Single Variable Calculus, Chapter 4, 4.7, Section 4.7, Problem 34

Suppose that a piece of wire 10m long is cut into two pieces. One piece is bent into a square and the other is bent into a circle. How should the wire be cut so that the total area enclosed is (a) a maximum? (b) a minimum?




Let $x$ be the perimeter of the square. Then $10 - x$ will be the circumference of the circle.
So, the perimeter of the square is equal to...

$
\begin{equation}
\begin{aligned}
x &= 4s ; \text{ where } s \text{ is the length of the side of the square...}\\
\\
s &= \frac{x}{4}
\end{aligned}
\end{equation}
$


Then, the circumference is...


$
\begin{equation}
\begin{aligned}
10 - x &= 2 \pi r\\
\\
r &= \frac{10-x}{2\pi}
\end{aligned}
\end{equation}
$


Then, the total area is

$
\begin{equation}
\begin{aligned}
A(x) &= (s)^2 + \pi r^2\\
\\
A(x) &= \left( \frac{x}{4} \right)^2 + \pi \left( \frac{10-x}{2\pi} \right)^2
\end{aligned}
\end{equation}
$


Taking the derivative with respect to $x$, we obtain...


$
\begin{equation}
\begin{aligned}
A'(x) &= 2 \left( \frac{x}{4} \right) \left( \frac{1}{4} \right) + 2\pi \left( \frac{10-x}{2 \pi} \right) \left( \frac{-1}{2\pi} \right)\\
\\
A'(x) &= \frac{x}{8} - \frac{5}{\pi} + \frac{x}{2 \pi }\\
\\
A'(x) &= \left( \frac{1}{2\pi} + \frac{1}{8} \right) x - \frac{5}{\pi}
\end{aligned}
\end{equation}
$


when $A'(x) =0$,

$
\begin{equation}
\begin{aligned}
0 &= \left( \frac{1}{2 \pi} + \frac{1}{8} \right) x - \frac{5}{\pi}\\
\\
x &= \frac{\frac{5}{\pi}}{\left(\frac{1}{2\pi} + \frac{1}{8}\right)} = \frac{40}{4 + \pi}
\end{aligned}
\end{equation}
$


If we evaluate the total area with $x=0$, $x= 10$ and $\displaystyle x = \frac{40}{4+\pi}$, then...

$
\begin{equation}
\begin{aligned}
A(0) &= \left( \frac{0}{4} \right)^2 + \pi \left( \frac{10-0}{2\pi} \right)^2 = \frac{25}{\pi} = 7.96\\
\\
A(10) &= \left( \frac{10}{4} \right)^2 + \pi \left( \frac{10-10}{2\pi} \right)^2 = 6.25\\
\\
A\left( \frac{40}{4+\pi} \right) &= \left( \frac{\frac{40}{4+\pi} }{4} \right)^2 + \pi \left( \frac{10\left( \frac{40}{4+\pi}\right)}{2\pi} \right)^2 = 3.50

\end{aligned}
\end{equation}
$


a.) We can conclude that the maximum area will be attained if the entire wire is used only in circle.
b.) The area is minimum when $\displaystyle x = \frac{40}{4+\pi}$