Question 1042718
Let {{{ L }}} = the length of fence parallel to the street
Let {{{ W }}} = the side perpendicular to the street
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Equation for total length of fencing:
{{{ 320 = L + 2W }}}
{{{ L = 320 - 2W }}}
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Let {{{ A }}} = the area of the lot
{{{ A = W*L }}}
{{{ A = W*( 320 - 2W ) }}}
{{{ A = -2W^2 + 320W }}}
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The W-value of the vertex ( in this case it's a peak ), is:
{{{ W[v] = -b/(2a) }}}, when the general form of the parabola is:
{{{ y = a*x^2 + b*x + c }}}
{{{ a = -2 }}}
{{{ b = 320 }}}
{{{ W[v] = -320/(2*( -2 )) }}}
{{{ W[v] = 80 }}}
and
{{{ A[max] = W*( 320 - 2W ) }}}
{{{ A[max] = 80*( 320 - 2*80 ) }}}
{{{ A[max] = 80*( 320 - 160 ) }}}
{{{ A[max] = 80*160 }}}
{{{ A[max] = 12800 }}} ft2
The largest area that can be inclosed is  12,800 ft2
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check the answer:
{{{ W[v] = 80 }}}
{{{ L[v] = 320 - 2W }}}
{{{ L[v] = 320 - 160 }}}
{{{ L[v] = 160 }}}
and
{{{ 320 = L + 2W }}}
{{{ 320 = 160 + 2*80 }}}
{{{ 320 = 320 }}}
OK
Here's the plot:
{{{ graph( 400, 400, -40, 200, -1500, 15000, -2x^2 + 320x ) }}}