Question 1129657


Area of rectangle is {{{6000}}}: 


{{{L*W=6000}}}


{{{W=6000/L}}}........eq.1




and  perimeter is {{{340 }}}:

{{{2(L+W)=340}}}

{{{L+W=340/2}}}

{{{L+W=170}}}

{{{W=170-L}}}.......eq.2



from eq.1 and eq.2


{{{6000/L=170-L}}}

{{{6000=170L-L^2}}}

{{{L^2-170L+6000=0}}}

{{{L^2-120L-50L+6000=0}}}

{{{(L^2-120L)-(50L-6000)=0}}}

{{{L(L-120)-50(L-120)=0}}}

{{{(L - 120) (L - 50) = 0}}}

solutions:

if {{{(L - 120) = 0}}}=>{{{L=120}}}
if {{{ (L - 50) = 0}}}=>{{{L=50}}}


so, {{{L=120}}} or {{{L=50}}}


then

{{{W=170-L}}}=>{{{W=170-120}}}=>{{{W=50}}} 
{{{W=170-L}}}=>{{{W=170-50}}}=>{{{W=120}}} 

=>{{{W=50}}}  or {{{W=120}}}  


since length is greater than width, then length of rectangle is {{{L=120}}} and the width is {{{W=50}}}