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

Commented by mathocean1 last updated on 09/Apr/20

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

Commented by mathocean1 last updated on 09/Apr/20

Answered by ajfour last updated on 12/Apr/20

$$A=\frac{1}{2}\left(2\sin \theta \right)(4+4+4\cos \theta )$$$$\Rightarrow A=8\sin \theta +4\sin \theta \cos \theta$$$$\frac{dA}{d\theta }=8\cos \theta +4\frac{d}{d\theta }\left(\sin \theta \cos \theta \right)$$$$=8\cos \theta +4\{\cos \theta \frac{d}{d\theta }\left(\sin \theta \right)+\sin \theta \frac{d}{d\theta }\left(\cos \theta \right)\}$$$$=8\cos \theta +4(\cos {}^2\theta -\sin {}^2\theta )$$$$\frac{dA}{d\theta }=8\cos \theta +4\cos {}^2\theta -4\sin {}^2\theta =0$$$$let\cos \theta =t$$$$8t+4t^2+4t^2-4=0$$$$2t^2+2t-1=0$$$$t=-\frac{1}{2}+\frac{\sqrt{3}}{2}$$$$\theta ={\cos }^{-1}\left(\frac{\sqrt{3}-1}{2}\right)\approx 68.5293°.$$

Commented by mathocean1 last updated on 12/Apr/20

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

Commented by mathocean1 last updated on 12/Apr/20

$$\mathrm{Thanks}\mathrm{sir}\mathrm{i}\mathrm{have}\mathrm{understood}...$$$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{great}.$$

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