Question 660402
A rectangular enclosure, fenced along 4 sides, and divided into three parts by two fences parallel to one of the sides, would look like this:
{{{drawing(300,150,-1,7,-1,3,
rectangle(0,0,6,2),
line(2,0,2,2),line(4,0,4,2),
locate(6.1,1.3,x), locate(2.9,-0.1,y),
arrow(2.8,-0.3,0,-0.3),arrow(3.2,-0.3,6,-0.3)
)}}}
If the length of each of the four parallel fence sections is {{{x}}} yards,
and the length of the other sections of fencing is {{{y}}} yards,
the area, {{{A}}}, of the whole enclosure, in square yards, is
{{{A=x*y}}}, which is supposed to equal 9200 yards.
The total length of the fencing, in yards, is
{{{4x+2y=800}}} --> {{{2x+y=400}}} --> {{{y=400-2x}}}
 
Substituting that expression for {{{y}}}, the area of the enclosure, in square yards, is
{{{A=x(400-2x)}}} --> {{{A=400x-2x^2}}}, which is supposed to equal 9200 yards.
That gives us the quadratic equation {{{400x-2x^2=9200}}}.
 
With a little algebra, we can rearrange it into a standard form:
{{{400x-2x^2=9200}}} --> {{{-2x^2+400x-9200=0}}},
and dividing everything by {{{-2}}} we get
{{{highlight(x^2-200x+4600=0)}}}, which I like even better.
 
Using the quadratic formula, {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
we solve that equation.
{{{x = (-(-200) +- sqrt( (-200)^2-4*1*4600 ))/(2*1) }}}
{{{x = (200 +- sqrt( 40000-18400 ))/2 }}}
{{{x = (200 +- sqrt(21600 ))/2 }}}
{{{sqrt(216)}}} = about {{{147}}} (rounding)
So a good approximation of the two solutions would be
{{{x=(200 +- 147)/2 }}}
 
One solution is {{{x=(200+147)/2}}} --> {{{x=347/2}}} --> {{{highlight(x=173.5)}}} ,
which substituted into {{{y=400-2x}}}
gives {{{y=400-2*173.5}}} --> {{{y=400-347}}} --> {{{highlight(y=53)}}}
 
The other solution is {{{x=(200-147)/2}}} --> {{{x=53/2}}} --> {{{highlight(x=26.5)}}} ,
which substituted into {{{y=400-2x}}}
gives {{{y=400-2*26.5}}} --> {{{y=400-53}}} --> {{{highlight(y=347)}}}