document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1189873>Question 1189873</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 1. Richard has available 400 yards of fencing and wishes to enclose a rectangular area.\r<BR>\n" );
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document.write( "a.) Express the area A of the rectangle as a function of the width x of the rectangle?\r<BR>\n" );
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document.write( "b.) What is the domain of A? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #821370 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=821370\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> a = l * w<BR>\n" );
document.write( "p = 2 * (l + w)\r<BR>\n" );
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document.write( "a is the area<BR>\n" );
document.write( "l is the length<BR>\n" );
document.write( "w is the width<BR>\n" );
document.write( "p is the perimeter\r<BR>\n" );
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document.write( "when p = 400, p = 2 * (l + w) becomes 400 = 2 * (l + w)<BR>\n" );
document.write( "divide both sides of this equation by 2 to get:<BR>\n" );
document.write( "200 = l + w<BR>\n" );
document.write( "solve for l to get:<BR>\n" );
document.write( "l = 200 - w\r<BR>\n" );
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document.write( "in the formula for area:<BR>\n" );
document.write( "when l = 200 - 2, a = l * w becomes:<BR>\n" );
document.write( "a = (200 - w) * w.<BR>\n" );
document.write( "simplify to get:<BR>\n" );
document.write( "a = 200 * w - w^2.\r<BR>\n" );
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document.write( "a must be greater than or equal to 0 (can't be negative).<BR>\n" );
document.write( "if a >= 0, then:<BR>\n" );
document.write( "200 * w - w^2 >= 0<BR>\n" );
document.write( "add w^2 to both sides of this equation to get:<BR>\n" );
document.write( "200 * w >= w^2<BR>\n" );
document.write( "divide both sides of this equation by w to get:<BR>\n" );
document.write( "200 >= w\r<BR>\n" );
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document.write( "if 200 >= w, then:<BR>\n" );
document.write( "w <= 200<BR>\n" );
document.write( "w must  also be >= 0, therefore:<BR>\n" );
document.write( "0 <= w <= 200\r<BR>\n" );
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document.write( "that's your domain.<BR>\n" );
document.write( "the area is your range.\r<BR>\n" );
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document.write( "the domain is the width which is greater than or equal to 0 and less than or equal to 200.<BR>\n" );
document.write( "that says the domain of the area is 0 <= width <= 200.<BR>\n" );
document.write( "that should be your solution.\r<BR>\n" );
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document.write( "the equation of area = 200 * width minus width squared can be graph by letting y = the area and x = the width.<BR>\n" );
document.write( "the formula for graphing becomes y = 200x - x^2.\r<BR>\n" );
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document.write( "the graph looks like this:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" src = \"http://theo.x10hosting.com/2022/012101.jpg\" >\r<BR>\n" );
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document.write( "you can see from the grpah that the area, represented by y, is not negative when 0 <= x <= 200, with x representing the width.\r<BR>\n" );
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document.write( "the coordinate points on the graph are in (x,y) format.<BR>\n" );
document.write( "x represents the width of the rectangle.<BR>\n" );
document.write( "y represents the area of the rectangle.<BR>\n" );
document.write( "at (100,10,000) the width is 100 and the area is 10,000.<BR>\n" );
document.write( "you don't see the length on the graph, but you can always find the length, because length + width will always be 200.<BR>\n" );
document.write( "therefore, when the width is 100, the length has to be 100.<BR>\n" );
document.write( "the area is equal to length * width = 100 * 100 = 10,000, as shown on the graph.<BR>\n" );
document.write( " </SPAN>\n" );
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