Question 662391

The perimeter {{{P}}} of a rectangle is {{{240ft}}}. 

{{{P=240ft}}}........1

 the area of the rectangle {{{A}}} is not to exceed {{{2700ft^2}}} 

{{{A<=2700ft^2}}}......2


{{{240ft=2(L+W)}}}

{{{240ft/2=L+W}}}

{{{120ft=L+W}}}

{{{L=120ft-W}}}......

{{{A<=2700ft^2}}}

{{{L*W<=2700ft^2}}}....plug in {{{L}}} and solve for {{{W}}}

{{{(120ft-W)*W<=2700ft^2}}}

{{{120Wft-W^2<=2700ft^2}}}

{{{-W^2+120Wft-2700ft^2<=0}}}


{{{W= (-120ft +- sqrt( (120ft)^2-4*(-1)*(2700ft^2) ))/(2*-1) }}}


{{{W= (-120ft +- sqrt( 14400ft^2-10800ft^2 ))/-2 }}}


{{{W= (-120ft +- sqrt(3600ft^2))/-2 }}}


{{{W= (-120ft +- 60ft)/-2 }}}


solutions:


{{{W= (-120ft + 60ft)/-2 }}}

{{{W= (-60ft)/-2 }}}

{{{highlight(W=30ft) }}}


or

{{{W= (-120ft - 60ft)/-2 }}}

{{{W= (-180ft)/-2 }}}

{{{highlight(W=90ft) }}}


now find {{{L}}}

{{{L=120ft-W}}}

{{{L=120ft-30ft}}}

{{{highlight(L=90ft)}}}


{{{L=120ft-W}}}

{{{L=120ft-90ft}}}

{{{highlight(L=30ft)}}}


so, {{{highlight(L=90ft)}}} and {{{highlight(W=30ft) }}} or vice versa

since {{{A<=2700ft^2}}}, any length {{{highlight(L<=90ft)}}} will be the

possible length of a side