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 $\displaystyle x = \frac{40}{4+\pi}$