Chapter 4.6, Problem 25E

i.

To determine

The maximum area enclosed by a piece of wire $10\text{m}$ is cut into two pieces, one piece is bent into a square and the other is bent into an equilateral triangle.

Answer to Problem 25E

The entire wire should be used for the square.

Explanation of Solution

Given information:

A piece of wire $10\text{m}$ is cut into two pieces, one piece is bent into a square and the other is bent into an equilateral triangle.

Formula Used:

Let $a$ be the side of the square and equilateral triangle. Then

Area of square is $A=a^2$

Side of the square is $a=\frac{\text{perimeter of square}}{4}$

Area of equilateral triangle is $A=\frac{\sqrt{3}}{4}{a}^2$

Side of an equilateral triangle is $a=\frac{\text{perimeter of equilateral triangle}}{4}$

And

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\text{'}}(a)=0$ and
2. $f^{\text{'}}(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\text{'}}(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\text{'}}(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima.
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\text{'}}(a)=0$ and
2. $f^{\text{'}}(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\text{'}}(x)<0$ at every point sufficiently close to and to the left of $a$ , and at $f^{\text{'}}(x)>0$ every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\text{'}}(a)=0$ and If $f^{\text{'}}(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.
4. • If $f^{″}(a)>0$ then $f$ has a local minimum at $x=a$
   • If $f^{″}(a)<0$ then $f$ has a local maximum at $x=a$

Calculation:

As per the given problem

Draw the diagram of a piece of wire $10\text{m}$ is cut into two pieces, one piece is bent into a square and the other is bent into an equilateral triangle.

  

Let $x$ be the length required to make the square and $y$ be the length required to make an equilateral triangle. Then

The side length of square is $\frac{x}{4}$

The side length of an equilateral triangle is $\frac{y}{3}$

Therefore,

The Length of wire

  $\begin{array}{l}x+y=10\\ \Rightarrow y=10-x\end{array}$

The total area enclosed by the both figure,

  $\begin{array}{l}A=\text{Area of square + Area of equilateral triangle}\\ A={\left(\text{side}\right)}^2\text{ + }\frac{\sqrt{3}}{4}\times {\left(\text{side}\right)}^2\\ A={\left(\frac{x}{4}\right)}^2\text{ + }\frac{\sqrt{3}}{4}\times {\left(\frac{y}{3}\right)}^2\\ A=\frac{x^2}{16}\text{ + }\frac{\sqrt{3}}{4}\times \frac{y^2}{9}\\ A=\frac{x^2}{16}\text{ + }\frac{\sqrt{3}{y}^2}{36}\end{array}$

Substitute $y=10-x$ , to get

  $A\left(x\right)=\frac{x^2}{16}\text{ + }\frac{\sqrt{3}}{36}\times {\left(10-x\right)}^2$

Recall that,

And

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\text{'}}(a)=0$ and
2. $f^{\text{'}}(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\text{'}}(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\text{'}}(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima.
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\text{'}}(a)=0$ and
2. $f^{\text{'}}(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\text{'}}(x)<0$ at every point sufficiently close to and to the left of $a$ , and at $f^{\text{'}}(x)>0$ every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\text{'}}(a)=0$ and If $f^{\text{'}}(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.
4. • If $f^{″}(a)>0$ then $f$ has a local minimum at $x=a$
   • If $f^{″}(a)<0$ then $f$ has a local maximum at $x=a$

Differentiate on both sides,

  $\begin{array}{l}{A}^{\prime }\left(x\right)=\frac{2x}{16}\text{ + }\frac{\sqrt{3}}{36}\times 2\times \left(10-x\right)\times -1\\ A^{\prime }\left(x\right)=\frac{2x}{16}-\frac{\sqrt{3}}{18}\left(10-x\right)\end{array}$

For maximum and minimum value of $A^{\prime }\left(x\right)=0$ ,

  $\begin{array}{l}\frac{x}{8}-\frac{\sqrt{3}}{18}\left(10-x\right)=0\\ \Rightarrow \frac{x}{8}=\frac{\sqrt{3}}{18}\left(10-x\right)\\ \Rightarrow 18x=80\sqrt{3}-8\sqrt{3}x\\ \Rightarrow 18x+8\sqrt{3}x=80\sqrt{3}\\ \Rightarrow x\left(18+8\sqrt{3}\right)=80\sqrt{3}\\ \Rightarrow x=\frac{80\sqrt{3}}{18+8\sqrt{3}}\end{array}$

Now evaluate the function to find the maximum and minimum values at all critical point $\left(x=0,x=4.35,x=10\right)$

At $x=0$

  $\begin{array}{l}A\left(0\right)=\frac{0}{16}\text{ + }\frac{\sqrt{3}}{36}\times {\left(10-0\right)}^2\\ A\left(0\right)=\frac{\sqrt{3}}{36}\times 100\\ \approx 4.81\end{array}$

At $x=4.35$

  $\begin{array}{l}A\left(4.35\right)=\frac{{\left(4.35\right)}^2}{16}\text{ + }\frac{\sqrt{3}}{36}\times {\left(10-4.35\right)}^2\\ =\frac{18.92}{16}\text{ + }\frac{\sqrt{3}}{36}\times {\left(5.65\right)}^2\\ \approx 2.71\\ \end{array}$

At $x=10$

  $\begin{array}{l}A\left(10\right)=\frac{{\left(10\right)}^2}{16}\text{ + }\frac{\sqrt{3}}{36}\times {\left(10-10\right)}^2\\ =\frac{100}{16}\text{ + }\frac{\sqrt{3}}{36}\times 0\\ \approx 6.25\\ \end{array}$

For maximum area, The entire wire should be used for the square

Conclusion:

The entire wire should be used for the square

ii.

To determine

The minimum area enclosed by a piece of wire $10\text{m}$ is cut into two pieces, one piece is bent into a square and the other is bent into an equilateral triangle

Answer to Problem 25E

For minimum area, $4.35\text{m}$ should be used for square and $5.65\text{m}$ should be used for equilateral triangle.

Explanation of Solution

Given information:

A piece of wire $10\text{m}$ is cut into two pieces, one piece is bent into a square and the other is bent into an equilateral triangle

Formula Used:

Let $a$ be the side of the square and equilateral triangle. Then

Area of square is $A=a^2$

Side of the square is $a=\frac{\text{perimeter of square}}{4}$

Area of equilateral triangle is $A=\frac{\sqrt{3}}{4}{a}^2$

Side of an equilateral triangle is $a=\frac{\text{perimeter of equilateral triangle}}{4}$

And

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\text{'}}(a)=0$ and
2. $f^{\text{'}}(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\text{'}}(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\text{'}}(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima.
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\text{'}}(a)=0$ and
2. $f^{\text{'}}(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\text{'}}(x)<0$ at every point sufficiently close to and to the left of $a$ , and at $f^{\text{'}}(x)>0$ every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\text{'}}(a)=0$ and If $f^{\text{'}}(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.
4. • If $f^{″}(a)>0$ then $f$ has a local minimum at $x=a$
   • If $f^{″}(a)<0$ then $f$ has a local maximum at $x=a$

Calculation:

As per the given problem

Draw the diagram of a piece of wire $10\text{m}$ is cut into two pieces, one piece is bent into a square and the other is bent into an equilateral triangle.

  

Let $x$ be the length required to make the square and $y$ be the length required to make an equilateral triangle. Then

The side length of square is $\frac{x}{4}$

The side length of an equilateral triangle is $\frac{y}{3}$

Therefore,

The Length of wire

  $\begin{array}{l}x+y=10\\ \Rightarrow y=10-x\end{array}$

The total area enclosed by the both figure,

  $\begin{array}{l}A=\text{Area of square + Area of equilateral triangle}\\ A={\left(\text{side}\right)}^2\text{ + }\frac{\sqrt{3}}{4}\times {\left(\text{side}\right)}^2\\ A={\left(\frac{x}{4}\right)}^2\text{ + }\frac{\sqrt{3}}{4}\times {\left(\frac{y}{3}\right)}^2\\ A=\frac{x^2}{16}\text{ + }\frac{\sqrt{3}}{4}\times \frac{y^2}{9}\\ A=\frac{x^2}{16}\text{ + }\frac{\sqrt{3}{y}^2}{36}\end{array}$

Substitute $y=10-x$ , to get

  $A\left(x\right)=\frac{x^2}{16}\text{ + }\frac{\sqrt{3}}{36}\times {\left(10-x\right)}^2$

Recall that,

And

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\text{'}}(a)=0$ and
2. $f^{\text{'}}(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\text{'}}(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\text{'}}(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima.
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\text{'}}(a)=0$ and
2. $f^{\text{'}}(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\text{'}}(x)<0$ at every point sufficiently close to and to the left of $a$ , and at $f^{\text{'}}(x)>0$ every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\text{'}}(a)=0$ and If $f^{\text{'}}(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.
4. • If $f^{″}(a)>0$ then $f$ has a local minimum at $x=a$
   • If $f^{″}(a)<0$ then $f$ has a local maximum at $x=a$

Differentiate on both sides,

  $\begin{array}{l}{A}^{\prime }\left(x\right)=\frac{2x}{16}\text{ + }\frac{\sqrt{3}}{36}\times 2\times \left(10-x\right)\times -1\\ A^{\prime }\left(x\right)=\frac{2x}{16}-\frac{\sqrt{3}}{18}\left(10-x\right)\end{array}$

For maximum and minimum value of $A^{\prime }\left(x\right)=0$ ,

  $\begin{array}{l}\frac{x}{8}-\frac{\sqrt{3}}{18}\left(10-x\right)=0\\ \Rightarrow \frac{x}{8}=\frac{\sqrt{3}}{18}\left(10-x\right)\\ \Rightarrow 18x=80\sqrt{3}-8\sqrt{3}x\\ \Rightarrow 18x+8\sqrt{3}x=80\sqrt{3}\\ \Rightarrow x\left(18+8\sqrt{3}\right)=80\sqrt{3}\\ \Rightarrow x=\frac{80\sqrt{3}}{18+8\sqrt{3}}\end{array}$

Now evaluate the function to find the maximum and minimum values at all critical point $\left(x=0,x=4.35,x=10\right)$

At $x=0$

  $\begin{array}{l}A\left(0\right)=\frac{0}{16}\text{ + }\frac{\sqrt{3}}{36}\times {\left(10-0\right)}^2\\ A\left(0\right)=\frac{\sqrt{3}}{36}\times 100\\ \approx 4.81\end{array}$

At $x=4.35$

  $\begin{array}{l}A\left(4.35\right)=\frac{{\left(4.35\right)}^2}{16}\text{ + }\frac{\sqrt{3}}{36}\times {\left(10-4.35\right)}^2\\ =\frac{18.92}{16}\text{ + }\frac{\sqrt{3}}{36}\times {\left(5.65\right)}^2\\ \approx 2.71\\ \end{array}$

At $x=10$

  $\begin{array}{l}A\left(10\right)=\frac{{\left(10\right)}^2}{16}\text{ + }\frac{\sqrt{3}}{36}\times {\left(10-10\right)}^2\\ =\frac{100}{16}\text{ + }\frac{\sqrt{3}}{36}\times 0\\ \approx 6.25\\ \end{array}$

For minimum area, $4.35\text{m}$ should be used for square and $\left(10-4.35=5.65\text{m}\right)$ should be used for equilateral triangle

Conclusion:

For minimum area, $4.35\text{m}$ should be used for square and $5.65\text{m}$ should be used for equilateral triangle