Question

A piece of wire $40\mathrm{cm}$ long is to be cut into two pieces.

One-piece will be bent to form a circle; the other will be bent to form a square.

$\left(a\right)$ Find the lengths of the two pieces that cause the sum of the area of the circle and the area of the square to be a minimum.

$\left(b\right)$ How could you make the total area of the circle and the square a maximum.

Answer 1

Step-1: Express the area of circle:

Let $x$ be the circumference of the circle.

Using the concept of the circumference of the circle and calculate the radius of the circle below :

$\begin{array}{rcl}x& =& 2\mathrm{\pi r}\left[\mathrm{Circumference}\mathrm{of}\mathrm{circle}\right]\\ \therefore \mathrm{r}& =& \frac{\mathrm{x}}{2\mathrm{\pi }}\end{array}$

Using the formula of the area of the circle and it can be expressed below :

$\begin{array}{rcl}{\text{A}}_c& =& {\mathrm{\pi r}}^2\left[\mathrm{Area}\mathrm{of}\mathrm{circle}\right]\\ {\mathrm{A}}_{\mathrm{c}}& =& \mathrm{\pi }{\left(\frac{\mathrm{x}}{2\mathrm{\pi }}\right)}^2\left[\mathrm{Substitute}\mathrm{r}=\frac{\mathrm{x}}{2\mathrm{\pi }}\right]\\ {\mathrm{A}}_{\mathrm{c}}& =& \mathrm{\pi }\times \frac{{\mathrm{x}}^2}{4{\mathrm{\pi }}^2}\\ \therefore {\mathrm{A}}_{\mathrm{c}}& =& \frac{{\mathrm{x}}^2}{4\mathrm{\pi }}\end{array}$

Step-2: Express area of square in terms of $x$:

Since the circumference of the circle is $x$, so the perimeter of the square is $\left(40-x\right)$.

Let $y$ be the side of the square.

Using the concept of the perimeter of the square and it can be expressed and solve for $y$ :

$\begin{array}{rcl}4y& =& 40-x\\ \therefore y& =& \frac{40-x}{4}\end{array}$

Using the formula of the area of the square and it can be expressed below :

$\begin{array}{rcl}{\text{A}}_s& =& {\left(\text{side}\right)}^2\left[\text{Area of square}\right]\\ & \Rightarrow & {\text{A}}_s={\left(y\right)}^2\\ & \Rightarrow & {\text{A}}_s={\left(\frac{40-x}{4}\right)}^2\left[\text{Substitute}y=\frac{40-x}{4}\right]\\ \therefore {\text{A}}_s& =& \frac{1600+x^2-80x}{16}\end{array}$

Step-3: Find the maximum or minimum value of $x$:

Since the total area is equal to the sum of the area of the circle and the area of the square, so it can be expressed below :

$$\begin{gathered} \text{A}={\text{A}}_c+{\text{A}}_s \\ \text{A}=\frac{x^2}{4\mathrm{\pi }}+\frac{1600+x^2-80x}{16} \end{gathered}$$

Since the total area is minimum, so take the derivative and set it equal to zero and solve for $x$ :

$\begin{array}{rcl}\frac{dA}{dx}& =& \frac{2x}{4\mathrm{\pi }}+\frac{0+2x-80}{16}\\ & \Rightarrow & 0=\frac{x}{2\mathrm{\pi }}+\frac{x-40}{8}\\ & \Rightarrow & \frac{x}{2\mathrm{\pi }}+\frac{x-40}{8}=0\\ & \Rightarrow & \frac{8x+2\mathrm{\pi x}-80\mathrm{\pi }}{16\mathrm{\pi }}=0\\ & \Rightarrow & 8x+2\mathrm{\pi x}-80\mathrm{\pi }=0\\ & \Rightarrow & \left(8+2\mathrm{\pi }\right)\mathrm{x}=80\mathrm{\pi }\\ & \Rightarrow & \mathrm{x}=\frac{80\mathrm{\pi }}{8+2\mathrm{\pi }}\\ \therefore \mathrm{x}& \approx & 17.6\end{array}$

Step-4: Find the maximum circumference of circle and perimeter of square:

Therefore, the circumference of the circle is about $17.6\mathrm{cm}$.

And the perimeter of the square is $40-17.6=22.4\mathrm{cm}$.

Also,

The total area of the circle and the square can be maximum by differentiating and equating it to zero.

Hence,

(a) The circumference of the circle is $\mathbf{17}\mathbf{.}\mathbf{6}\mathbf{}\mathbf{cm}$ and the perimeter of the square is $\mathbf{22}\mathbf{.}\mathbf{4}\mathbf{}\mathbf{cm}$.

(b) The total area of the circle and the square can be maximum by differentiating and equating it to zero.