Question 1096893
Join the 3 sections together along their {{{ W }}} sides
Each section is an {{{ L }}} by {{{ W }}} rectangle
Adding up the side lengths, let {{{ K }}} = 
total length of fencing needed
{{{ K = 4W + 6L }}}
( note that 2 of the {{{W}}} sides are internal sides )
Let {{{ A }}} = total area
{{{ A = 3L*W }}}
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{{{ K = 4000 }}}
{{{ 4W + 6L = 4000 }}} ft
{{{ L = ( 4000 - 4W ) / 6 }}}
{{{ L = ( 2000 - 2W ) / 3 }}}
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Substituting:
{{{ A = ( 2000 - 2W )*W }}}
{{{ A = -2W^2 + 2000W }}}
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The W-value of {{{ A[max] }}} is:
{{{ W[max] = -b/(2a) }}}
{{{ a = -2 }}}
{{{ b = 2000 }}}
{{{ W[max] = -2000/(2*(-2)) }}}
{{{ W[max] = 500 }}}
Plug this result back into equation
{{{ A[max] = -2*500^2 + 2000*500 }}}
{{{ A[max] = -500000 + 1000000 }}}
{{{ A[max] = 500000 }}}
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The maximum area is 500,000 ft2
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check:
{{{ 3L = A/W }}}
{{{ 3L = 500000/500 }}}
{{{ 3L = 1000 }}}
and
{{{ K = 4W + 6L }}}
{{{ K = 4*500 + 6*(1000/3) }}}
{{{ K = 2000 + 2000 }}}
{{{ K = 4000 }}} ( given )
{{{ 4000 = 4000 }}}
OK
Check the math & get
another opinion if needed