Question 810649
Rectangle dimensions would be like, x and y.  One more term of either x or y occurs as the divider fence length.  You are not clear about running this subdividor accross the middle.  Diagonally, or perpendicularly?  


I am choosing the diagonal, because now no need to look at choice of either x or y.  The total fence length using variables would be {{{2x+2y+diagonal=3000}}}.


That diagonal is {{{x^2+y^2=d^2}}}
{{{d=sqrt(x^2+y^2)}}}


So fencing length is {{{2x+2y+sqrt(x^2+y^2)=3000}}} and the area is {{{xy=A}}}


Solve the fence length equation for either x or y.
sqrt(x^2+y^2)=3000-2x-2y
x^2+y^2=(3000-2(x+y))^2
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Excuse me.  This is easier to do on paper than in text form typed into a website page.  


The plan is solve the fencing equation for either x or y, substitute this into the Area equation or function, and then find the derivative; and set equal to zero and solve for the variable.  Then calculate the other variable.