<BR>\n" );
document.write( "add 6x^2 to both sides of that and you get 6x^2 = 18<BR>\n" );
document.write( "divide both sides of that by 6 and you get x^2 = 3<BR>\n" );
document.write( "take the square root of that and you get x = plus or minus sqrt(3).\r<BR>\n" );
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document.write( "the value of x is plus or minus sqrt(3).\r<BR>\n" );
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document.write( "the height of your rectangle is y = 9-x^2<BR>\n" );
document.write( "the width of your rectangle is 2x.<BR>\n" );
document.write( "the maximum area is when x = plus or minus sqrt(3).<BR>\n" );
document.write( "when x = sqrt(3), your area is (9 - sqrt(3)^2) * 2 * sqrt(3)) which is equal to 20.78460969.\r<BR>\n" );
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document.write( "you can also find this graphically by graphing y = (9 - x^2) * 2 * x.\r<BR>\n" );
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document.write( "the graph of that equation is shown below:\r<BR>\n" );
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document.write( "that's the graph of the area of your rectangle and it shows that the maximum area is when x = 1.732 which is the rounded version of sqrt(3).\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" src = \"http://theo.x10hosting.com/2015/110503.jpg\" alt=\"$$$\" </>\r<BR>\n" );
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