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Question Number 199459 by cortano12 last updated on 04/Nov/23

What minimum value f(x,y)=x^2 +y^2 −z^2 when x+2y+4z=21

$$\:\:\mathrm{What}\:\mathrm{minimum}\:\mathrm{value}\: \\ $$$$\:\:\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{z}^{\mathrm{2}} \:\mathrm{when}\: \\ $$$$\:\:\mathrm{x}+\mathrm{2y}+\mathrm{4z}=\mathrm{21} \\ $$

Answered by Frix last updated on 04/Nov/23

x=a y=2a z=−4a ⇒ x=−((21)/(11)) y=−((42)/(11)) z=((84)/(11)) x^2 +y^2 −z^2 =−((441)/(11))

$${x}={a} \\ $$$${y}=\mathrm{2}{a} \\ $$$${z}=−\mathrm{4}{a} \\ $$$$\Rightarrow \\ $$$${x}=−\frac{\mathrm{21}}{\mathrm{11}} \\ $$$${y}=−\frac{\mathrm{42}}{\mathrm{11}} \\ $$$${z}=\frac{\mathrm{84}}{\mathrm{11}} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −{z}^{\mathrm{2}} =−\frac{\mathrm{441}}{\mathrm{11}} \\ $$

Answered by mr W last updated on 04/Nov/23

F=x^2 +y^2 −z^2 −λ(x+2y+4z−21) (∂F/∂x)=2x−λ=0 ⇒x=(λ/2) (∂F/∂y)=2y−2λ=0 ⇒y=λ (∂F/∂z)=−2z−4λ=0 ⇒z=−2λ (λ/2)+2λ−8λ=21 ⇒λ=−((42)/(11)) ⇒x=−((21)/(11)), y=−((42)/(11)), z=((84)/(11)) f(x,y,z)_(min) =((21^2 +42^2 −84^2 )/(11^2 ))=−((441)/(11))

$${F}={x}^{\mathrm{2}} +{y}^{\mathrm{2}} −{z}^{\mathrm{2}} −\lambda\left({x}+\mathrm{2}{y}+\mathrm{4}{z}−\mathrm{21}\right) \\ $$$$\frac{\partial{F}}{\partial{x}}=\mathrm{2}{x}−\lambda=\mathrm{0}\:\Rightarrow{x}=\frac{\lambda}{\mathrm{2}} \\ $$$$\frac{\partial{F}}{\partial{y}}=\mathrm{2}{y}−\mathrm{2}\lambda=\mathrm{0}\:\Rightarrow{y}=\lambda \\ $$$$\frac{\partial{F}}{\partial{z}}=−\mathrm{2}{z}−\mathrm{4}\lambda=\mathrm{0}\:\Rightarrow{z}=−\mathrm{2}\lambda \\ $$$$\frac{\lambda}{\mathrm{2}}+\mathrm{2}\lambda−\mathrm{8}\lambda=\mathrm{21}\:\Rightarrow\lambda=−\frac{\mathrm{42}}{\mathrm{11}} \\ $$$$\Rightarrow{x}=−\frac{\mathrm{21}}{\mathrm{11}},\:{y}=−\frac{\mathrm{42}}{\mathrm{11}},\:{z}=\frac{\mathrm{84}}{\mathrm{11}} \\ $$$${f}\left({x},{y},{z}\right)_{{min}} =\frac{\mathrm{21}^{\mathrm{2}} +\mathrm{42}^{\mathrm{2}} −\mathrm{84}^{\mathrm{2}} }{\mathrm{11}^{\mathrm{2}} }=−\frac{\mathrm{441}}{\mathrm{11}} \\ $$

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