Question 891964
w for width and L for length;
{{{2w+2L=50}}} and {{{wL>=50}}}.


Using first the perimeter equation, w+L=25
{{{w=25-L}}}
Substitute.
{{{(25-L)L>=50}}}
{{{25L-L^2>=50}}}
{{{-L^2+25L-50>=0}}}
{{{highlight_green(L^2-25L+50<=0)}}}


roots for the equation
{{{L=(25+- sqrt(25^2-4*50))/2}}}
{{{L=(25+- sqrt(425))/2}}}
{{{L=(25+- 5*sqrt(17))/2}}}


The parabola with L has a minimum based on coefficient on {{{L^2}}} being a positive 1.  This means that L cannot be between the two roots.  The inequality is satisfied for {{{highlight(0<L<=(25-5sqrt(17))/2)}}}  or for {{{highlight((25+5sqrt(17))/2<=L<50)}}}.