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Question Number 130582 by bramlexs22 last updated on 27/Jan/21

A metal rain gutter is to have 3−inch and  a 3−inch bottom the sides making an equal  angle θ with the bottom. What should   θ be in order to maximize carrying capasity  of the gutter ?

$$\mathrm{A}\:\mathrm{metal}\:\mathrm{rain}\:\mathrm{gutter}\:\mathrm{is}\:\mathrm{to}\:\mathrm{have}\:\mathrm{3}−\mathrm{inch}\:\mathrm{and} \\ $$$$\mathrm{a}\:\mathrm{3}−\mathrm{inch}\:\mathrm{bottom}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{making}\:\mathrm{an}\:\mathrm{equal} \\ $$$$\mathrm{angle}\:\theta\:\mathrm{with}\:\mathrm{the}\:\mathrm{bottom}.\:\mathrm{What}\:\mathrm{should}\: \\ $$$$\theta\:\mathrm{be}\:\mathrm{in}\:\mathrm{order}\:\mathrm{to}\:\mathrm{maximize}\:\mathrm{carrying}\:\mathrm{capasity} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{gutter}\:? \\ $$

Answered by EDWIN88 last updated on 27/Jan/21

The carrying capasity of the gutter  is maximized when the area  of the vertical end of the gutter  is maximized.  A=3(3sin θ)+2((1/2))(3cos θ)(3sin θ)  A=9sin θ+9cos θsin θ  (dA/dθ)=9(2cos^2 θ+cos θ−1)=0   when θ=(π/3)⇒A=((27(√3))/4)≈ 11.7  The carrying capasity is max  when θ=(π/3)

$${The}\:{carrying}\:{capasity}\:{of}\:{the}\:{gutter} \\ $$$${is}\:{maximized}\:{when}\:{the}\:{area} \\ $$$${of}\:{the}\:{vertical}\:{end}\:{of}\:{the}\:{gutter} \\ $$$${is}\:{maximized}. \\ $$$${A}=\mathrm{3}\left(\mathrm{3sin}\:\theta\right)+\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\mathrm{3cos}\:\theta\right)\left(\mathrm{3sin}\:\theta\right) \\ $$$${A}=\mathrm{9sin}\:\theta+\mathrm{9cos}\:\theta\mathrm{sin}\:\theta \\ $$$$\frac{{dA}}{{d}\theta}=\mathrm{9}\left(\mathrm{2cos}\:^{\mathrm{2}} \theta+\mathrm{cos}\:\theta−\mathrm{1}\right)=\mathrm{0} \\ $$$$\:{when}\:\theta=\frac{\pi}{\mathrm{3}}\Rightarrow{A}=\frac{\mathrm{27}\sqrt{\mathrm{3}}}{\mathrm{4}}\approx\:\mathrm{11}.\mathrm{7} \\ $$$${The}\:{carrying}\:{capasity}\:{is}\:{max} \\ $$$${when}\:\theta=\frac{\pi}{\mathrm{3}} \\ $$

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