Chapter 4.6, Problem 56E

To determine

To calculate: A rain gutter is to be constructed from a metal sheet of width 30 cm by bending up one-third of a sheet on each side through an angle $\theta$ . How should $\theta$ be chosen so that the gutter will carry the maximum amount of water?

Answer to Problem 56E

The gutter can carry the maximum amount of water when $\theta =\frac{\pi }{3}$

Explanation of Solution

Given information:

A rain gutter is to be constructed from a metal sheet of width 30 cm by bending up one-third of a sheet on each side through an angle $\theta$

Formula used:

Pythagorean Theorem: The sum of the squares on the legs of the right angled triangle is equal to the square on the side opposite to the right angle triangle. That is:

  

  ${\left(H\right)}^2={\left(P\right)}^2+{\left(B\right)}^2$

Trigonometric ratio:

  $\sin \theta =\frac{p}{h},\cos \theta =\frac{b}{h}$

Let $h$ be the height and $b$ be the base of the triangle and rectangle respectively, then

The area of triangle is $=\frac{1}{2}bh$

The area of rectangle $=bh$

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

Calculation:

As per the given problem

Draw the diagram of a rain gutter is to be constructed from a metal sheet of width 30 cm by bending up one-third of a sheet on each side through an angle $\theta$

  

The amount of water that the gutter can carry is proportional to the area of the a cross section of the gutter.

Let $h$ is the height that the tip of one bent-up segments rise above the base, then the cross-sectional area is

  $\begin{array}{l}A=\text{area of recangle +2}\times \text{area of right angle triangle}\\ =bh+2\times \frac{1}{2}bh\end{array}$

Recall that,

Trigonometric ratio:

  $\sin \theta =\frac{p}{h},\cos \theta =\frac{b}{h}$

  $\begin{array}{l}\sin \theta =\frac{h}{10}\text{and}\cos \theta =\frac{b}{10}\\ \Rightarrow h=10\sin \theta \text{and}b=10\cos \theta \end{array}$

Recall that,

Let $h$ be the height and $b$ be the base of the triangle and rectangle respectively, then

The area of triangle is $=\frac{1}{2}bh$

The area of rectangle $=bh$

  $\begin{array}{l}A=\text{area of recangle +2}\times \text{area of right angle triangle}\\ =bh+2\times \frac{1}{2}bh\end{array}$

Substitute $h=10\sin \theta \text{and}b=10\cos \theta$ and simplified

  $\begin{array}{l}A=\text{area of recangle +2}\times \text{area of right angle triangle}\\ =10\times 10\sin \theta +10\sin \theta \times 10\cos \theta \\ A\left(\theta \right)=100\sin \theta +100\sin \theta \cos \theta \\ =100\sin \theta +50\sin 2\theta \end{array}$

Recall that,

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

Differentiate with respect to $\theta$

  $\begin{array}{l}{A}^{\prime }\left(\theta \right)=100\cos \theta +100\cos 2\theta \\ =100\left(\cos \theta +\cos 2\theta \right)\\ =100\left(\cos \theta +2{\cos }^2\theta -1\right)\end{array}$

Solve for $A^{\prime }\left(\theta \right)=0$ , and simplified

  $\begin{array}{l}100\left(\cos \theta +2{\cos }^2\theta -1\right)=0\\ \Rightarrow \cos \theta +2{\cos }^2\theta -1=0\\ \Rightarrow \left(2\cos \theta -1\right)\left(\cos \theta -1\right)=0\\ \Rightarrow \cos \theta =\frac{1}{2}\text{or}\text{cos}\theta \text{=1}\\ \Rightarrow \theta \text{=}\frac{\pi }{3}\text{or}\theta =\pi \\ \end{array}$

Evaluate the $A\left(\theta \right)$ at the critical point and end point

At $\theta =0$

  $A\left(0\right)=0$

At $\theta =\frac{\pi }{3}$

  $\begin{array}{l}A\left(\frac{\pi }{3}\right)=100\times \frac{\sqrt{3}}{2}+100\times \frac{\sqrt{3}}{2}\times \frac{1}{2}\\ \approx 129.9\end{array}$

At $\theta =\pi$

  $A\left(\pi \right)=0$

Therefore, the gutter can carry the maximum amount of water when $\theta =\frac{\pi }{3}$

Conclusion:

Thus the gutter can carry the maximum amount of water when $\theta =\frac{\pi }{3}$