Wanda Kane

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?

lalilulelo2k3eq

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

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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