# Thread: Cutting Wire into Two Pieces

1. ## Cutting Wire into Two Pieces

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.

(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.

(b) What is the domain of the area A?

2. Hello, magentarita!

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

(a) Express the total area $\displaystyle A$ enclosed by the pieces of wire as a function
of the length $\displaystyle x$ of a side of the equilateral triangle.
Let $\displaystyle x$ = side of the triangle.
The area of an equilateral triangle of side $\displaystyle x$ is: .$\displaystyle A_t \:=\:\frac{\sqrt{3}}{4}x^2$

The triangle uses $\displaystyle 3x$ of the wire.
This leaves $\displaystyle 10-3x$ for the circumference of the circle.

Circumference formula: .$\displaystyle C \:=\:2\pi r$

So we have: .$\displaystyle 10-3x \:=\:2\pi r \quad\Rightarrow\quad r \:=\:\frac{10-3x}{2\pi}$

Area formula: .$\displaystyle A \:=\:\pi r^2$

So we have: .$\displaystyle A_c \;=\;\pi\left(\frac{10-3x}{2\pi}\right)^2 \quad\Rightarrow\quad A_c\;=\;\frac{(10-3x)^2}{4\pi}$

Therefore: . $\displaystyle \boxed{A \;=\;\frac{\sqrt{3}}{4}x^2 + \frac{(10-3x)^2}{4\pi}}$

(b) What is the domain of $\displaystyle A$ ?

Domain: .$\displaystyle 0 \,\leq\,x\,\leq\,3\tfrac{1}{3}$

3. ## Soroban....

Originally Posted by Soroban
Hello, magentarita!

Let $\displaystyle x$ = side of the triangle.
The area of an equilateral triangle of side $\displaystyle x$ is: .$\displaystyle A_t \:=\:\frac{\sqrt{3}}{4}x^2$

The triangle uses $\displaystyle 3x$ of the wire.
This leaves $\displaystyle 10-3x$ for the circumference of the circle.

Circumference formula: .$\displaystyle C \:=\:2\pi r$

So we have: .$\displaystyle 10-3x \:=\:2\pi r \quad\Rightarrow\quad r \:=\:\frac{10-3x}{2\pi}$

Area formula: .$\displaystyle A \:=\:\pi r^2$

So we have: .$\displaystyle A_c \;=\;\pi\left(\frac{10-3x}{2\pi}\right)^2 \quad\Rightarrow\quad A_c\;=\;\frac{(10-3x)^2}{4\pi}$

Therefore: . $\displaystyle \boxed{A \;=\;\frac{\sqrt{3}}{4}x^2 + \frac{(10-3x)^2}{4\pi}}$

Domain: .$\displaystyle 0 \,\leq\,x\,\leq\,3\tfrac{1}{3}$

You are fantastic. I love your replies. I am going to take calculus 1 soon. Are you interested in helping me as I go through calculus 1?