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MargotAnais92
@MargotAnais92
May 2019
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Bonjour,
J'aurai besoin d'aide svp
Merci d'avance et bonne journée à vous.
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ProfdeMaths1
Verified answer
V(x)=AM*AP*AI
=x²(6-x)
=-x³+6x²
V'(x)=-3x²+12x
=(3x)(4-x)
V'(x)=0 si (3x)(4-x)=0 soit x=0 ou x=4
V'(x)>0 si (3x)(4-x)>0 soit 0<x<4
donc V est croissante sur [0;4] puis décroissante sur [4;6]
ainsi V admet un
maximum
en
x=4
ce volume maximal vaut V(4)=-4³+6*4²=
32 cm³
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Verified answer
V(x)=AM*AP*AI=x²(6-x)
=-x³+6x²
V'(x)=-3x²+12x
=(3x)(4-x)
V'(x)=0 si (3x)(4-x)=0 soit x=0 ou x=4
V'(x)>0 si (3x)(4-x)>0 soit 0<x<4
donc V est croissante sur [0;4] puis décroissante sur [4;6]
ainsi V admet un maximum en x=4
ce volume maximal vaut V(4)=-4³+6*4²=32 cm³