document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=109832>Question 109832</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A developer wants to enclose a rectangular lot that borders a city street. If the developer has 248 ft. of fence and does not fence the side bordering the street, what is the largest area that can be enclosed. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #80028 by </B><A HREF=/tutors/aboutme.mpl?userid=bucky><B>bucky(2189)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=bucky><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=80028\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let's begin by picturing what the developer will do. Taking his roll of fencing, he will<BR>\n" );
document.write( "start at the edge of the street and will go perpendicular to the street for some unknown distance.<BR>\n" );
document.write( "Call that distance x. Then he will turn the fence line 90 degrees and go parallel to the<BR>\n" );
document.write( "street some unknown distance. Call that distance y. The he will turn the fence line by<BR>\n" );
document.write( "90 degrees again and will head back to the street. If you are picturing this correctly<BR>\n" );
document.write( "you will see that the distance to return to the edge of the street must again be x, the same<BR>\n" );
document.write( "distance as he originally went away from the street.<BR>\n" );
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document.write( "Now we can see that in using up all 248 feet of fencing, he goes a distance x away from the<BR>\n" );
document.write( "street, a distance y parallel to the street, and a distance x back to the street. <BR>\n" );
document.write( "We can write this is equation form as:<BR>\n" );
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document.write( "x + y + x = 248<BR>\n" );
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document.write( "and combine the two x terms to reduce it to:<BR>\n" );
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document.write( "2x + y = 248<BR>\n" );
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document.write( "Then we can solve for y in terms of x by subtracting 2x from both sides of this equation<BR>\n" );
document.write( "to get:<BR>\n" );
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document.write( "y = 248 - 2x<BR>\n" );
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document.write( "So in terms of x we now have that one side of the rectangle is x and the other dimension<BR>\n" );
document.write( "of the rectangle is 248 - 2x.<BR>\n" );
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document.write( "The area (A) of this rectangle is just the product of these two dimensions. (Area of a <BR>\n" );
document.write( "rectangle is the product of length times width).<BR>\n" );
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document.write( "So for this rectangle we can write the Area equation as:<BR>\n" );
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document.write( "A = x(248 - 2x)<BR>\n" );
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document.write( "At this point we can do some \"thought analysis.\"  If you multiply out these two terms you<BR>\n" );
document.write( "will get the quadratic equation A = 248x - 2x^2. The graph of a quadratic equation <BR>\n" );
document.write( "is a parabola, and in this case the minus sign on the x^2 term tells you that as you move<BR>\n" );
document.write( "from left to right on the graph, the parabola rises to a peak and then drops down the<BR>\n" );
document.write( "further you go ... similar to the arc traced by a football pass.<BR>\n" );
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document.write( "Since we are graphing the Area (vertical axis) against x (horizontal axis) we can tell<BR>\n" );
document.write( "that at some point on the graph it comes from below the x-axis, crosses it and goes positive.<BR>\n" );
document.write( "How can we tell? Easy. The area has to be a positive value, so the graph must be above the<BR>\n" );
document.write( "x-axis somewhere. The area rises to a peak and then starts to drop off as x keeps increasing.<BR>\n" );
document.write( "At some larger value of x the graph again crosses the x-axis on its way down.<BR>\n" );
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document.write( "Where the graph crosses the x-axis the value of the area (A) must be zero. Therefore,<BR>\n" );
document.write( "our equation becomes:<BR>\n" );
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document.write( "0 = x(248 - 2x)<BR>\n" );
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document.write( "Notice that this equation will be true if either factor equals zero. So our graph crosses<BR>\n" );
document.write( "the x-axis when x = 0 (on its way up) and again when (248 - 2x) = 0 (on its way down).<BR>\n" );
document.write( "Solving the second factor for x :<BR>\n" );
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document.write( "248 - 2x = 0<BR>\n" );
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document.write( "Subtract 248 from both sides:<BR>\n" );
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document.write( "-2x = -248<BR>\n" );
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document.write( "Then divide both sides by -2 to get:<BR>\n" );
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document.write( "x = -248/-2 = 124<BR>\n" );
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document.write( "So the graph crosses the x-axis when x = 0 and when x = 124. Because the parabola<BR>\n" );
document.write( "is symmetrical the peak will be reached halfway between these two values, and the midway<BR>\n" );
document.write( "value of x between x = 0 and x = 124 occurs when x = 124/2 = 62 feet.<BR>\n" );
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document.write( "So the maximum area will occur when x (the distance from the edge of the street to the<BR>\n" );
document.write( "first corner in the fence) is 62 feet. Going back to our original \"picture\" of the fence,<BR>\n" );
document.write( "we can get that the length of the fence (248 feet) is now comprised of 62 feet perpendicular<BR>\n" );
document.write( "to the street, then y feet parallel to the street, and finally another 62 feet to get<BR>\n" );
document.write( "back to the street edge. So y must be 248 minus the two x-sides or 248 - 2(62) which turns<BR>\n" );
document.write( "out to be 124 feet. The area enclosed by the fence will be 62 times 124 which is 7688 sq ft.<BR>\n" );
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document.write( "You can check this out by supposing that x was 61 feet. If you work it out, the length of<BR>\n" );
document.write( "the fence parallel to the street will be 126 feet, and the area enclosed will therefore<BR>\n" );
document.write( "be 61 times 126 or 7686 square feet ... a little smaller than the area if x is 62 feet.<BR>\n" );
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document.write( "Similarly, if x is 63 feet, then the length of the fence parallel to the street will work<BR>\n" );
document.write( "out to be 122 feet, and the area enclosed will be 63 times 122 which will again result in 7686 <BR>\n" );
document.write( "square feet. Again this area is slightly smaller than the area we got when x was 62 feet.  <BR>\n" );
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document.write( "Now that we have done this analysis (mainly to get a feel for what is going on here) we<BR>\n" );
document.write( "can return to the equation:<BR>\n" );
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document.write( "A = 248x - 2x^2<BR>\n" );
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document.write( "Set the area equal to zero and rearrange this equation into the quadratic form of:<BR>\n" );
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document.write( "-2x^2 + 248x = 0<BR>\n" );
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document.write( "From the quadratic formula we know that the first term for the answer to x is:<BR>\n" );
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document.write( "-b/(2*a) <BR>\n" );
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document.write( "In this standard form we can see a = -2 and b = 248. Plugging these values into the term<BR>\n" );
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document.write( "-b/(2*a)<BR>\n" );
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document.write( "results in:<BR>\n" );
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document.write( "-(248)/(2*-2) = -248/-4 = 62<BR>\n" );
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document.write( "So using our quadratic equation and -b/(2*a) also gives us the answer 62 feet. Just another<BR>\n" );
document.write( "way of doing the problem.<BR>\n" );
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document.write( "In summary, the maximum area that can be enclosed is 7688 sq ft.<BR>\n" );
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document.write( "Hope all this discussion isn't too confusing and gives you some sense of the problem and <BR>\n" );
document.write( "how to work it.<BR>\n" );
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document.write( " </SPAN>\n" );
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