Chapter 2.6, Problem 12AYU

To determine

To find:

a. To 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.

Answer to Problem 12AYU

Solution:

a. $=\frac{\sqrt{3}}{4}{x}^2+\frac{100+9x^2-60x}{4\pi }$

Explanation of Solution

Given:

A 10m long wire is cut into two pieces. One piece is shaped as an equilateral triangle and other piece is shaped as a circle.

Calculation:

a. To 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.

The side of the equilateral triangle $=x$.

Therefore, Area of the equilateral triangle $A_1=\frac{\sqrt{3}}{4}{x}^2$.

Total length of the wire which made equilateral triangle is $3x$.

The perimeter of the circle $\text{= }2\pi r=\left(10-3x\right)$.

$r=\frac{10-3x}{2\pi }$

Area of the circle $A_2=\pi r^2=\frac{{\left(10-3x\right)}^2}{4\pi }$.

Therefore, the total area enclosed by the wire $=A_1+A_2$.

$=\frac{\sqrt{3}}{4}{x}^2+\frac{{\left(10-3x\right)}^2}{4\pi }$

$=\frac{\sqrt{3}}{4}{x}^2+\frac{100+9x^2-60x}{4\pi }$

To determine

To find:

b. To the domain of $A$.

Answer to Problem 12AYU

Solution:

b. $\text{Domain of}A=\left\{x|0<x<3.3\right\}$

Explanation of Solution

Given:

A 10m long wire is cut into two pieces. One piece is shaped as an equilateral triangle and other piece is shaped as a circle.

Calculation:

b. To the domain of $A$:

Since, perimeter $\left(10-3x\right)$ has to be positive, the domain of

$A\left(x\right)=\left\{x|0<x<3.333\right\}$

To determine

To find:

c. To graph $A=A\left(x\right)$ and to find the value of $x$ is $A$ smallest.

Answer to Problem 12AYU

Solution:

c. $x=2.9$

Explanation of Solution

Given:

A 10m long wire is cut into two pieces. One piece is shaped as an equilateral triangle and other piece is shaped as a circle.

Calculation:

c. To graph $A=A\left(x\right)$ and to find the value of $x$ is $A$ smallest.

From the graph it can be seen that when $x=2.9,A$attains its minimum value $4.18$.

Therefore, $A$ is smallest when $x=2.9$.