Question 1072946
Look at an isosceles triangle with sides, m, m, and 2n.
So that 
{{{m=a}}}
{{{n=b/2}}}
The perimeter is then,
{{{P=m+m+2n}}}
{{{P=2m+2n}}}
The perimeter is set to a constant length, 2L.
{{{2m+2n=2L}}}
{{{m+n=L}}}
{{{n=L-m}}}
The area would then be,
{{{A=(1/2)(Base)(Height)}}}
{{{A=(1/2)n(sqrt(m^2-n^2))}}}
Substituting for n,
{{{A=(1/2)(L-m)(sqrt(m^2-(L-m)^2))}}}
{{{A=(1/2)(L-m)(sqrt(2mL-L^2))}}}
So to find the maximum area, take the derivative of A with respect to a and set it equal to zero.
{{{dA/da=(L(2L-3m))/(2(sqrt(2mL-L^2)))}}}
{{{(L(2L-3m))=0}}}
One solution,
{{{2L-3m=0}}}
{{{3m=2L}}}
{{{m=(2/3)L}}}
and
{{{n=L-(2/3)L}}}
{{{n=(1/3)L}}}
So in terms of a and b, 
{{{m=a}}}
and
{{{n=b/2}}}
So substituting,
{{{a=(2/3)L}}}
and
{{{b/2=(1/3)L}}}
{{{b=(2/3)L}}}
So the ratio is,
{{{a/b=((2/3)L)/((2/3)L)}}}
{{{highlight(a/b=1)}}}