Question 668721
This is like fencing in a rectangle, then removing one side
Let {{{ A }}} = the area enclosed
Let {{{ y }}} = the side parallel to the street
Let {{{ x }}} = the length of a side perpendicular to the street
The amount of fencing is {{{ y + 2x }}}
(1) {{{ y + 2x = 328 }}}
The area enclosed is 
(2) {{{ A = x*y }}}
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(1) {{{ y = 328 - 2x }}}
By substitution:
(2) {{{ A = x*( 328 - 2x ) }}}
(2) {{{ A = -2x^2 + 328x }}}
Because of the minus sign in front of the squared term, the equation
has a maximum and not a minimum
(2) {{{ A = x*( -2x + 328 ) }}}
The maximum occurs halfway between the 2 roots
One root is at {{{ x = 0 }}}
and to find the other root,
{{{ -2x + 328 = 0 }}}
{{{ 2x = 328 }}}
{{{ x = 164 }}}
Half way between is {{{ x = 82 }}}
and, {{{ y = 328 - 2x }}}
{{{ y = 328 - 2*82 }}}
{{{ y = 328 - 164 }}}
{{{ y = 164 }}}
and, at the maximum,
(2) {{{ A = x*y }}}
(2) {{{ A = 82*164 }}}
(2) {{{ A = 13448 }}} ft2
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Here's the plot with {{{ A }}} on the vertical axis
and {{{ x }}} on the horizontal axis
{{{ graph( 400, 400, -50, 200, -3000, 15000, -2x^2 + 328x ) }}}