Q. 12 - Precalculus Enhanced with Graphing Utilities

Question

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

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 equilateral triangle.

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 $\mathrm{A}\left(\mathrm{x}\right)=\frac{\sqrt{3}}{4}{\mathrm{x}}^2+\frac{100-60\mathrm{x}+9{\mathrm{x}}^2}{4\mathrm{\pi }}$.

Part (b): The domain of A is $\left(0,\frac{10}{3}\right)$.

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

A is smallest at $x\approx 2.08$.

Part (a) Step 1. Given information.

Assume x m to be the length of the given equilateral triangle.

Perimeter$\left(p\right)=3x m$

From the given figure,

Circumference of the circle formed is $10-3x$m.

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

$$\begin{gathered} 2\mathrm{\pi r}\left(\mathrm{x}\right)=10-3\mathrm{x} \\ \mathrm{r}\left(\mathrm{x}\right)=\frac{10-3\mathrm{x}}{2\mathrm{\pi }} \end{gathered}$$

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

Consider the given question,

Total area A(x)  enclosed by the pieces of wire is the sum of the areas enclosed by the triangle and circle,

$$\begin{gathered} A\left(x\right)=\frac{\sqrt{3}}{4}{x}^2+{\mathrm{\pi r}}^2 \\ \mathrm{A}\left(\mathrm{x}\right)=\frac{\sqrt{3}}{4}{\mathrm{x}}^2+\mathrm{\pi }{\left(\frac{10-3\mathrm{x}}{2\mathrm{\pi }}\right)}^2 \\ \mathrm{A}\left(\mathrm{x}\right)=\frac{\sqrt{3}}{4}{\mathrm{x}}^2+\frac{100-60\mathrm{x}+9{\mathrm{x}}^2}{4\mathrm{\pi }} \end{gathered}$$

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

Consider the given question,

$\mathrm{r}\left(\mathrm{x}\right)=\frac{10-3\mathrm{x}}{2\mathrm{\pi }}$. Also $r\left(x\right)>0$. Then,

$$\begin{gathered} \frac{10-3\mathrm{x}}{2\mathrm{\pi }}>0 \\ 10-3x>0 \\ x<\frac{10}{3} \end{gathered}$$

x  cannot be negative. As length cannot be negative.

Therefore, the domain is $\left(0,\frac{10}{3}\right)$.

Part (c) Step 1. Plot the function.

On plotting the function, we get,

From the graph, we can say that A is smallest at $x\approx 2.08$.