Answer
$\frac{144}{4+\pi }\text{ in}$.
Work Step by Step
Let $x$ be the length of the first string; then $\left( 36-x \right)$ will be the length of the second string.
The area of the circle is $\pi {{r}^{2}}$and the area of the square is ${{s}^{2}}$.
It is known that the circumference of the circle is x.
Thus,
$\begin{align}
& x=2\pi r \\
& r=\frac{x}{2\pi } \\
\end{align}$
The perimeter of the square is $\left( 36-x \right)$.
$\begin{align}
& \left( 36-x \right)=4s \\
& s=\frac{36-x}{4} \\
\end{align}$
The sum of the areas of the circle and the square is,
$\begin{align}
& A=\pi {{r}^{2}}+{{s}^{2}} \\
& =\pi {{\left( \frac{x}{2\pi } \right)}^{2}}+{{\left( \frac{\left( 36-x \right)}{4} \right)}^{2}} \\
& =\pi \left( \frac{{{x}^{2}}}{4{{\pi }^{2}}} \right)+\left( \frac{1}{16} \right){{\left( 36-x \right)}^{2}} \\
& =\frac{{{x}^{2}}}{4\pi }+\left( \frac{1}{16} \right)\left( 1296-72x+{{x}^{2}} \right)
\end{align}$
Further simplify,
$\begin{align}
& A=\frac{{{x}^{2}}}{4\pi }+\left( \frac{1}{16}\cdot {{x}^{2}} \right)-\left( \frac{1}{16}\cdot 72x \right)+\left( \frac{1}{16}\cdot 1296 \right) \\
& =\frac{{{x}^{2}}}{4\pi }+\left( \frac{{{x}^{2}}}{16} \right)-\left( \frac{72x}{16} \right)\left( \frac{1296}{16} \right)
\end{align}$
Now, multiply and divide the first term by 4 to make the denominator common with the second term:
$\left( \frac{{{x}^{2}}}{4\pi } \right)=\left( \frac{4{{x}^{2}}}{16\pi } \right)$
Multiply and divide the ${{x}^{2}}$terms by $\pi $ to make the denominator common with the first term:
$\left( \frac{{{x}^{2}}}{16} \right)=\left( \frac{\pi {{x}^{2}}}{16\pi } \right)$
The area becomes,
$\begin{align}
& A=\frac{4{{x}^{2}}}{16\pi }+\left( \frac{\pi {{x}^{2}}}{16\pi } \right)-\left( \frac{72x}{16} \right)\left( \frac{1296}{16} \right) \\
& =\left( \frac{\left( 4+\pi \right)}{16\pi } \right){{x}^{2}}-\left( \frac{72}{16} \right)x+\left( \frac{1296}{16} \right)
\end{align}$
Compare with the equation of a parabola $A=a{{x}^{2}}+bx+c$.
$a=\frac{4+\pi }{16\pi },b=-\frac{72}{16}$
Area is maximum at vertex.
Therefore, the vertex is,
$\begin{align}
& x=\frac{-b}{2a} \\
& =-\frac{\frac{-72}{16}}{2\left( \frac{4+\pi }{16\pi } \right)} \\
& =-\frac{\frac{-72}{16}}{\left( \frac{4+\pi }{8\pi } \right)}
\end{align}$
Further simplify:
$\begin{align}
& x=\frac{72}{16}\cdot \left( \frac{8\pi }{4+\pi } \right) \\
& =\frac{36\pi }{4+\pi }
\end{align}$
The length of the second piece of string is,
$\begin{align}
& 36-x=36-\frac{36\pi }{4+\pi } \\
& =\frac{144-36\pi +36\pi }{4+\pi } \\
& =\frac{144}{4+\pi }
\end{align}$
The length of the first piece of string to make the circle is $x=\frac{36\pi }{4+\pi }$.
The length of second piece of string to make the square is $36-x=\frac{144}{4+\pi }$.
Therefore, the length of the piece for making a circle is $x=\frac{36\pi }{4+\pi }$ and for making a square is
$36-x=\frac{144}{4+\pi }$.
