document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1183556>Question 1183556</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The line x/3 + y/4 = 1 forms a triangle with the x and y axes in the first quadrant. Lines passing through the point (1,1), which is inside <BR>\n" );
document.write( "this triangle, are free to pivot around it inside this triangle up to the x- and y-intercepts of the line x/3 + y/4 = 1.  What is the maximum or <BR>\n" );
document.write( "minimum area of the region that can be enclosed by the line  x/3 + y/4 = 1, any of the pivoting lines around the point (1,1), and the x or y axis?\r<BR>\n" );
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document.write( "Thank you in advance.\r<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #814016 by </B><A HREF=/tutors/aboutme.mpl?userid=robertb><B>robertb(5830)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=robertb><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=robertb><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=814016\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The line passing through the point (1,1) with slope <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=s ALIGN=MIDDLE  ALT=\"s\"   DISABLED_style= \"max-width:90%; height:auto;\" > is represnted by the equation \r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y-1=s%28x-1%29 ALIGN=MIDDLE  ALT=\"y-1=s%28x-1%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >.\r<BR>\n" );
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document.write( "the x-intercept of the line is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x+=+1-1%2Fs ALIGN=MIDDLE  ALT=\"x+=+1-1%2Fs\"   DISABLED_style= \"max-width:90%; height:auto;\" >. (Obtained by letting y=0 and solving for x.)   Based on the specifics of the problem, <BR>\n" );
document.write( "the range of values of <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=s ALIGN=MIDDLE  ALT=\"s\"   DISABLED_style= \"max-width:90%; height:auto;\" > is the closed interval [-3, -1/2].\r<BR>\n" );
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document.write( "The area of the region enclosed by the line x/3 + y/4 = 1, any of the pivoting lines around the point (1,1), and the x and/or y axis is then given by \r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A+=+int%28%284-%284%2F3%29%2Ax%29+-+%28s%28x-1%29%2B1%29%2C+dx%2C+0%2C+1-1%2Fs%29+%2B+int%284-%284%2F3%29%2Ax%2C+dx%2C+1-1%2Fs%2C+3%29+ ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "=<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28%283%2Bs%29x+-+%284%2F3+%2Bs%29%2A%28x%5E2%2F2%29%29%5B0%5D%5E%281-1%2Fs%29+%2B+%284%2A%28x-x%5E2%2F6%29%29%5B1-1%2Fs%5D%5E3 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "= <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%281%2F2%29%2A%28%28s-1%29%5E2%2Fs%29%2B6 ALIGN=MIDDLE  ALT=\"%281%2F2%29%2A%28%28s-1%29%5E2%2Fs%29%2B6\"   DISABLED_style= \"max-width:90%; height:auto;\" >  ===> <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A+=+%281%2F2%29%2A%28%28s-1%29%5E2%2Fs%29%2B6 ALIGN=MIDDLE  ALT=\"A+=+%281%2F2%29%2A%28%28s-1%29%5E2%2Fs%29%2B6\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "we now look for the extrema of this area function.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=dA%2Fds+=+%28s%5E2+-+1%29%2F%282s%5E2%29+=+0 ALIGN=MIDDLE  ALT=\"dA%2Fds+=+%28s%5E2+-+1%29%2F%282s%5E2%29+=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" > ===> <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=s+=+%2B-+1 ALIGN=MIDDLE  ALT=\"s+=+%2B-+1\"   DISABLED_style= \"max-width:90%; height:auto;\" >.\r<BR>\n" );
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document.write( "Choose s = -1 because 1 is not in the range of values of s, i.e., [-3, -1/2].\r<BR>\n" );
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document.write( "If <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-3+%3C=+s+%3C+-1 ALIGN=MIDDLE  ALT=\"-3+%3C=+s+%3C+-1\"   DISABLED_style= \"max-width:90%; height:auto;\" >, <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=dA%2Fds+%3E+0 ALIGN=MIDDLE  ALT=\"dA%2Fds+%3E+0\"   DISABLED_style= \"max-width:90%; height:auto;\" >;\r<BR>\n" );
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document.write( "If <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-1+%3C+s+%3C=+-1%2F2 ALIGN=MIDDLE  ALT=\"-1+%3C+s+%3C=+-1%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" >, <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=dA%2Fds+%3C+0 ALIGN=MIDDLE  ALT=\"dA%2Fds+%3C+0\"   DISABLED_style= \"max-width:90%; height:auto;\" >, hence by the first derivative test \r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A ALIGN=MIDDLE  ALT=\"A\"   DISABLED_style= \"max-width:90%; height:auto;\" > has a local maximum at <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=s+=+-1 ALIGN=MIDDLE  ALT=\"s+=+-1\"   DISABLED_style= \"max-width:90%; height:auto;\" >.  \r<BR>\n" );
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document.write( "The absolute extrema are obtained by comparing the function value at s = -1 with those of the function at s = -1/2 and s = -3 <BR>\n" );
document.write( "(the endpoints of the closed interval [-3, -1/2]).\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A%28-1%29+=+%281%2F2%29%2A%28%28-2%29%5E2%2F%28-1%29%29+%2B+6+=+highlight%284%29 ALIGN=MIDDLE  ALT=\"A%28-1%29+=+%281%2F2%29%2A%28%28-2%29%5E2%2F%28-1%29%29+%2B+6+=+highlight%284%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > <--- maximum area\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A%28-3%29+=+%281%2F2%29%2A%28%28-3-1%29%5E2%2F%28-3%29%29+%2B+6+=+-8%2F3+%2B+6+=+highlight%2810%2F3%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" > <---minimum area\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A%28-1%2F2%29+=+%281%2F2%29%2A%28%28-1%2F2+-1%29%5E2%2F%28-1%2F2%29%29+%2B+6+=+-9%2F4+%2B+6+=+15%2F4 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" >.\r<BR>\n" );
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document.write( "Therefore the max area occurs when s = -1, defined by the line <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y-1+=+-1%2A%28x-1%29 ALIGN=MIDDLE  ALT=\"y-1+=+-1%2A%28x-1%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >, having enclosed area of 4.  <BR>\n" );
document.write( "The min area occurs when s = -3, defined by the line <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y-1+=+-3%28x-1%29 ALIGN=MIDDLE  ALT=\"y-1+=+-3%28x-1%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >, having enclosed area of 10/3.\r<BR>\n" );
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