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Question Number 88218 by mathocean1 last updated on 09/Apr/20

Commented by mathocean1 last updated on 09/Apr/20

This is face view of an evacuation canal. It has a trapeze form. 4m represents its small base. 1) Determinate θ ∈ [60;90] such as the capacity of canal is maximal.

$$\mathrm{This}\:\mathrm{is}\:\mathrm{face}\:\mathrm{view}\:\mathrm{of}\:\mathrm{an}\:\mathrm{evacuation}\: \\ $$$$\mathrm{canal}.\:\mathrm{It}\:\mathrm{has}\:\mathrm{a}\:\mathrm{trapeze}\:\mathrm{form}.\:\mathrm{4m} \\ $$$$\mathrm{represents}\:\mathrm{its}\:\mathrm{small}\:\mathrm{base}. \\ $$$$\left.\mathrm{1}\right)\:\mathrm{Determinate}\:\:\theta\:\in\:\left[\mathrm{60};\mathrm{90}\right]\:\mathrm{such}\:\mathrm{as}\:\mathrm{the}\: \\ $$$$\mathrm{capacity}\:\mathrm{of}\:\mathrm{canal}\:\mathrm{is}\:\mathrm{maximal}. \\ $$

Commented by mathocean1 last updated on 09/Apr/20

Answered by ajfour last updated on 12/Apr/20

A=(1/2)(2sin θ)(4+4+4cos θ) ⇒ A=8sin θ+4sin θcos θ (dA/dθ)=8cos θ+4(d/dθ)(sin θcos θ) = 8cos θ+4{cos θ(d/dθ)(sin θ)+sin θ(d/dθ)(cos θ)} = 8cos θ+4(cos^2 θ−sin^2 θ) (dA/dθ)= 8cos θ+4cos^2 θ−4sin^2 θ=0 let cos θ=t 8t+4t^2 +4t^2 −4=0 2t^2 +2t−1=0 t=−(1/2)+((√3)/2) θ=cos^(−1) ((((√3)−1)/2))≈ 68.5293° .

$${A}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2sin}\:\theta\right)\left(\mathrm{4}+\mathrm{4}+\mathrm{4cos}\:\theta\right) \\ $$$$\Rightarrow\:\:{A}=\mathrm{8sin}\:\theta+\mathrm{4sin}\:\theta\mathrm{cos}\:\theta \\ $$$$\:\:\:\frac{{dA}}{{d}\theta}=\mathrm{8cos}\:\theta+\mathrm{4}\frac{{d}}{{d}\theta}\left(\mathrm{sin}\:\theta\mathrm{cos}\:\theta\right) \\ $$$$\:\:=\:\mathrm{8cos}\:\theta+\mathrm{4}\left\{\mathrm{cos}\:\theta\frac{{d}}{{d}\theta}\left(\mathrm{sin}\:\theta\right)+\mathrm{sin}\:\theta\frac{{d}}{{d}\theta}\left(\mathrm{cos}\:\theta\right)\right\} \\ $$$$\:\:=\:\mathrm{8cos}\:\theta+\mathrm{4}\left(\mathrm{cos}\:^{\mathrm{2}} \theta−\mathrm{sin}\:^{\mathrm{2}} \theta\right) \\ $$$$\frac{{dA}}{{d}\theta}=\:\mathrm{8cos}\:\theta+\mathrm{4cos}\:^{\mathrm{2}} \theta−\mathrm{4sin}\:^{\mathrm{2}} \theta=\mathrm{0} \\ $$$${let}\:\mathrm{cos}\:\theta={t} \\ $$$$\mathrm{8}{t}+\mathrm{4}{t}^{\mathrm{2}} +\mathrm{4}{t}^{\mathrm{2}} −\mathrm{4}=\mathrm{0} \\ $$$$\mathrm{2}{t}^{\mathrm{2}} +\mathrm{2}{t}−\mathrm{1}=\mathrm{0} \\ $$$${t}=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\theta=\mathrm{cos}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}}\right)\approx\:\mathrm{68}.\mathrm{5293}°\:. \\ $$

Commented by mathocean1 last updated on 12/Apr/20

please sir can you explain the 2^(nd ) line.

$$\mathrm{please}\:\mathrm{sir}\:\mathrm{can}\:\mathrm{you}\:\mathrm{explain}\:\mathrm{the}\:\mathrm{2}^{\mathrm{nd}\:} \:\mathrm{line}. \\ $$

Commented by mathocean1 last updated on 12/Apr/20

Thanks sir i have understood... it′s great.

$$\mathrm{Thanks}\:\mathrm{sir}\:\mathrm{i}\:\mathrm{have}\:\mathrm{understood}... \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{great}. \\ $$

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