Question 1021424
There are three borders using the fencing.
{{{2L+W=80}}}
The area is then,
{{{A=L*W}}}
Substituting from the perimeter,
{{{A=L(80-2L)}}}
It's a quadratic, convert to vertex form.
{{{A(L)=80L-2L^2}}}
{{{A(L)=-2L^2+80L}}}
{{{A(L)=-2(L^2-40L)}}}
{{{A(L)=-2(L^2-40L+400)+2*400}}}
{{{A(L)=-2(L-20)^2+800}}}
So the maximum occurs when {{{L=20}}}{{{m}}} and the value is {{{A(20)=800}}}{{{m^2}}}
Then,
{{{20*W=800}}}
{{{W=40}}}{{{m}}}