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Question Number 194257 by tri26112004 last updated on 01/Jul/23

Know x,y,z ∈ R^+  such that:  2x + 4y + 7z = 2xyz  Find Min(x+y+z)¿

$${Know}\:{x},{y},{z}\:\in\:{R}^{+} \:{such}\:{that}: \\ $$$$\mathrm{2}{x}\:+\:\mathrm{4}{y}\:+\:\mathrm{7}{z}\:=\:\mathrm{2}{xyz} \\ $$$${Find}\:{Min}\left({x}+{y}+{z}\right)¿ \\ $$

Commented by Frix last updated on 01/Jul/23

x=3∧y=(5/2)∧z=2 ⇒answer is ((15)/2)

$${x}=\mathrm{3}\wedge{y}=\frac{\mathrm{5}}{\mathrm{2}}\wedge{z}=\mathrm{2}\:\Rightarrow\mathrm{answer}\:\mathrm{is}\:\frac{\mathrm{15}}{\mathrm{2}} \\ $$

Commented by tri26112004 last updated on 01/Jul/23

Can you explain it¿

$${Can}\:{you}\:{explain}\:{it}¿ \\ $$

Answered by mr W last updated on 01/Jul/23

F=x+y+z+λ(2x+4y+7z−2xyz)  (∂F/∂x)=1+λ(2−2yz)=0 ⇒(1/(2λ))+1=yz  (∂F/∂y)=1+λ(4−2xz)=0 ⇒(1/(2λ))+2=xz  (∂F/∂z)=1+λ(7−2xy)=0 ⇒(1/(2λ))+(7/2)=xy  let ξ=(1/(2λ))  yz=ξ+1  xz=ξ+2  xy=ξ+(7/2)  (2/(yz))+(4/(xz))+(7/(xy))=2  (2/(ξ+1))+(4/(ξ+2))+(7/(ξ+(7/2)))=2  2ξ^3 −25ξ−28=0  (ξ−4)(2ξ^2 +8ξ+7)=0  ⇒ξ=4 >−1 ✓  ⇒ξ=((−4±(√2))/2) <−1 ⇒rejected  yz=5   ...(i)  xz=6   ...(ii)  xy=((15)/2)   ...(iii)  ⇒x=(√((6×((15)/2))/5))=3  ⇒y=(√((5×((15)/2))/6))=(5/2)  ⇒z=(√((5×6)/((15)/2)))=2  ⇒(x+y+z)_(min) =3+(5/2)+2=((15)/2)

$${F}={x}+{y}+{z}+\lambda\left(\mathrm{2}{x}+\mathrm{4}{y}+\mathrm{7}{z}−\mathrm{2}{xyz}\right) \\ $$$$\frac{\partial{F}}{\partial{x}}=\mathrm{1}+\lambda\left(\mathrm{2}−\mathrm{2}{yz}\right)=\mathrm{0}\:\Rightarrow\frac{\mathrm{1}}{\mathrm{2}\lambda}+\mathrm{1}={yz} \\ $$$$\frac{\partial{F}}{\partial{y}}=\mathrm{1}+\lambda\left(\mathrm{4}−\mathrm{2}{xz}\right)=\mathrm{0}\:\Rightarrow\frac{\mathrm{1}}{\mathrm{2}\lambda}+\mathrm{2}={xz} \\ $$$$\frac{\partial{F}}{\partial{z}}=\mathrm{1}+\lambda\left(\mathrm{7}−\mathrm{2}{xy}\right)=\mathrm{0}\:\Rightarrow\frac{\mathrm{1}}{\mathrm{2}\lambda}+\frac{\mathrm{7}}{\mathrm{2}}={xy} \\ $$$${let}\:\xi=\frac{\mathrm{1}}{\mathrm{2}\lambda} \\ $$$${yz}=\xi+\mathrm{1} \\ $$$${xz}=\xi+\mathrm{2} \\ $$$${xy}=\xi+\frac{\mathrm{7}}{\mathrm{2}} \\ $$$$\frac{\mathrm{2}}{{yz}}+\frac{\mathrm{4}}{{xz}}+\frac{\mathrm{7}}{{xy}}=\mathrm{2} \\ $$$$\frac{\mathrm{2}}{\xi+\mathrm{1}}+\frac{\mathrm{4}}{\xi+\mathrm{2}}+\frac{\mathrm{7}}{\xi+\frac{\mathrm{7}}{\mathrm{2}}}=\mathrm{2} \\ $$$$\mathrm{2}\xi^{\mathrm{3}} −\mathrm{25}\xi−\mathrm{28}=\mathrm{0} \\ $$$$\left(\xi−\mathrm{4}\right)\left(\mathrm{2}\xi^{\mathrm{2}} +\mathrm{8}\xi+\mathrm{7}\right)=\mathrm{0} \\ $$$$\Rightarrow\xi=\mathrm{4}\:>−\mathrm{1}\:\checkmark \\ $$$$\Rightarrow\xi=\frac{−\mathrm{4}\pm\sqrt{\mathrm{2}}}{\mathrm{2}}\:<−\mathrm{1}\:\Rightarrow{rejected} \\ $$$${yz}=\mathrm{5}\:\:\:...\left({i}\right) \\ $$$${xz}=\mathrm{6}\:\:\:...\left({ii}\right) \\ $$$${xy}=\frac{\mathrm{15}}{\mathrm{2}}\:\:\:...\left({iii}\right) \\ $$$$\Rightarrow{x}=\sqrt{\frac{\mathrm{6}×\frac{\mathrm{15}}{\mathrm{2}}}{\mathrm{5}}}=\mathrm{3} \\ $$$$\Rightarrow{y}=\sqrt{\frac{\mathrm{5}×\frac{\mathrm{15}}{\mathrm{2}}}{\mathrm{6}}}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\Rightarrow{z}=\sqrt{\frac{\mathrm{5}×\mathrm{6}}{\frac{\mathrm{15}}{\mathrm{2}}}}=\mathrm{2} \\ $$$$\Rightarrow\left({x}+{y}+{z}\right)_{{min}} =\mathrm{3}+\frac{\mathrm{5}}{\mathrm{2}}+\mathrm{2}=\frac{\mathrm{15}}{\mathrm{2}} \\ $$

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