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

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

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

Answered by EDWIN88 last updated on 27/Jan/21

$$Thecarryingcapasityofthegutter$$$$ismaximizedwhenthearea$$$$oftheverticalendofthegutter$$$$ismaximized.$$$$A=3\left(3\sin \theta \right)+2\left(\frac{1}{2}\right)\left(3\cos \theta \right)\left(3\sin \theta \right)$$$$A=9\sin \theta +9\cos \theta \sin \theta$$$$\frac{dA}{d\theta }=9(2\cos {}^2\theta +\cos \theta -1)=0$$$$when\theta =\frac{\pi }{3}\Rightarrow A=\frac{27\sqrt{3}}{4}\approx 11.7$$$$Thecarryingcapasityismax$$$$when\theta =\frac{\pi }{3}$$

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