Question Number 88218 by mathocean1 last updated on 09/Apr/20
Commented by mathocean1 last updated on 09/Apr/20
$$\mathrm{This}\:\mathrm{is}\:\mathrm{face}\:\mathrm{view}\:\mathrm{of}\:\mathrm{an}\:\mathrm{evacuation}\: \\ $$$$\mathrm{canal}.\:\mathrm{It}\:\mathrm{has}\:\mathrm{a}\:\mathrm{trapeze}\:\mathrm{form}.\:\mathrm{4m} \\ $$$$\mathrm{represents}\:\mathrm{its}\:\mathrm{small}\:\mathrm{base}. \\ $$$$\left.\mathrm{1}\right)\:\mathrm{Determinate}\:\:\theta\:\in\:\left[\mathrm{60};\mathrm{90}\right]\:\mathrm{such}\:\mathrm{as}\:\mathrm{the}\: \\ $$$$\mathrm{capacity}\:\mathrm{of}\:\mathrm{canal}\:\mathrm{is}\:\mathrm{maximal}. \\ $$
Commented by mathocean1 last updated on 09/Apr/20
Answered by ajfour last updated on 12/Apr/20
$${A}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2sin}\:\theta\right)\left(\mathrm{4}+\mathrm{4}+\mathrm{4cos}\:\theta\right) \\ $$$$\Rightarrow\:\:{A}=\mathrm{8sin}\:\theta+\mathrm{4sin}\:\theta\mathrm{cos}\:\theta \\ $$$$\:\:\:\frac{{dA}}{{d}\theta}=\mathrm{8cos}\:\theta+\mathrm{4}\frac{{d}}{{d}\theta}\left(\mathrm{sin}\:\theta\mathrm{cos}\:\theta\right) \\ $$$$\:\:=\:\mathrm{8cos}\:\theta+\mathrm{4}\left\{\mathrm{cos}\:\theta\frac{{d}}{{d}\theta}\left(\mathrm{sin}\:\theta\right)+\mathrm{sin}\:\theta\frac{{d}}{{d}\theta}\left(\mathrm{cos}\:\theta\right)\right\} \\ $$$$\:\:=\:\mathrm{8cos}\:\theta+\mathrm{4}\left(\mathrm{cos}\:^{\mathrm{2}} \theta−\mathrm{sin}\:^{\mathrm{2}} \theta\right) \\ $$$$\frac{{dA}}{{d}\theta}=\:\mathrm{8cos}\:\theta+\mathrm{4cos}\:^{\mathrm{2}} \theta−\mathrm{4sin}\:^{\mathrm{2}} \theta=\mathrm{0} \\ $$$${let}\:\mathrm{cos}\:\theta={t} \\ $$$$\mathrm{8}{t}+\mathrm{4}{t}^{\mathrm{2}} +\mathrm{4}{t}^{\mathrm{2}} −\mathrm{4}=\mathrm{0} \\ $$$$\mathrm{2}{t}^{\mathrm{2}} +\mathrm{2}{t}−\mathrm{1}=\mathrm{0} \\ $$$${t}=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\theta=\mathrm{cos}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}}\right)\approx\:\mathrm{68}.\mathrm{5293}°\:. \\ $$
Commented by mathocean1 last updated on 12/Apr/20
$$\mathrm{please}\:\mathrm{sir}\:\mathrm{can}\:\mathrm{you}\:\mathrm{explain}\:\mathrm{the}\:\mathrm{2}^{\mathrm{nd}\:} \:\mathrm{line}. \\ $$
Commented by mathocean1 last updated on 12/Apr/20
$$\mathrm{Thanks}\:\mathrm{sir}\:\mathrm{i}\:\mathrm{have}\:\mathrm{understood}… \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{great}. \\ $$