Question 1163322
<pre>{{{drawing(400,5600/19,-4,34,-4,24,

locate(7,0,u),locate(22,0,u), 
locate(7,21.57,u),locate(22,21.57,u),
locate(.5,11,x),locate(15.5,11,x),locate(30.5,11,x),


line(0,0,30,0), line(30,0,30,20), line(30,20,0,20), line(0,0,0,20), line(0,0,30,0), line(15,0,15,20) )}}}


{{{Let}}}{{{y}}}{{{""=""}}}{{{area}}}{{{""=""}}}{{{length*width}}}{{{"=""}}}{{{(2u)(x)}}}{{{""=""}}}{{{2ux}}}

{{{fencing}}}{{{""=""}}}{{{3x+4u}}}{{{""=""}}}{{{120}}}
                          {{{4u}}}{{{"=""}}}{{{120-3x}}}
                           {{{u}}}{{{"=""}}}{{{120/4-3x/4}}}
                           {{{u}}}{{{"=""}}}{{{30-expr(3/4)x}}}

                          {{{y}}}{{{""=""}}}{{{2ux}}}
                          {{{y}}}{{{""=""}}}{{{2(30-expr(3/4)x)x)}}}
                         {{{y}}}{{{""=""}}}{{{2x(30-expr(3/4)x)}}}
                         {{{y}}}{{{""=""}}}{{{60x-expr(3/2)x}}}
                         {{{y}}}{{{""=""}}}{{{-expr(3/2)x+60x}}}

The graph of the area with respect to the width x has this graph,
which is a parabola:

{{{graph(200,200,-.5,41,-1,610,(-3/2)x^2+60x))}}}

The maximum value is the y-value at the vertex. We find the vertex of a
parabola by using the vertex formula for the x-coordinate of the vertex:

{{{-b/2a}}}{{{""=""}}}{{{(-60^"")/(2(-3/2))}}}{{{(-60)/(-3)}}}{{{""=""}}}{{{20}}}

So the maximum area is when the width x is 20 meters and the length is 2u

                           {{{u}}}{{{"=""}}}{{{30-expr(3/4)x}}}
                           {{{u}}}{{{"=""}}}{{{30-expr(3/4)(20)}}}
                           {{{u}}}{{{"=""}}}{{{30-15}}}
                           {{{u}}}{{{"=""}}}{{{15}}}{{{meters}}}
                          {{{2u}}}{{{"=""}}}{{{30}}}{{{meters}}}

The two-pen corral has maximum area when its width is 20 meters and its 
length is 30 meters. [Incidentally that maximum area is the y-coordinate
of the vertex, which you can get by substitution and finding y. It's 600 m². 

Edwin</pre>