Question 1065428
Let {{{ W }}} = the width
Let {{{ L }}} = the length
Using formula for perimeter:
{{{ 2W + 2L = 212 }}}
{{{ W + L = 106 }}}
{{{ L = 106 - W }}}
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Let {{{ A }}} = the area
{{{ A = W*( 106 - W ) }}}
{{{ A = -W^2 + 106W }}}
This is a parabola with the vertex a maximum,
and not a minimum
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Formula for vertex:
{{{ W[v] = -b/(2a) }}}
{{{ a = -1 }}}
{{{ b = 106 }}}
{{{ W[v] = -106/( 2*(-1)) }}}
{{{ W[v] = 53 }}}
and
{{{ L[v] = 106 - W }}}
{{{ L[v] = 106 - 53 }}}
{{{ L[v] =  53 }}}
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The enclosed area is maximum when it is 53 yds x 53 yds
check:
{{{ 4*53 = 212 }}} yds of fencing
OK
Here's a plot of {{{ A }}} and {{{ W }}}
{{{ graph( 400, 400, -15, 150, -350, 3500, -x^2 + 106x ) }}}