if you notice the picture below
if we take "x" amount for the triangle, then the circle will be using "b-x", since the wire is "b" long
now, that's just their perimeter only
anyhow. ..the sum of their lengths will then be [tex]\bf A_t=\cfrac{x^2\sqrt{3}}{36}+\cfrac{(b-x)^2}{4\pi }[/tex]
so... take the derivative of that areas sum, and check the critical points for a minimum, using the first-derivative test
keeping mind that, "b" is a constant, since that matters for taking the derivative
