Question Number 101258 by bemath last updated on 01/Jul/20

minimum value f(x,y)=x^2 +y^2   with constrain g(x,y)= x^2 y−16

Commented byjohn santu last updated on 01/Jul/20

f(x,y,λ)=x^2 +y^2 +λ(x^2 y−16)  (∂f/∂x) = 2x+λ(2xy)=0 →λ=((−2x)/(2xy)) =− (1/y)  (∂f/∂y) = 2y+λx^2 =0 →λ=−((2y)/x^2 )  ⇔−(1/y) = ((−2y)/x^2 ) ⇒y^2 =(1/2)x^2   (∂f/∂λ) = x^2 y−16=0 →x^2 = ((16)/y)  we get y^2 =(1/2)(((16)/y)) ⇒y^3 = 8  y = 2 ∧x = ± 2(√2)   { ((for (2(√2),2) ⇒f(2(√2) ,2)= 12)),((for (−2(√2) ,2)⇒f(−2(√2) ,2)=12)) :}  minimum value is 12 ★

Commented bybramlex last updated on 02/Jul/20

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Answered by mr W last updated on 01/Jul/20

y=((16)/x^2 )  f(x)=x^2 +(((16)/x^2 ))^2   (df/dx)=2x+2(((16)/x^2 ))(−((2×16)/x^3 ))=0  x−(2^9 /x^5 )=0  ⇒x^2 =8  f_(min) =8+(((16)/8))^2 =12

Answered by 1549442205 last updated on 07/Jul/20

f(x,y)=h(y)=((16)/y)+y^2 ⇒h′(y)=−((16)/y^2 )+2y=0  ⇔2y^3 −16=0⇔y=2,h ε(y)=((32)/y^3 )+2  hε(2)=6>0⇒h_(min) (y)=h(2)=12  ⇒f(x,y)_(min) =h_(min) (y)=12 when y=2  x^2 =((16)/2)=8⇔x=±(√2)  Thus,f_(min) (x,y)=12 when (x;y)∈{(−2(√2);2);(2(√2);2)}  other way:  f(x,y)=g(x)=x^2 +(((16)/x^2 ))^2 =x^2 +((256)/x^4 )  =(x^2 /2)+(x^2 /2)+((256)/x^4 )≥3 ^3 (√((x^2 /2).(x^2 /2).((256)/x^4 )))=  =3 ^3 (√(64))=3×4=12.  Equality ocurrs if and only if    { (((x^2 /2)=((256)/x^4 ))),((x^2 y=16)) :} ⇔ { ((x^2 =8)),((y=2)) :} ⇔(x,y)∈{(−2(√2);2);(2(√2);2)}  Thus,f_(min) (x,y)=g_(min) (x)=12when  (x,y)∈{(−2(√2);2);(2(√2);2)}