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Question Number 57283 by ajfour last updated on 01/Apr/19

Commented by ajfour last updated on 01/Apr/19

$$\mathrm{Find}\mathrm{maximum}\mathrm{area}\mathrm{that}\mathrm{can}\mathrm{be}$$$$\mathrm{cut}\mathrm{out}\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{of}\mathrm{a}\mathrm{circle}\mathrm{and}$$$$\mathrm{rectangle}\mathrm{from}\mathrm{an}\mathrm{equilateral}$$$$\mathrm{triangular}\mathrm{sheet}\mathrm{of}\mathrm{paper},\mathrm{edges}\mathrm{a}.$$

Answered by mr W last updated on 01/Apr/19

$$letx=widthofrectangle$$$$heigthofrectangleis(1-\frac{x}{a})\frac{\sqrt{3}a}{2}$$$$Arearectangle=(1-\frac{x}{a})\frac{\sqrt{3}ax}{2}$$$$radiusofcircleis\frac{\sqrt{3}x}{6}$$$$Areaofcircle=\frac{\pi x^2}{12}$$$$A_{◻+}=(1-\frac{x}{a})\frac{\sqrt{3}ax}{2}+\frac{\pi x^2}{12}=\frac{\sqrt{3}ax}{2}-(\frac{\sqrt{3}}{2}-\frac{\pi }{12})x^2$$$$\frac{dA}{dx}=\frac{\sqrt{3}a}{2}-2(\frac{\sqrt{3}}{2}-\frac{\pi }{12})x=0$$$$(6\sqrt{3}-\pi )x=3\sqrt{3}a$$$$\Rightarrow x=\frac{3\sqrt{3}a}{6\sqrt{3}-\pi }=0.717a$$$$A_{max}=...$$

Commented by ajfour last updated on 02/Apr/19

$$\mathrm{thanks}\mathrm{sir}.\mathrm{its}\mathrm{alright}.$$

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