Question 697179
The original area is {{{ 30*20 = 600 }}} m^2
The area of patch left is {{{ .6*600 = 360 }}}
Let {{{ x }}} = the uniform width of mowed part
The unmowed patch is {{{ 30 - 2x }}} by {{{ 20 - 2x }}}
{{{ ( 30 - 2x )*( 20 - 2x ) = 360 }}}
{{{ 600 - 40x - 60x + 4x^2 = 360 }}}
{{{ 4x^2 - 100x + 240 = 0 }}}
{{{ x^2 - 25x + 60 = 0 }}}
Use quadratic formula
{{{ x = (-b +- sqrt( b^2 - 4*a*c )) / (2*a) }}}
{{{ a = 1 }}}
{{{ b = -25 }}}
{{{ c = 60 }}}
{{{ x = (-(-25) +- sqrt( (-25)^2 - 4*1*60 )) / (2*1) }}}
{{{ x = ( 25 +- sqrt( 625 - 240 )) / 2 }}}
{{{ x = ( 25 +- sqrt( 385 )) / 2 }}}
{{{ x = ( 25 +- 19.6214) / 2 }}}
{{{ x = 5.3786 / 2 }}}
{{{ x = 2.6893 }}}
The width of the strip is 2.6893 m
check answer:
{{{ ( 30 - 2x )*( 20 - 2x ) = 360 }}}
{{{ ( 30 - 5.3786 )*( 20 - 5.3786 ) = 360 }}}
{{{ 24.6214 * 14.6214 = 360 }}}
{{{ 359.9993 = 360 }}}
close enough