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Question Number 138004 by benjo_mathlover last updated on 09/Apr/21

What is the minimum value of x+y+z, subject to the condition xyz = a³?\n

Answered by EDWIN88 last updated on 09/Apr/21

AM−GM for x, y, z ≥ 0 ⇒ ((x+y+z)/3) ≥ ((xyz))^(1/3) ; x+y+z ≥ 3 (a^3 )^(1/3) minimum x+y+z = 3a

AM−GM forx,y,z⩾0⇒x+y+z3⩾xyz3;x+y+z⩾3a33 minimumx+y+z=3a

Answered by bobhans last updated on 09/Apr/21

f(x,y,z,λ)=x+y+z+λ(xyz−a^3 ) (∂f/∂x) = 1+λyz=0 (∂f/∂y)=1+λxz = 0 (∂f/∂z)=1+λxy=0 ⇔ −(1/(yz))=−(1/(xz))=−(1/(xy)) we get xy=xz=yz ; x=y=z from condition xyz = a^3 we get x=y=z=a minimum x+y+z = 3a

f(x,y,z,λ)=x+y+z+λ(xyz−a3) ∂f∂x=1+λyz=0 ∂f∂y=1+λxz=0 ∂f∂z=1+λxy=0 ⇔−1yz=−1xz=−1xy wegetxy=xz=yz;x=y=z fromconditionxyz=a3 wegetx=y=z=a minimumx+y+z=3a

Commented bymr W last updated on 09/Apr/21

how to explain this example: a=1 x=−10, y=−(1/(10)), z=1 xyz=(−10)(−(1/(10)))(1)=1=a^3 but x+y+z=−10−(1/(10))+1=−9(1/(10))<3a=3 i.e. 3a is not minimum

howtoexplainthisexample: a=1 x=−10,y=−110,z=1 xyz=(−10)(−110)(1)=1=a3 butx+y+z=−10−110+1=−9110<3a=3 i.e.3aisnotminimum

Commented bymr W last updated on 09/Apr/21

i think x+y+z has no minimum or maximum under the condition that xyz=a^3 .

ithinkx+y+zhasnominimumor maximumundertheconditionthat xyz=a3.

Commented bymr W last updated on 09/Apr/21

i just wanted to show that x=y=z doesn′t lead to minimum of x+y+z. with x=−10 it is only an example. i can also take an other example: x=1, y=−100, z=−(1/(100)). i fulfill xyz=1=a^3 but x+y+z=−99(1/(100)) which is smaller than the minimum 3a=3 you calculated.

ijustwantedtoshowthatx=y=z doesn′tleadtominimumofx+y+z. withx=−10itisonlyanexample. icanalsotakeanotherexample: x=1,y=−100,z=−1100.ifulfill xyz=1=a3 butx+y+z=−991100whichissmaller thantheminimum3a=3youcalculated.

Commented bybobhans last updated on 09/Apr/21

why you get x=−10? from λ it clear x=y=z

whyyougetx=−10?fromλit clearx=y=z

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