Question 1097035
Let the length along be irrigation side by Y and the width be X.
The fence would then have a perimeter of {{{P=2X+Y=80}}}
The area enclosed would be {{{A=XY}}}
From the perimeter equation,
{{{Y=80-2X}}}
Substituting into the area equation,
{{{A=X(80-2X)=-2X^2+80X}}}
To find the maximum area, convert to vertex form,
{{{A(X)=-2(X^2-40X)}}}
{{{A(X)=-2(X^2-40X+400)+800}}}
{{{A(X)=-2(X-20)^2+800}}}
So the maximum area of {{{800}}}{{{m^2}}} occurs when {{{X=20}}}{{{m}}},
So then,
{{{Y=80-2(20)}}}
{{{Y=80-40}}}
{{{Y=40}}}{{{m}}}