Question 1051143
.
Find the dimensions of the rectangular corral split into 2 pens of the same size producing the greatest possible 
enclosed area given 600 feet of fencing. (Assume that the length is greater than or equal to the width.)
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<pre><TABLE>
  <TR>
  <TD>
I prepared the Figure on the right to show how I see and understand &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
the condition.

In the Figure, L means the length and W means the width of each pen.

So, we have 3 pieces of fencing of the length L each and 4 pieces 
of fencing of the length W each.

Then we have this equation 

3L + 4W = 600,

from which we have W = {{{(600 - 3L)/4}}}.
 </TD>
  <TD>
{{{drawing( 280, 160,  0.5, 7.5, 0.5, 4.5, 
            line( 1.0, 1.0, 6.0, 1.0), 
            line( 1.0, 4.0, 6.0, 4.0),
            line( 1.0, 1.0, 1.0, 4.0),
            line( 6.0, 1.0, 6.0, 4.0),

            line( 1.0, 2.5, 6.0, 2.5),

            locate(3.4, 1.0, L),
            locate(3.4, 2.9, L),
            locate(3.4, 4.4, L),

            locate(0.6, 2.0, W),
            locate(0.6, 3.4, W),

            locate(6.2, 2.0, W),
            locate(6.2, 3.4, W),

            locate(2.7, 2.1, coral_2),
            locate(2.7, 3.5, coral_1)

)}}}


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<B>Figure</B>. 

 </TD>
 </TR>
</TABLE>Next, the combined area of the two corals is 

A = L*2W = {{{L*2(600 - 3L)/4}}} = {{{(1/2)*(600L - 3L^2)}}} = {{{-(3/2)L^2 + 300L}}},

and we have to find the length L in a way to maximize the area A, i.e. maximize the quadratic function

A = {{{-(3/2)L^2 + 300L)}}}.

    Now let me remind you that, if you have a quadratic function f(x) = {{{ax^2 + bx + c}}} of the general form, 
    then it reaches the maximum/minimum at x = {{{-b/2a}}}.

For our situation,  a = {{{-3/2}}}  and  b = 300.

Therefore, the maximum is at L = - {{{(-300)/(2*(3/2))}}} = {{{(300*2)/(2*3)}}} = 100.

Thus the area get a maximum at L = 100 feet.

Then W = {{{(600 - 3L)/4}}} = {{{(600 - 3*100)/4}}} = 75 feet.

<U>Answer</U>.  The area is maximal at L = 100 feet and W = 75 feet.
         Then the area of one coral is 100*75 = 7500 square feet.
         The combined area of the two corals is twice this value, i.e. 15000 square feet.
</pre>

<TABLE> 
  <TR>
  <TD>The plot below confirms this solution.


{{{graph( 330, 330, -50, 300, -11000, 20000,
          -(3/2)x^2 + 300x
)}}}


Plot f(L) = {{{-(3/2)L^2 + 300L)}}}

  </TD>
  </TR>
</TABLE>