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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

$${AM}−{GM} \\ $$ $$\:{for}\:{x},\:{y},\:{z}\:\geqslant\:\mathrm{0}\:\Rightarrow\:\frac{{x}+{y}+{z}}{\mathrm{3}}\:\geqslant\:\sqrt[{\mathrm{3}}]{{xyz}}\:;\:{x}+{y}+{z}\:\geqslant\:\mathrm{3}\:\sqrt[{\mathrm{3}}]{{a}^{\mathrm{3}} } \\ $$ $${minimum}\:{x}+{y}+{z}\:=\:\mathrm{3}{a} \\ $$

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

$${f}\left({x},{y},{z},\lambda\right)={x}+{y}+{z}+\lambda\left({xyz}−{a}^{\mathrm{3}} \right) \\ $$ $$\frac{\partial{f}}{\partial{x}}\:=\:\mathrm{1}+\lambda{yz}=\mathrm{0} \\ $$ $$\frac{\partial{f}}{\partial{y}}=\mathrm{1}+\lambda{xz}\:=\:\mathrm{0} \\ $$ $$\frac{\partial{f}}{\partial{z}}=\mathrm{1}+\lambda{xy}=\mathrm{0} \\ $$ $$\Leftrightarrow\:−\frac{\mathrm{1}}{{yz}}=−\frac{\mathrm{1}}{{xz}}=−\frac{\mathrm{1}}{{xy}} \\ $$ $${we}\:{get}\:{xy}={xz}={yz}\:;\:{x}={y}={z} \\ $$ $${from}\:{condition}\:{xyz}\:=\:{a}^{\mathrm{3}} \\ $$ $${we}\:{get}\:{x}={y}={z}={a} \\ $$ $${minimum}\:{x}+{y}+{z}\:=\:\mathrm{3}{a} \\ $$

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

$${how}\:{to}\:{explain}\:{this}\:{example}: \\ $$ $${a}=\mathrm{1} \\ $$ $${x}=−\mathrm{10},\:{y}=−\frac{\mathrm{1}}{\mathrm{10}},\:{z}=\mathrm{1} \\ $$ $${xyz}=\left(−\mathrm{10}\right)\left(−\frac{\mathrm{1}}{\mathrm{10}}\right)\left(\mathrm{1}\right)=\mathrm{1}={a}^{\mathrm{3}} \\ $$ $${but}\:{x}+{y}+{z}=−\mathrm{10}−\frac{\mathrm{1}}{\mathrm{10}}+\mathrm{1}=−\mathrm{9}\frac{\mathrm{1}}{\mathrm{10}}<\mathrm{3}{a}=\mathrm{3} \\ $$ $${i}.{e}.\:\mathrm{3}{a}\:{is}\:{not}\:{minimum} \\ $$

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 .

$${i}\:{think}\:{x}+{y}+{z}\:{has}\:{no}\:{minimum}\:{or} \\ $$ $${maximum}\:{under}\:{the}\:{condition}\:{that} \\ $$ $${xyz}={a}^{\mathrm{3}} . \\ $$

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.

$${i}\:{just}\:{wanted}\:{to}\:{show}\:{that}\:{x}={y}={z} \\ $$ $${doesn}'{t}\:{lead}\:{to}\:{minimum}\:{of}\:{x}+{y}+{z}. \\ $$ $${with}\:{x}=−\mathrm{10}\:{it}\:{is}\:{only}\:{an}\:{example}. \\ $$ $${i}\:{can}\:{also}\:{take}\:{an}\:{other}\:{example}: \\ $$ $${x}=\mathrm{1},\:{y}=−\mathrm{100},\:{z}=−\frac{\mathrm{1}}{\mathrm{100}}.\:{i}\:{fulfill} \\ $$ $${xyz}=\mathrm{1}={a}^{\mathrm{3}} \\ $$ $${but}\:{x}+{y}+{z}=−\mathrm{99}\frac{\mathrm{1}}{\mathrm{100}}\:{which}\:{is}\:{smaller} \\ $$ $${than}\:{the}\:{minimum}\:\mathrm{3}{a}=\mathrm{3}\:{you}\:{calculated}. \\ $$

Commented bybobhans last updated on 09/Apr/21

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

$${why}\:{you}\:{get}\:{x}=−\mathrm{10}?\:{from}\:\lambda\:{it} \\ $$ $${clear}\:{x}={y}={z} \\ $$

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