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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+y2z2whenx+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=4ax=2111y=4211z=8411x2+y2z2=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+y2z2λ(x+2y+4z21)Fx=2xλ=0x=λ2Fy=2y2λ=0y=λFz=2z4λ=0z=2λλ2+2λ8λ=21λ=4211x=2111,y=4211,z=8411f(x,y,z)min=212+422842112=44111

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