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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

Whatminimumvaluef(x,y)=x2+y2−z2whenx+2y+4z=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=ay=2az=−4a⇒x=−2111y=−4211z=8411x2+y2−z2=−44111

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=x2+y2−z2−λ(x+2y+4z−21)∂F∂x=2x−λ=0⇒x=λ2∂F∂y=2y−2λ=0⇒y=λ∂F∂z=−2z−4λ=0⇒z=−2λλ2+2λ−8λ=21⇒λ=−4211⇒x=−2111,y=−4211,z=8411f(x,y,z)min=212+422−842112=−44111

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