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Exercise: Maxima and Minima · Q29

Q.Find two positive numbers xx and yy such that x+y=60x + y = 60 and xy3xy^3 is maximum.

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Given x+y=60  ⟹  y=60−xx+y=60 \implies y=60-x. Let P(x)=xy3=x(60−x)3P(x)=xy^3=x(60-x)^3, for 0<x<600<x<60.

P′(x)=(60−x)3+x⋅3(60−x)2(−1)=(60−x)2[(60−x)−3x]=(60−x)2(60−4x).P'(x)=(60-x)^3+x\cdot3(60-x)^2(-1)=(60-x)^2\big[(60-x)-3x\big]=(60-x)^2(60-4x).

P′(x)=0  ⟹  (60−x)2=0P'(x)=0 \implies (60-x)^2=0 (giving x=60x=60, an endpoint where y=0y=0 and P=0P=0 -- not an interior maximum) or 60−4x=0  ⟹  x=1560-4x=0 \implies x=15. …

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