Question 999931
The contribution of length from the square is {{{L[1]=4x}}}.
The contribution of length from the circle is {{{L[2]=2pi*r}}}
{{{L[1]+L[2]=130}}}
{{{4x+2pi*r=130}}}
{{{2x+pi*r=65}}}
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For the square,
{{{A[s]=x^2}}}
For the circle,
{{{A[c]=pi*r^2}}}
So then, 
{{{A=A[s]+A[c]}}}
{{{A=x^2+pi*r^2}}}
From the length above,
{{{pi*r=65-2x}}}
{{{r=(65-2x)/pi}}}
{{{pi^2r^2=(65-2x)^2}}}
{{{pi^2r^2=4x^2-260x+4225}}}
{{{pi*r^2=(4x^2-260x+4225)/pi}}}
Substituting into the area formula,
{{{A=x^2+(4x^2-260x+4225)/pi}}}
{{{A=(pi*x^2)/pi+(4x^2-260x+4225)/pi}}}
{{{A=((4+pi)x^2-260x+4225)/pi}}}
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Stationary value is where the derivative of A with respect to x equals zero.
{{{2(4+pi)x-260=0}}}
{{{x=260/(2(4+pi))}}}
{{{x=130/(4+pi)}}}