Question 926521
Let {{{ s }}} = the length of a side
The length of the other side is {{{ ( 200 - 2s ) / 2  = 100 - s }}}
The area is:
{{{ A = s*( 100 - s ) }}}
{{{ A = -s^2 + 100s }}}
The maximum is at:
{{{ s[max] = -b/(2a) }}}
{{{ a = -1 }}}
{{{ b = 100 }}}
{{{ -b/(2a) = -100 / ( -2) }}}
{{{ -b/(2a) = 50 }}}
{{{ s[max] = 50 }}}
-----------------
{{{ 4*50 = 200 }}}, so the yard must be a perfect square.
The maximum area is:
{{{ A[max] = 50^2 }}}
{{{ A[max] = 2500 }}}
-------------------
You can check this by making {{{ s = 49 }}}, find {{{ A }}}
then make {{{ s = 51 }}} find {{{ A }}} ( very good check )