Question 1019815
{{{R}}}= radius of the circle
{{{x}}}= side of the triangle
{{{L}}} length of the wire before cutting it
 
{{{2pi*R}}}= length of wire used for the circle
{{{3x=L-2pi*R}}}= length of wire used for the triangle

{{{3x=L-2pi*R}}}<-->{{{x=(L-2pi*R)/3}}}
 
Area of the triangle ={{{sqrt(3)x^2/4=(sqrt(3)/4)((L-2pi*R)/3)^2=sqrt(3)(L-2pi*R)^2/36}}}
Area of the circle ={{{pi*R^2}}}
Total area ={{{y=pi*R^2+sqrt(3)(L-2pi*R)^2/36}}}
{{{y=pi*R^2+sqrt(3)(L^2-4L*pi*R+4pi^2*R^2)/36}}}
{{{y=pi*R^2+sqrt(3)L^2/36-4sqrt(3)L*pi*R/36+4sqrt(3)pi^2*R^2/36}}}
{{{y=pi*R^2+sqrt(3)L^2/36-sqrt(3)L*pi*R/9+sqrt(3)pi^2*R^2/9}}}
{{{y=(pi+sqrt(3)pi^2/9)*R^2-(sqrt(3)L*pi/9)*R+sqrt(3)L^2/36}}}
That is a quadratic function.
Quadratic functions of the form {{{y=ax^2+bx+c}}} with {{{a>0}}}
have a minimum at {{{x=(-b)/2a}}} .
In the case of {{{y=(pi+sqrt(3)pi^2/9)*R^2-(sqrt(3)L*pi/9)*R+sqrt(3)L^2/36}}} ,
{{{x=R}}} , {{{a=(pi+sqrt(3)pi^2/9)>0}}} , and {{{b=-sqrt(3)L*pi/9}}} ,
so the minimum is at
{{{R=(sqrt(3)L*pi/9)/(2(pi+sqrt(3)pi^2/9))=(sqrt(3)L)/(2(9+sqrt(3)pi))=about0.59968}}}