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Question Number 203527 by Mastermind last updated on 21/Jan/24
Find the maximum value of the function  f(x,y)=x^2 y^2 z^2  subject to the condition that  x^2 +y^2 +z^2 =c^2 , where c is the constant.    Thank you in advance
Findthemaximumvalueofthefunctionf(x,y)=x2y2z2subjecttotheconditionthatx2+y2+z2=c2,wherecistheconstant.Thankyouinadvance
Answered by aleks041103 last updated on 21/Jan/24
x^2 =X  y^2 =Y  z^2 =Z  c^2 =C  max of XYZ if X+Y+Z=C, X,Y,Z,C>0  geom.mean≤arithm.mean  ⇒((XYZ))^(1/3) ≤((X+Y+Z)/3)=(C/3) (equality if X=Y=Z)  ⇒XYZ≤(C^3 /(27))  ⇒max f(x,y,z)=x^2 y^2 z^2    if   x^2 +y^2 +z^2 =c^2   is max(f)=(c^6 /(27)) for ∣x∣=∣y∣=∣z∣=((∣c∣)/( (√3)))
x2=Xy2=Yz2=Zc2=CmaxofXYZifX+Y+Z=C,X,Y,Z,C>0geom.meanarithm.meanXYZ3X+Y+Z3=C3(equalityifX=Y=Z)XYZC327maxf(x,y,z)=x2y2z2ifx2+y2+z2=c2ismax(f)=c627forx∣=∣y∣=∣z∣=c3
Commented by Mastermind last updated on 21/Jan/24
Thank you so much but could you please  help me with full details    Thank you once again
ThankyousomuchbutcouldyoupleasehelpmewithfulldetailsThankyouonceagain
Commented by aleks041103 last updated on 21/Jan/24
basically we use the inequality  GM≤AM  where GM is the geometric mean  and AM is the arithmetic mean
basicallyweusetheinequalityGMAMwhereGMisthegeometricmeanandAMisthearithmeticmean
Answered by AST last updated on 21/Jan/24
c^2 =x^2 +y^2 +z^2 ≥3((x^2 y^2 z^2 ))^(1/3) ⇒x^2 y^2 z^2 ≤(c^6 /(27))
c2=x2+y2+z23x2y2z23x2y2z2c627

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