Question 467901
Let the width of the garden = {{{W}}} m
Let the length of the garden = {{{L}}} m
given:
(1) {{{ W*L = 5000 }}} m2
(2) {{{ 2W + 2L = 300 }}} m
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(2) {{{ W + L = 150 }}}
and
(1) {{{ L = 5000/W }}}
so, by substitution:
(2) {{{ W + 5000/W = 150 }}}
(2) {{{ W^2 + 5000 = 150W }}}
(2) {{{ W^2 - 150W + 5000 = 0 }}}
Use quadratic formula:
{{{W = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{ a = 1 }}}
{{{ b = -150 }}}
{{{ c = 5000 }}}
{{{W = (-(-150) +- sqrt( (-150)^2-4*1*5000 ))/(2*1) }}} 
{{{W = ( 150 +- sqrt( 22500 - 20000 ))/2 }}} 
{{{W = ( 150 +- sqrt( 2500 ))/2 }}} 
{{{W = ( 150 +- 50)/2 }}} 
{{{ W = 75 + 25 }}}
{{{ W = 100 }}}
and
{{{ W = 75 - 25 }}}
{{{ W = 50 }}}
and
(2) {{{ W + L = 150 }}}
(2) {{{ 50 + 100 = 150 }}}
The dimensions are 50 x 100 m2
check:
(1) {{{ W*L = 5000 }}} m2
(1) {{{ 50*100 = 5000 }}}
(1) {{{ 5000 = 5000 }}}
OK