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Question Number 212146 by CrispyXYZ last updated on 03/Oct/24
Find maximum without derivative  x(6−x)(x−3)^2  (3<x<6)
$$\mathrm{Find}\:\mathrm{maximum}\:\mathrm{without}\:\mathrm{derivative} \\ $$$${x}\left(\mathrm{6}−{x}\right)\left({x}−\mathrm{3}\right)^{\mathrm{2}} \:\left(\mathrm{3}<{x}<\mathrm{6}\right) \\ $$
Answered by Frix last updated on 03/Oct/24
Let x=3+3sin t  x(6−x)(x−3)^2 =81(sin^2  t −sin^4  t)=  =((81)/8)(1−cos 4t)  0≤((81)/8)(1−cos 4t)≤((81)/4)
$$\mathrm{Let}\:{x}=\mathrm{3}+\mathrm{3sin}\:{t} \\ $$$${x}\left(\mathrm{6}−{x}\right)\left({x}−\mathrm{3}\right)^{\mathrm{2}} =\mathrm{81}\left(\mathrm{sin}^{\mathrm{2}} \:{t}\:−\mathrm{sin}^{\mathrm{4}} \:{t}\right)= \\ $$$$=\frac{\mathrm{81}}{\mathrm{8}}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{4}{t}\right) \\ $$$$\mathrm{0}\leqslant\frac{\mathrm{81}}{\mathrm{8}}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{4}{t}\right)\leqslant\frac{\mathrm{81}}{\mathrm{4}} \\ $$

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