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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?
What is the minimum value of x+y+z, subject to the condition xyz = a³?
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−GMforx,y,z⩾0⇒x+y+z3⩾xyz3;x+y+z⩾3a33minimumx+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=−1xywegetxy=xz=yz;x=y=zfromconditionxyz=a3wegetx=y=z=aminimumx+y+z=3a
Commented by mr 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=1x=−10,y=−110,z=1xyz=(−10)(−110)(1)=1=a3butx+y+z=−10−110+1=−9110<3a=3i.e.3aisnotminimum
Commented by mr 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+zhasnominimumormaximumundertheconditionthatxyz=a3.
Commented by mr 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=zdoesn′tleadtominimumofx+y+z.withx=−10itisonlyanexample.icanalsotakeanotherexample:x=1,y=−100,z=−1100.ifulfillxyz=1=a3butx+y+z=−991100whichissmallerthantheminimum3a=3youcalculated.
Commented by bobhans last updated on 09/Apr/21
why you get x=−10? from λ it clear x=y=z
whyyougetx=−10?fromλitclearx=y=z

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