Wanda Kane

## Answered question

2022-01-06

Arain gutter is to be constructed from a metal sheet of width 30 cm by bending up one-third of the sheet on each side through an angle 6. How should \theta be chosen so that the gutter will carry the maximum amount of water?

### Answer & Explanation

lalilulelo2k3eq

Beginner2022-01-07Added 38 answers

The amount of water that the gutter can carry is proportional to the area of a cross section of the gutter.
If h is the height that the tip of one of the bent-up segments rises above the base, then the cross-sectional area is
$A=10h+2\left(\begin{array}{c}\frac{1}{2}bh\end{array}\right)$.
Now, by the definition of the sine,
$\mathrm{sin}\theta =\frac{h}{10}⇒h=10\mathrm{sin}\theta$
Likewise,
$\mathrm{cos}\theta =\frac{b}{10}⇒b=10\mathrm{cos}\theta$
Therefore,
$A\left(\theta \right)=10\left(10\mathrm{sin}\theta \right)+\left(10\mathrm{sin}\theta \right)\left(10\mathrm{cos}\theta \right)=100\mathrm{sin}\theta +100\mathrm{sin}\theta \mathrm{cos}\theta$.
This is the function we're trying to maximize, so differentiate and find critical points:
${A}^{\prime }\left(\theta \right)=100\left(\mathrm{cos}\theta -\left(1-{\mathrm{cos}}^{2}\theta \right)+{\mathrm{cos}}^{2}\theta \right)=100\left(2{\mathrm{cos}}^{2}\theta +\mathrm{cos}\theta -1\right)=100\left(2\mathrm{cos}\theta -1\right)\left(\mathrm{cos}\theta +1\right)$.
Therefore, ${A}^{\prime }\left(\theta \right)=0$ when
$100\left(2\mathrm{cos}\theta -1\right)\left(\mathrm{cos}\theta +1\right)=0$
of, equivalently, when either
$\mathrm{cos}\theta =\frac{1}{2}\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}\mathrm{cos}\theta =-1$.
Thus $\theta =\frac{\pi }{3}\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}\theta =\pi$.
Now, \theta can range from 0 to $\pi$, so we just plug in critical points and endpoints into the area function:
$A\left(0\right)=0$
$A\left(\begin{array}{c}\frac{\pi }{3}\end{array}\right)=100\cdot \frac{\sqrt{3}}{2}+100\cdot \frac{\sqrt{3}}{2}\cdot \frac{1}{2}=\frac{300\sqrt{3}}{4}\approx 129.9$
$A\left(\pi \right)=0$
Therefore, the gutter can carry the maximum amount of water when $\theta =\frac{\pi }{3}$

Raymond Foley

Beginner2022-01-08Added 39 answers

Answer: The amount of water that the gutter can carry is proportional to the area of a cross section of the gutter. If h is the height that the tip of one of the bent-up segments rises above the base, then the cross-sectional area is
$A=10h+2\left(\begin{array}{c}\frac{1}{2}bh\end{array}\right)$
Now, by the definition of the sine,
$\mathrm{sin}\theta =\frac{h}{10},soh=10\mathrm{sin}\theta$.
Likewise,
$\mathrm{cos}\theta =\frac{b}{10},sob=10\mathrm{cos}\theta$.
Therefore,
$A\left(\theta \right)=10\left(10\mathrm{sin}\theta \right)+\left(10\mathrm{sin}\theta \right)\left(10\mathrm{cos}\theta \right)=100\mathrm{sin}\theta +100\mathrm{sin}\theta \mathrm{cos}\theta$.
This is the function were

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