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Question Number 208976 by efronzo1 last updated on 30/Jun/24

Answered by mr W last updated on 30/Jun/24

say x^2 +y^2 =u, xy=v  x^2 +y^2 ≥2(√(x^2 y^2 ))=2∣xy∣  ⇒u≥2∣v∣    x^2 +y^2 +xy=1  ⇒u+v=1  (x^2 +y^2 )xy+4=s, say  ⇒uv=s−4  u, v are roots of z^2 −z+s−4=0  ⇒u,v=((1±(√(17−4s)))/2)  ((1+(√(17−4s)))/2)≥2×((1−(√(17−4s)))/2)  (√(17−4s))≥1  ⇒s≤((38)/9)   → maximum  ((1+(√(17−4s)))/2)≥−2×((1−(√(17−4s)))/2)  (√(17−4s))≤3  ⇒s≥2  → minimum

sayx2+y2=u,xy=vx2+y22x2y2=2xyu2vx2+y2+xy=1u+v=1(x2+y2)xy+4=s,sayuv=s4u,varerootsofz2z+s4=0u,v=1±174s21+174s22×1174s2174s1s389maximum1+174s22×1174s2174s3s2minimum

Answered by Frix last updated on 30/Jun/24

x^2 +xy+y^2 =1 is an ellipse with center  ((0),(0) )  The extreme values of f(x, y)=x^3 y+xy^3 +4 occur  at the vertices of the ellipse.  Rotating by α=(π/4):  x=(((√2)(u+v))/2)∧y=(((√2)(v−u))/2)  ⇒  Ellipse: (u^2 /2)+((3v^2 )/2)=1 ⇔ v=±(1/( (√3)))(√(2−u^2 ))  Vertices at  (((±(√2))),(0) ) and  ((0),((±((√6)/3))) )  f(u, v)=−(u^4 /2)+(v^4 /2)+4  ⇒ 2≤f≤((38)/9)

x2+xy+y2=1isanellipsewithcenter(00)Theextremevaluesoff(x,y)=x3y+xy3+4occurattheverticesoftheellipse.Rotatingbyα=π4:x=2(u+v)2y=2(vu)2Ellipse:u22+3v22=1v=±132u2Verticesat(±20)and(0±63)f(u,v)=u42+v42+42f389

Answered by dimentri last updated on 30/Jun/24

   ⊎2

2

Answered by A5T last updated on 30/Jun/24

Let y=f(x); x^2 +y^2 +xy=1⇒f′(x)=((−2x−y)/(x+2y))  (d/dx)(x^3 y+xy^3 +4)=0⇒f′(x)=((−3x^2 y−y^3 )/(x^3 +3xy^2 ))  f′(x)=((−2x−y)/(x+2y))=((−3x^2 y−y^3 )/(x^3 +3xy^2 ))⇒x^4 +xy^3 =x^3 y+y^4   ⇒(x^2 −y^2 )(x^2 +y^2 )=xy(x^2 −y^2 )  ⇒x^2 =y^2 ...(I) or x^2 +y^2 =xy...(II)  I:x^2 =y^2 ⇒y=+_− x⇒ { ((y=x⇒3x^2 =1⇒x=+_− (1/( (√3)))=y)),((y=−x⇒x^2 =1⇒x=+_− 1=+^− y)) :}  II:x^2 +y^2 =xy⇒2xy=1  ⇒xy(x^2 +y^2 )+4=(xy)^2 +4=(1/4)+4=((17)/4)  Testing (x,y)=(+_− (1/( (√3))),+_− (1/( (√3)))),(+_− 1,+^− 1)  gives the maximum and minimum at   (x,y)=(+_− (1/( (√3))),+_− (1/( (√3)))) and (+_− 1,+^− 1) respectively  ⇒2≤x^3 y+xy^3 +4≤((38)/9)

Lety=f(x);x2+y2+xy=1f(x)=2xyx+2yddx(x3y+xy3+4)=0f(x)=3x2yy3x3+3xy2f(x)=2xyx+2y=3x2yy3x3+3xy2x4+xy3=x3y+y4(x2y2)(x2+y2)=xy(x2y2)x2=y2...(I)orx2+y2=xy...(II)I:x2=y2y=+x{y=x3x2=1x=+13=yy=xx2=1x=+1=+yII:x2+y2=xy2xy=1xy(x2+y2)+4=(xy)2+4=14+4=174Testing(x,y)=(+13,+13),(+1,+1)givesthemaximumandminimumat(x,y)=(+13,+13)and(+1,+1)respectively2x3y+xy3+4389

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