Q. 11 - Precalculus Enhanced with Graphing Utilities

Question

A wire $10$ meters long is to be cut into two pieces. One piece will be shaped as a square, and the other piece will be shaped as a circle. See the figure.

Part (a): Express the total area A enclosed by the pieces of wire as a function of the length x  of a side of the square.

Part (b): What is the domain of A?

Part (c): Graph $A=A\left(x\right)$. For what value of x is A smallest?

Step-by-Step Solution

Answer

Part (a): Total area A enclosed by the pieces of wire as a function is $\mathrm{A}\left(\mathrm{x}\right)={\mathrm{x}}^2+\frac{25-20\mathrm{x}+4{\mathrm{x}}^2}{\mathrm{\pi }}$.

Part (b): The domain of A is $\left(0,\frac{5}{2}\right)$.

Part (c): On plotting the function, we get,

The value of x  for which A is smallest at $x\approx 1.4$m.

Part (a) Step 1. Given information.

Assume x m to be the length of side of the square.

Perimeter$\left(p\right)=4x$

From the given figure,

Circumference of the circle formed is $10-4x$m.

Assume r(x) to be the radius of the circle formed. Then it can be written,

$$\begin{gathered} 2\mathrm{\pi r}\left(\mathrm{x}\right)=10-4\mathrm{x} \\ \mathrm{r}\left(\mathrm{x}\right)=\frac{10-4\mathrm{x}}{2\mathrm{\pi }} \\ \mathrm{r}\left(\mathrm{x}\right)=\frac{5-2\mathrm{x}}{\mathrm{\pi }} \end{gathered}$$

Part (a) Step 2. Calculate the total area.

Consider the given question,

$$\begin{gathered} A\left(x\right)=x^2+{\mathrm{\pi r}}^2 \\ \mathrm{A}\left(\mathrm{x}\right)={\mathrm{x}}^2+\mathrm{\pi }{\left(\frac{5-2\mathrm{x}}{\mathrm{\pi }}\right)}^2 \\ \mathrm{A}\left(\mathrm{x}\right)={\mathrm{x}}^2+\frac{25-20\mathrm{x}+4{\mathrm{x}}^2}{\mathrm{\pi }} \end{gathered}$$

Part (b) Step 1. Find the domain of A .

Consider the given question,

$\mathrm{r}\left(\mathrm{x}\right)=\frac{5-2\mathrm{x}}{\mathrm{\pi }}$. Also $r\left(x\right)>0$. Then,

$$\begin{gathered} \frac{5-2\mathrm{x}}{\mathrm{\pi }}>0 \\ 5-2x>0 \\ x<\frac{5}{2} \end{gathered}$$

x  cannot be negative. As length cannot be negative.

Therefore, the domain is $\left(0,\frac{5}{2}\right)$.

Part (c) Step 1. Plot the function.

On plotting the function, we get,

From the graph, we can say that A is smallest at $x\approx 1.4$m.